Monty Hall Problem

In the words of Cecil, he "screwed this one up" on the first reply.

Interestingly, the prominent mathematician Paul Erdos gave a similar answer to this question; the contestant should be indifferent to switching as the odds are still 50-50. From the book "The Man Who Loved Only Numbers", by Paul Hoffman.

When told that the contestant doubles his chances by switching, he hounded his colleagues for a proof from "the Book" - a sort of idealized book containing the best proofs of all mathematical theorems. He was finally satisfied when his friend Ron Graham explained the groundrules in a way similar to Cecil's second reply.

Come on, people.. forget this "human behavior" stuff for a minute. It's a rather clean-cut and logical problem of basic probability. On one reply which is included in the article as I write this, a reader refers to a problem thus:


"There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal?"


According to Cecil:


"The second question is much the same. The possible gender combinations for two children are:

(1) Child A is female and Child B is male.

(2) Child A is female and Child B is female.

(3) Child A is male and Child B is female.

(4) Child A is male and Child B is male.

We know one child is female, eliminating choice #4. In 2 of the remaining 3 cases, the female child's sibling is male. QED."


I'm afraid you can't have your Q.E.D. yet, Cecil.. In this branch of basic probability, the order of choices does not matter. You have two sets up there, #1 and #3, which denote a female child and a male sibling--but you count each seperately. Try to look at it this way. You have one constant, let's make that "child A". Child A is female. What are the odds that child B is male? As the mother hands down an X chromosome, and the father hands down either an X or a Y.. the child has two possible genders, male or female. We are looking for one particular solution--therefore, the odds are 1 in 2.


Cecil, if you respect your readers, please either admit your mistake or prove me wrong.

KWoodlock,
Try to look at it this way. If you decide to discard the order of the children you get three options:

1) 2 female
2) 1 male, 1 female
3) 2 male

As anyone familiar with basic math and probability knows, the probabilities of these three events are not equal. Options 1 and 3 have a probability of 1/2x1/2 = 1/4. And option 2 has a probability of 2x1/2x1/2 = 1/2. We know one child is female, so we can eliminate choice 3. Of the two remaining choices, only choice 2 indicates that they also have a son. To find the conditional probability we divide the probability of an event occuring by the probability that the condition occurs. Here we divide the probability of choice 2 by the probability of choices 1 and 2: (1/2)/(3/4) = 2/3. QED.

Interestingly, if you change the wording of the question subtly the answer changes. FOr example: Given that the first child is female, what is the probability that the couple also has a son? 1/2.

Actually, Cecil's right again. I remember this problem when it also showed up in Marilyn's column. The key is this: if you are told "one of the children is female", then he's right. If you are told something specific that identifies the child, then you're right. In your case, you are using "Child A is female" to specifically identify the child.

Just like the Monty Hall problem, this one is testable by experiment. Get a quarter and a penny. Flip the coins until at least one comes up heads, then write down what the other coin came up. 2/3 of the time, the other one will be tails.

Now change it a little bit -- decide that the quarter is the most important coin (KWoodlock's equivalent of holding "Child A" constant) and flip both coins until the quarter comes up heads. Now the chances of the other coin being tails drop to 1/2.

Note an interesting twist -- you must agree BEFOREHAND on what criteria will be used to identify the "first" coin. Otherwise, I could always say either "the quarter came up heads" or "the penny came up heads" depending on the circumstance, and the chances go right back to 2/3 that the other is tails.

Ah, what a thread to jump into. While I’m not going to touch the Monty Hall problem (Cecil has that correct), there is a problem with Jordan Drachman’s first variation, although I suspect it’s just a typo from all the times it’s probably been reentered in various places. The problem states that there’s a card in a hat that has an equal probability of being either the ace of spades or the king of spades. The problem then tells you add an ace and that you randomly removed an ace from the hat. Adding something to the hat and removing something from the hat does not change the probability of the first card being either the ace or king. The original card had a 50% chance of being the ace and a 50% chance of being the king. The odds of the original card in the hat being an ace is 1/2. Of course, the odds of the “remaining” card being an ace is 2/3 if the first card removed from the hat is an ace.

I was doubtful myself about the male/female kid question (and am now embarrassed for it), so I wrote a program to check. Look at www.shalott.com/monty.asp for an empirical test.

Source available upon request.

Here's another way of looking at the Mony Hall problem. You have three choices to start with, which means you have a 1/3 chance of picking the right door and a 2/3 chance of picking the wrong door.

Now the way the game is set up, if you picked the wrong door the first time, you will be given a second choice which will always be the right door.

So looking at the odds, you should always go with your second choice and in the long run you will get the right door twice as many times.

This is why I hate probability questions. They usually are confusing only because of the way they are stated. The question from KWoodlock's post says:

"There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal?"

I read this question as "A family has two children, you are told the sex of one of the children, what are the odds that the other child is of the opposite sex?" Obviously the answer to my question is 1:1 but the answer to the above question is 2:1.

The only reason the original question is confusing is because most people probably have the same interpretation of the problem as I do. In other words the difficulty lies not in understanding the probabilities involved but in interpreting the question "correctly."

Geoff - who likes his math problems well-stated

When I first saw this thread, I thought Cecil was on crack (I actually didn't see the original article.) I think it's confusing, as others have pointed out, because of the phrasing. I convinced myself when I considered the idea that there are four combinations of two-child families, all of equal odds. The key is that the boy-girl/girl-boy combination collapses into one "girl-and-boy" combination. So...by specificing that one child is a girl, the probability of the boy-boy combination is zero, leaving the three others: girl-girl, boy-girl, and girl-boy. Thus, assuming the probabilities of the original four combinations still remain equal (despite the boy-boy combination having a probability of zero now), there are two out of three equally-likely combinations with the other child being a boy.


Thus, the probability statement is that within the population of families with two children, 2/3 of those with a female child also have a male child and 1/3 of those have two female children.

------------------

In as much as the question involving the playing cards and the hat, I definitly think the answer is 50%. You can prove it using the same logic as the monty hall/curtain case. You have equal chances, 50%, with two subsets, king and ace. Each of these has a choice as you draw a card.
You have four choices total

1) Ace 1 + Ace 2 where you draw Ace 1--25%
2) Ace 1 + Ace 2 where you draw Ace 2--25%

3) King + Ace 2 where you draw King--25%
4) King + Ace 2 where you draw Ace 2--25%

1) and 2) are in the same subset. Total=50%
3) and 4) are in the same subset. Total=50%

When Monty showed us the wrong curtian, he effectively eliminated one of the choices of a subset(curtain 3). Thereby making the only remaining choice of that subset(curtain 2)take on the value of the entire subset. Thus, curtain 2 = 66%, curtain 1 = 33%. In the card problem, drawing an ace does the same thing. It eliminates one of the choices of the king subset, namely drawing the king, which makes the only remaining choice of that subset take on the value of the subset(I swear I'll go nuts if I have to write this word again.) , i.e. 50%. Thus, there is a 50% chance, overall, that the original card was a king, and a 50% chance it was an ace.

Larsy,
When you pick the ace out of the hat you effectively eliminate choice #3 from your list. This means that there are only three possibilities. Of the three possibilities, in two of them the remaining card is an ace and in one of them it is a king. Therefore, there is a 2/3 probability that if you choose an ace, the remaining card is also an ace.

TheDude

Another variant of the same problem -- the 3 coins. There are 3 coins in a hat, one with two heads, one with two tails, and one with a head and a tail. When you draw a coin at random and place it on the table, you see a head. What's the probability that the other side of this coin is a head?

Answer is again 2/3.

The exchanged regarding the family with two children (one female), while interesting, is not a proper mathematical model of the original Monty Hall question.

A correct model is as follows:
A family has 3 children. One, and only one, is male, though you don't know which.

Your options:
1) Male, Female, Female
2) Female, Male, Female
3) Female, Female, Male

It is revealed to you that Child #3 is female. What are the odds of Child #1 being male? 1:2

The "1000 ticket lottery" example can also be used to illustrate the principle. If one person scratches off their ticket, should you trade? No, because any ticket you trade for has a 1:999 chance of winning. If half the people scratch off their tickets, should you trade for someone else'd ticket then? No, because all of the remaining tickets have an equal 1:500 chance of winning. If all but two tickets are scratched off, should you trade your ticket for the other one? No, because that ticket has the same 1:2 chance of winning.
(This is a rather different situation then the original question of whether you should *sell* your ticket, however. In a million dollar lottery with 1000 tickets each ticket has an value of 1/1000th of a million, or $1000. Once you're down to two tickets the value has gone up to 1/2 a million, or $500000. Sell your ticket over the value and you make a profit.)

Monty revealing that the prize isn't behind Door #3 (IE- Child #3 is female) DOES increase the chances of Door #2 having the prize behind it. However, the revelation also increases the chances that it's behind Door #1. Therefore the chances of the prize being behind Door 1 or 2 are equal. Stick with Door #1 or switch, it doesn't make a difference.

Aufwiederlesen
1010011010 ¦¬)

TheDude
Yes, you eliminate choice 3, I stated that, however, what you missed was that that does not redefine the problem. The subset remains the same. Yes, there are now only three choices, however, they are of unequal probabilities. One choice has a 50% chance, and the other two choices have 25% chances.
Its the same thing as the monty hall case, changing the internal probabilities of the subset does not change the probability of the entire subset.

December,
I don't think the problem is the same. In your example you start with three possibilities of equal probability, in the card problem, there are four to start. In fact, in your problem, wether a head or a tail is observed makes absolutly no difference whatsoever to the answer. You could say the same about both. Your example was much more straightforward than the others.

Larsy

Larsy

The correct answer to the cards in a hat problem is 2/3. Maybe this can be made more clear by carefully defining the problem and by offering a modification.

The problem states that there’s a card in a hat that has an equal probability of being either the ace of spades or the king of spades. The problem then says that you added an ace and then you randomly removed an ace. The question is what the odds are that the remaining card is an ace.

First of all, when the problem says, that you "randomly removed an ace," it means that you removed a card at random, which turned out to be an ace.

Now, change the problem by assuming that after you randomly removed an ace, you returned it to the hat and again chose a card at random, which again turned out to be an ace. This is called "sampling with replacement." Keep repeating this process.

If you continue to sample with replacement and you always get an ace, then you become more and more certain that there's no king in that hat.

Hope this helps...

There is a simple experiment you can do at home, with a friend, in order to prove the right answer to the Monty Hall problem:

Take three cards, say a joker and two aces.

Mix them up so that your friend doesn't know which are which, but keep track of them yourself.

Lay them out face down so that you know where each card is.

Have you friend try to pick the joker by pointing to one of the three cards.

After your friend has made his pick, turn over one of the other cards so that you always reveal an ace (you can do this, because you know which is which. Then give your friend the chance to switch from his original choice (which remains face down) to the other card which also remains face down.

At first, it seems as if, with one ace revealed, there is now a fifty-fifty chance that either of the remaining cards could be the joker; therefore, one would conclude that it makes no difference whether your friend switches cards or not.

When played out, however, after only ten or twenty tries, something else will start to become apparent, something that seems too obvious to be worth mentioning but which is vital: the only time it is correct for your friend to stick with his original choice is when the original choice was correct (i.e., he pointed to the joker.)

Seems pretty obvious, right? But here's the point: the original choice will only be correct one out of three times. (That's because the original choice was made when there were three cards turned down, and only one card was the correct one.)This means that if your friend always sticks with his original choice, then only one time out of three will be pick the joker; two times out of three he will have picked one of the two aces and stuck with the wrong choice. Therefore, he will be wrong two-thirds of the time and right one-third of the time.

On the other hand, if your friend switches cards after you have revealed an ace, then he will pick the joker two-thirds of the time. This is because there were two chances that the joker was one of the unpicked cards, and by always turning over an ace you eliminated one incorrect choice.

It's as if you said, "You can keep your first card, or you can have both of the other two cards." Clearly, picking two cards rather than one doubles your chances. Turning over a m ace merely confuses the issue, because at a glance it appears as if you're randomly eliminating one choice. However, it is not random: of the two remaining cards, if the joker is card A, you reveal card B; if the Joker is card B, you reveal card A. Whichever card you reveal, it doesn't alter the fact that there was a two-thirds chance that one of the unpicked cards was the joker.

Or think of it another way: Obviously your friend has a one-in-three chance of picking the joker on the first try. It should be equally obvious that, no matter what card he picks, there will always be at least one ace for you to turn over. When you turn this ace up to reveal it to your friend, have you somehow magically affected his original choice after the fact, making it so that his first selection had a fifty-fifty chance of being correct?

Comments by 1010011010 illustrate how slight changes in conditions can affect probability problems. Wording is often imprecise, where the solver is supposed to fill in the missing portions. The model suggested by 1010011010 doesn't quite match the intended conditons of the Monte Mall problem

The original Monte Hall problem was imprecisely stated. A complete wording should have specified that:
-- Mr. Hall knew which door had a car behind it
-- He chose to open a door that he knew had a goat behind it, and
-- His decision of whether or not to open a door was independent of whether the contestent's orginal choice of door was a winner or a loser.

One confusing aspect of the problem is that the 3rd condition was not specified in the original problem, nor is it necessarily true on the actual Monte Hall show.

The way I convinced myself of the answer to the Monty Hall problem was to look at an extreme case:

Let's say Monty shows you 1000 doors. The grand prize is hidden randomly behind one of them and the rest are empty. You are asked to choose one door. You choose door #1. Now Monty, "We're going to show you what's behind 998 of the other doors." He opens door #2 - empty... door #3 - empty... door #4 - empty... (yada yada)... door #732 - empty... door #733 - "Ummm, let's just skip door #733 for now"... door #734 - empty... door #735 - empty... (yada yada)... door #999 - empty... door #1000 - empty. Now Monty says, "Would you like to keep what's behind door #1, or trade for what's behind door #733?" That's a no-brainer!



------------------
"For what a man had rather were true, he more readily believes" - Francis Bacon

Let's break this down into all possible scenarios... and beat it into the ground.

W = winning door
L = losing door

Possible setups: WLL, LWL, LLW.

1) Setup WLL, you pick 1. Monty reveals 3, you stick with 1.
2) Setup WLL, you pick 1. Monty reveals 3, you switch to 2.
3) Setup WLL, you pick 1. Monty reveals 2, you stick with 1.
4) Setup WLL, you pick 1. Monty reveals 2, you switch to 3.
5) Setup WLL, you pick 2. Monty reveals 3, you stick with 2.
6) Setup WLL, you pick 2. Monty reveals 3, you switch to 1.
7) Setup WLL, you pick 3. Monty reveals 2, you stick with 3.
8) Setup WLL, you pick 3. Monty reveals 2, you switch to 1.
9) Setup LWL, you pick 1. Monty reveals 3, you stick with 1.
10) Setup LWL, you pick 1. Monty reveals 3, you switch to 2.
11) Setup LWL, you pick 2. Monty reveals 1, you stick with 2.
12) Setup LWL, you pick 2. Monty reveals 1, you switch to 3.
13) Setup LWL, you pick 2. Monty reveals 3, you stick with 2.
14) Setup LWL, you pick 2. Monty reveals 3, you switch to 1.
15) Setup LWL, you pick 3. Monty reveals 1, you stick with 3.
16) Setup LWL, you pick 3. Monty reveals 1, you switch to 2.
17) Setup LLW, you pick 1. Monty reveals 2, you stick with 1.
18) Setup LLW, you pick 1. Monty reveals 2, you switch to 3.
19) Setup LLW, you pick 2. Monty reveals 1, you stick with 2.
20) Setup LLW, you pick 2. Monty reveals 1, you switch to 3.
21) Setup LLW, you pick 3. Monty reveals 1, you stick with 3.
22) Setup LLW, you pick 3. Monty reveals 1, you switch to 2.
23) Setup LLW, you pick 3. Monty reveals 2, you stick with 3.
24) Setup LLW, you pick 3. Monty reveals 2, you switch to 1.

Winning Scenarios: 1, 3, 6, 8, 10, 11, 13, 16, 18, 20, 21, 23 or 12:24 or 1:2

Winning Scenarios where you switch: 6, 8, 10, 16, 18, 20 or 6:24 or 1:4

Winning Scenarios where you stick: 1, 3, 11, 13, 21, 23 or 6:24 or 1:4

1:4 = 1:4

I reiterate: Switch or Stay, it doesn't make a difference.

Aufwiederlesen
1010011010 ¦¬)

I thought it would be necessary to explain the "1000 doors" version as well. (Don't worry, I'm not going to break it down)

The only game that is relevant is the last game in which you have the choice between 2 doors, one of which is the winning door.

Regardless of the number of doors to begin with it always narrows down to:
1) you've got the winning door and you stay
2) you've got the winning door and you switch
3) you've got the losing door and you stay
4) you've got the losing door and you switch

It doesn't matter is it's door #1 and #2 or if it's #1 and #733. The chances of you winning are the same whether you switch or stay.

Aufwiederlesen
1010011010 ¦¬)

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"All I say here is by way of discourse and nothing by the way of advice. I should not speak so strongly if it were my due to be believed." ~ Montaigne

Binary person, your fallacy is in multiplying the scenarios where keeping the door wins by two. Just because Monty can has the option of giving you either of the two losing doors when you have picked the winning one does not make each of those occurrences equally likely!

Which one of the two losing doors Monty gives you is irrelevant, and should not be included in your enumeration.

1010011010, since you went to all the trouble of showing so many cases, I feel a little embarassed to correct you. However, this is The Straight Dope, and truth rules...

The key to this problem is that your Cases 5, 6, 7, and 8 are each twice as likely as 1, 2, 3, or 4. The assumption is that MH will ALWAYS choose to open a door with a goat. When you have chosen a goat, and he has only one such door available, this situation has the same total probability as do MH's 2 options when you have chosen the door with a car.

E.g., Case 5 corresponds to Case 1 plus Case 3. In Case 5, you chose Door 2 and MH opened Door 3. (The car is behind Door 1.) Now,imagine that Case 5 also includes Case 5a, where the contestent selected Door 2, and MH would have opened door #1, but there was a car behind it, so he had to switch his choice and open Door #3.

Regarding the Monty Hall problem, Marilyn vos Savant was wrong and your original answer was correct. The key is in the statement that the host knows what is behind each door, consequently, he can always show a door with no prize behind it. Assuming your original choice is door #1, there are three four possiblities as to which door can be opened and which contains the prize. These possibilities are as follows:
Door #1 has prize and door #2 is opened
Door #1 has prize and door #3 is opened
Door #2 has prize and door #3 is opened
Door #3 has prize and door #2 is opened
As you can see, of the four possibilities, two have Door #1 having the prize and two have door #1 not having the prize. Therefore, the odds are even. The argument about choosing one out of a million or one card out of a deck of 52, then revealing all except the the original choice and one other is not a valid comparison. It only appears that it is because revealling one out of three leaves two.

Also, concerning the two children problem, you state that the possibilities are:
A (female) B (female)
A (female) B (male)
A (male) B (female)
A (male) B (male)

More correctly they should be stated by which child enters first, that adds four more possibilities:
B (female) A (female)
B (female) A (male)
B (male) A (female)
B (male) A (male)
Now, you can see that if the first child is female, there are four possiblities, in two of which the second child is female and in two of which the second child is male. 50-50.

Or, conversely, using your four possibilities, in both of the last two the first child is male, and you should eliminate both of them, again leaving a 50-50 probability as to the gender of the second child.

Robert D Schader
Green Bay, WI

Robert D Schader certainly displays self-confidence, contradicting the (self-proclaimed) smartest man in the world and smartest women in the world!

I do agree with Robert's 4 possible cases. However, the the 1st and 2nd cases each have probability 1/6 and the 3rd and 4th cases each have probability 1/3.

The simplest way I can put it:

The probability of you picking a loser door initially is 2/3 (clearly). If you pick a loser, you will always win by switching because Monty will have helpfully eliminated the other loser. So if you follow the strategy of always switching you will win 2/3 of the time.

The always-switch strategy only loses when you picked the winning door with your first choice, and clearly that will happen only 1/3 of the time.

jen and dec, admirable effort. You might have had bad arguments but you were fighting for the right side. Mandrake has convinced me.
I have seen the err of my ways
Possible scenarios:
1) you chose right, you stay
2) you chose right, you switch
3) you chose wrong1, you stay
4) you chose wrong1, you switch
5) you chose wrong2, you stay
6) you chose wrong2, you switch

Winning scenarios: 1, 4, 6 or 1:2
Winning scenarios where you switch: 4,6 or 1:3
Winning scenarios where you stay: 1 or 1:6

IE- you're twice as likely to win when you switch!
<<switches>>

Aufwiederlesen
1010011010 ¦¬)

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"All I say here is by way of discourse and nothing by the way of advice. I should not speak so boldly if it were my due to be believed." ~ Montaigne

Note that my intention was NOT to convince you. Sufficient materials for that, including previous incarnations of Manduck's explanation, had already been posted.

My intention was to point out the fallacy in your particular argument, which seemed on first sight to be logically phrased. Usually people who bother to enumerate come up with the correct answer, which made your enumeration interesting to me.

Manduck2 -- Must congratulate you. Your explanation is the simplest and clearest I've seen.

Let me try to explain this one other way. First I'll list all of the possible outcomes.
The first number is the door that the prize is behind, and the second number is the choice made by the contestant.
PC---PC---PC
11---12---13
21---22---23
31---32---33
Each of these possiblities have the same probability, 1 in 9.
Now the host, knowing where the prize is, selects a door that does not contain the prize and shows it to the contestant.
The contestant chooses a door, removing 6 of the possiblities, then the host opens a door(not containing the prize), removing 1 of the remaining possibilities. That leaves 2 possibilities, both of which have the same probability, one where the original choice was correct and one where it was not.

RD --You wrote:

The contestant chooses a door, removing 6 of the possiblities, then the host opens a door(not containing the prize), removing 1 of the remaining possibilities. That leaves 2 possibilities, both of which have the same probability, one where the original choice was correct and one where it was not.

The trick is that these remaining cases do NOT have the same probability. You have to also factor in the probabiklity that MH opened a particular door.

E.g., assume that you chose door #1 and MH opened Door #3.

If the car was behind Door #1, then MH had a 50% chance of opening #2 and 50% chance of opening #3. If the car was behind Door #2, then MH was 100% certain to open Door #3. For this reason, when MH opens Door #3, it is twice as likely that the car is behind Door #2.

RD - Look at the outcomes you listed:

PC---PC---PC
11---12---13
21---22---23
31---32---33

Note that in 3 of those 9 cases (11, 22 and 33) your first choice was correct. In the other 6 cases, you would have to switch doors to win the car. So in 6 of the 9 cases the switching strategy wins, while not switching only wins in 3 of the cases.

Let's assume that the contestant chooses door #1. Now there is a 1/2 probability that MH will open door #2 and a 1/2 probability that he will open door #3.

MH opens door #2 - probability 1/2
Prize behind Door #1-probability (1/2)*(1/3)
Prize behind Door #3-probability (1/2)*(2/3)

MH opens door #3 - probability 1/2
Prize behind Door #1-probability (1/2)*(1/3)
Prize behind Door #3-probability (1/2)*(2/3)

Probability of Door #1 being correct
(1/2)*(1/3)+(1/2)*(1/3)=1/3

Probability of Door #2 being correct
(1/2)*(2/3)=1/3

Probability of Door #3 being correct
(1/2)*(2/3)=1/3

Opening Door #2, for example, removes 1/3 of the possibilities, but the probabilities for Door #1 and Door #3 being correct remain in a one to one ratio.

-----

One of the arguments I read was that showing a non-prize door was equivalent to offering both of the other doors for your one. If this were what was happening then it would indeed be to your advantage to switch. This,
however is not the case. The host is just removing from consideration a door that he knows does not have the prize. He can do this 100% of the time. If your choice was correct, then he has a choice as to which door he opens, if your choice was not correct then there is only one door he can open.

Hypothetical situation:
You consider choosing door #3, but at the last momment change to door #1. Now the host reveals door #2, showing that it does not contain a prize.

The probability of the host opening door #2 is the same had your choice been door #3 or door #1.

What you are saying is that it is now to your advantage to switch back to door #3, but if you had not change your mind, and originally choosen door #3, it would now be to your advantage to switch to door #1. Clearly these can not both be true.

RD - You state that if you choose door #1, there is a 50% chance that MH will open door #3. This is not necessarily true - if the prize is behind door #2, MH is forced to open door #3, i.e. the probability is 100%.

Let's break it down again. Assume the prize is behind door #3. The possibilities now are:

You pick door #1 (probability = 1/3), MH opens door #2 (probability = 1): total probability = 1/3

You pick door #2 (P = 1/3), MH opens door #1 (P = 1): total P = 1/3

You pick door #3 (P = 1/3), MH opens door #1 (P = 1/2): total P = 1/6

You pick door #3 (P = 1/3), MH opens door #2 (P = 1/2): total P = 1/6.

The first two cases above are the ones that you would win if you switched, and their probabilities add up to 2/3.

Manduck, after careful reconsideration, I am forced to admit that you are indeed correct.

[thatsmrberns2u]

It doesn't matter whether the game show host always opens the door or not. Switching doors provides no advantage. The way the question is worded though should indicate to you that the game is that the host opens a door every time. If it was optional for the host to do that, then they would have to say that the host "opened" another door, not that he "opens" another door. Even if the game WAS that it is optional for the host to open a door, the question is about a particular instance where the host does open the door, so you don't even need to consider all that. You guys are way off. Anyway, it doesn't matter, here's why:

Without a door being opened, you have 3 choices, in 1 of which you win, so your chances of winning are 1/3 for each choice.
With a door opened, you have 2 choices, in 1 of which you win, so your chances of winning are 1/2 for each choice.

If the host opens a non-winning door, you're not learning any more about the door you didn't choose than you are about the door you did choose. Someone said that having the
host open a door and opening the other door you didn't originally choose is like opening 2 out of 3 doors, whereas if you keep your original door, you're only opening one door. Take a closer look, in the latter case 2 out of 3 doors are also open. People who claim
to have proven an advantage one way or another by doing experiments or by writing
programs are either totally incompetant or are lying.

Marilyn is wrong quite often (see http://www.wiskit.com/marilyn), and I suspect she does it on purpose to generate responses and hence publicity.

------------------

[Stratocaster]

I believe there is still some specious logic afoot in the analyses of this problem, Cecil's responses included. I refer specifically to the notion that Monty Hall's pre-knowledge of what's behind the doors is a factor that distinguishes this from the lottery ticket quandary. While it adds a certain Machiavellian intrigue to the proceedings, I'm not sure it's relevant.

It's just not clear to me why this should matter at all (I'll avoid the smug certainty that has come back to haunt virtually every person who has commented on this important issue). Stated more emphatically, if this fact is apropos, I am completely misunderstanding the logical conclusion, which is why I bring it up at all (since the conclusions are otherwise "accurate").

Cecil himself asserts that this dilemma is tantemount to being offered all the unselected doors in return for your original choice. If this is true, why would that fact change because this option emerged randomly and not from the schemings of the evil Hall?

I believe that even if Monty randomly selected one of the unchosen doors, faced with an infinite series of the subset of trials where the knucklehead hasn't selected the actual correct choice--that's what we have here--I'm still better off switching. The same is true for the lottery ticket if I calculate the benefit derived from this strategy over the course of an infinite series of trials.

Am I loopy?

[Lagged2Death]

thatsmrberns2u said:

People who claim to have proven an advantage one way or another by doing experiments or by writing programs are either totally incompetant or are lying.

Them's fightin' words, Mr. Berns. And on your first post, too - tsk, tsk.

I was going to stay out of this, and just read this one. I've had arguments about this subject before. But about seven years ago, when this subject was a little less stale, I did, in fact, write a computer program to play this game a few thousand times, and test the always-switch and never-switch strategies. So your offhand remark becomes a personal insult! I must respond!

The program did show that the always-switch strategy wins 2/3 of the time and the never-switch strategy 1/3 of the time. And I wrote the program as a skeptic, mind you - I was rooting for a 50/50 result. But I'm convinced now.

It was a very simple program, so even if I am "incompetant" (which is usually spelled "incompetent", by the way - what an unfortunate word to misspell) I'm sure I managed to get an accurate result, as have many others. And why in the world would anyone lie about such a thing?

If you're sure that ALL of the experiments and computer programs testing the hypothesis are flawed, why not do one yourself? If you turn out to be right, you'll turn the worlds of math and physics upside down.

The mistake in your reasoning is a common one; you assume that when the number of closed doors changes from 3 to 2, the 1/3 probability of your original choice being the correct one auto-magically changes to 1/2. It doesn't. Nuclear war wouldn't change it. Monty going postal wouldn't change it. The contestant suddenly getting abducted by aliens wouldn't change it. And certainly, opening some silly door isn't going to change it. That's really the heart of the argument; the door you initially pick is probably the wrong door, and opening more doors does NOT change that. No matter how many doors get opened later, your first pick was probably wrong. It's simple to see that if the door you picked is probably wrong, then the doors you didn't pick probably include the prize. Opening doors doesn't change that, either, which is why switching usually wins.

To believe anything else is pretty close to believing in time travel. If the odds of your first pick being right change (from 1/3 to 1/2), then the events of the present (opening doors) are effecting the events of the past (your first pick)! In this universe, most of us like our causes to come BEFORE our effects. Opening doors after you make your choice doesn't effect the odds that your choice was right or wrong. Making a new choice, however, can have rewards aplenty.

[Dyerseve63]

OK, i know that this question has been completely beaten to death, but i have a very simple way of looking at it.(IN 5 SIMPLE POINTS)
1) When you flip a penny, the chances of it being "heads" is 1/2. The second time, the chances are still 1/2, regardless of the first outcome.
2) in a lottery game with 10,000 possible outcomes (0000-9999) the odds of one ticket being a winner are 1/10,000. If you play 1234, every day for one year, and it does not win, the odds are still 1/10,000 that your ticket is a winner.
3) the odds of door #1 winning is 1/3. The odds of door #2 winning is 1/3. The odds of door #3 winning is 1/3.
4) when door #3 is eliminated, the odds of door #1 winning remain 1/3. The odds of door #2, also remain 1/3. They are equal.
5) WHAT DOOR YOU CHOOSE DOES NOT MATTER BECAUSE YOU HAVE THE SAME CHANCES OF WINNING

[Boris B]

How can those of us at the Always Switch School convince you? People from the Information Has No Value School have lined up one by one and argued with Manduck and the rest, and eventually given up the argument, only to be replaced by someone else.

You have a two-thirds chance of picking a losing door. Monty always opens a losing door, which is not the same as the one you are standing in front of. Monty does not open a door at random (in which case it would have a 33% chance of being the winner, allowing you to just pick the car ... not a very great challenge). Thus Monty has added information about which doors are losers. If you know one door is a winner, knowing that another is a loser adds information, which is highly prized by those of us outside the Information Has No Value School. So assuming you're wrong to begin with, you will win if you switch.

It's a 67% likely assumption. Assuming that you are right, upon guessing, is a 33% likely assumption.

Or are you guys seriously arguing that you have a 50-50 chance of picking the winning door out of the three closed doors?

[Boris B]

Originally posted by Dyerseve63:
3) the odds of door #1 winning is 1/3. The odds of door #2 winning is 1/3. The odds of door #3 winning is 1/3.
4) when door #3 is eliminated, the odds of door #1 winning remain 1/3. The odds of door #2, also remain 1/3. They are equal.
5) WHAT DOOR YOU CHOOSE DOES NOT MATTER BECAUSE YOU HAVE THE SAME CHANCES OF WINNING

If door #3 is eliminated, that means it has a 0% of being the winner, not a 33% chance. Your first guess has a 33% chance of being right, not a 50% chance of being right. Switching to the door Monty did not open gives an overall 67% chance of winning.

[Lagged2Death]

Quoth Boris B:

How can those of us at the Always Switch School convince you? People from the Information Has No Value School have lined up one by one and argued with Manduck and the rest, and eventually given up the argument, only to be replaced by someone else.

Hear hear. What is it about this problem that makes so many so willing to argue, yet unwilling to actually try it out? It's not as if you need a particle accelerator to set up an experiment. Any yutz can try this out with no more equipment than pencil, paper, and a six-sided die. And the results will be extremely convincing. Even if the logic of the problem isn't convincing, watching the always-switch strategy win twice as often ought to do the trick.

[Lagged2Death]

3) the odds of door #1 winning is 1/3. The odds of door #2 winning is 1/3. The odds of door #3 winning is 1/3.
4) when door #3 is eliminated, the odds of door #1 winning remain 1/3. The odds of door #2, also remain 1/3. They are equal.
5) WHAT DOOR YOU CHOOSE DOES NOT MATTER BECAUSE YOU HAVE THE SAME CHANCES OF WINNING

You were doing great until you got to the second part of step 4 there. Supposing what you said in step four was true, it's plain to see that there would be only a 2/3 chance that a prize even existed! (2 doors * 1/3 chance of each door winning = 2/3 chance that some door is a winner.) That's not the game we're playing here - in our game, there is always (1/1 chance) a prize somewhere. The odds that each unopened door is a winner must add up to 1.

If this is really imporant to you, (step 5 sure seems to be a passionate one) try it out with pencil and paper. Have a friend play Monty, if that helps you.

[Sam Stone]

Or try this: Play the game with 1000 doors. You pick one, and Monty always opens the other 998 that don't have a prize left behind them, leaving one left. Do you really think there is no advantage to switching? Thinking so is tantamount to believing that you'll always have a 50/50 chance of picking the right door out of 1000.

The problem can indeed be reduced to, "You can keep the door you picked, or exchange it for ALL the others." This is the case because out of all the others, Monty eliminates all the incorrect choices for you.

The question gets more interesting if Monty only chooses to open one of the other doors some of the time. Then it becomes a matter of game theory.

[CurtC]

dhanson wrote:
The question gets more interesting if Monty only chooses to open one of the other doors some of the time. Then it becomes a matter of game theory.And this is where it really gets interesting. At least more so than trying to convince people who post to a public board shooting from the hip, without bothering to put pencil to paper first.

The way the problem is usually stated, you don't know that Monty will always offer you the choice, or what his motivations might be. If you look at the problem from a game theory perspective and assume that you want to get the good prize and Monty wants you to get a year's supply of chicken feathers, then if you follow a pattern such as always switching when you have the choice, Monty will start offering the choice only when you've picked it right on the first try. Since Monty has the advantage in that he knows where the prize is, the best you can hope for is having a 1/3 probability of winning, which happens if you stick.

If you sometimes stick and sometimes switch, then Monty will offer you the opportunity to switch only when it is advantageous to him, resulting in less than a 1/3 win rate for you. You should always stick with your first choice. And if you always stick, then Monty will not even bother to offer you the choice.

That's how it would happen in a competitive game. Of course, his real goal is to make the show exciting, so he'd probably offer the switch in a way to make it average out to 50/50.

Note to poor readers: I am in no way suggesting that the probability, assuming he always offers the chance to switch, will be any other answer but switch by 2/3 to 1/3.

[Boris B]

Hey, Irishman, that's a good idea. This whole thing kind of reminds me of certain gambling card games. The ones like Blackjack and Stud Poker where you have to guess what cards your opponents have based on the ones you can see, but your guess changes each time you see more.

So I guess five-card stud is a better example. Your opponent has an Ace and a down card showing; you have two Kings. He'll win is his down card is an Ace, which makes you nervous until more Aces appear in other people's hands. Did the appearance of those Aces change the identify of the down card? No, it just changed your best guess.

I thought of another way to explain this, not that Lagged and Manduck and the rest of us haven't already posted a lot. You guess at a door without opening it; it has a 33% chance of being right. Since there's exactly one prize, the remaining two doors have a combined 67% chance of being right. Monty opens one, subtracting 0% from final door's probability. The final door has a 67% chance of being right; your original guess keeps its 33%.

[Irishman]

Here is another thread on this topic.
http://boards.straightdope.com/ubb/Forum1/HTML/000465.html

Anyone who still doubts the outcome, try it yourself.

Take a deck of cards, have a friend play "Monty". He gets to see all three cards, and knows which is the ace (car), vs. the two jokers (goats).

You pick a card, then he must reveal a card. (Note: he is not allowed to circumvent the game by revealing the ace.) Thus you know he will reveal a joker.

Or play with the whole (52 card) deck, with the Ace of Spades as the winning card. You pick one, Monty turns over 50 remaining non-ace cards. Do you switch?

Play for money - you'll learn the lesson better that way.

[rorschach]

Before anyone says anything, yes the probability of winning is 2/3 if you switch.
However lagged2death wrote &lt;quote&gt;What is it about this problem that makes so many so willing to argue, yet unwilling to actually try it out?&lt;/quote&gt; Perhaps that it is that statistical analysis is in no way equivalent to logical proof. This problem is amenable to both, however only one shows that it is correct the other merely suggest that it is probably so.
BTW further kudos to manduck that is one of the simplest, most elegant explanations I have ever seen. Despite prior experience of humans, I still couldn't believe people were still disagreeing after that.

[Little Nemo]

I agree, Rorschach, that a experimental run will not "prove" the problem in a mathematical sense. But hopefully, it would shake some of the doubters out of the certainty they seemed locked into. Once they see the evidence that switching is the winning strategy, they might consider the logic that backs this up. And if not, at least they'll have stopped pestering us for awhile.

[prokopowicz]

I'm not trying to convince anyone of anything regarding the Monty Hall problem.
Just play the game for money and you will see that Marilyn was right.

I want to offer another version of the somewhat simliar sibling problem that has a
slight twist which makes clear how important subtle concrete details are
in stating the problem.

Suppose you meet a girl on the street and learn she has one sibling. What is
the probability that the sibling is a boy? 50-50. Why? Here are the possible
scenarios.

1) boy1, boy2
2) boy-young, girl-old
3) boy-old, girl-young
4) girl1, girl2

On the street, you met either girl-old, girl-young, girl1, or girl2. If you met
girl-young or girl-old, the sibling is a boy. If you met girl1 or girl2, the sibling is
a boy.

[prokopowicz]

I'm sorry for posting so carelessly. Also sorry for beating a dead horse.

[norrinradd]

i didn't take a whole lot of statistics at university, but i took enough to have encountered this little scenario before. basically, your opinion on this issue is determined on how you answer the following question:

do you believe in the law of averages?

according to the law of averages, if the first child is female, then the other must be male. if you keep flipping heads, you're bound to flip tails soon, right?

well... the classic response is to treat each instance as a discrete event and say no. this avoids the classic fallacy of confusing performance and probability... if you keep flipping heads, maybe there's a greater probability of heads, right?

thing is, as pedestrian as the law of averages is, there is really is some substance to it, even with smaller samples than you would think. just ask a casino. in other words, marilyn is technically right... although in practicality, it can go either way for a while.

[Nightime]

Cecil himself asserts that this dilemma is tantemount to being offered all the unselected doors in return for your original choice. If this is true, why would that fact change because this option emerged randomly and not from the schemings of the evil Hall?



This is the toughest part of the problem. Why does it matter if Monty knows what is behind the door?


First, think of the original problem. The initial reaction of most people is that the deliberate opening of an empty door tells you nothing, because no matter what it is always possible to open a empty door.

This is actually very close to being right - it tells you nothing about the door you already chose, because indeed, there was no chance that one of the other doors was not empty, so there was no chance Monty could not choose to open an empty door.

However, if you learned nothing about your chosen door, then the chances that it is correct stay the same. You still have a 1/3 chance of having chosen correctly.



But what if it was completely random? What if any of the three doors could have been opened?

In that case, if an empty door other than the one you chose is opened, you DO learn something about your choice. It becomes more likely to be correct.

This is because, by random chance, 1/3 of the time the winning door will be opened. Since it was not, we must be in one of the other 2 scenarios, 1 of which has your door being correct. A 1/2 chance.





And now the scenario which I don't think has been brought up in any of these threads before (though I was unable to search for all of them).

What if Monty chooses randomly, but only among the doors you did not choose, and opens an empty door?


It seems to me that in this case, you would actually be better off staying with your original choice.


If one of the other doors was the winner, then half the time Monty would unveil the winner.

But if both the other doors are losers, then every time he will open a loser.

Therefore, 2/3 of the time, if Monty opens an empty door, your original choice was correct!



I don't think this has been brought up in previous threads, and I hope someone can confirm or disprove this last line of thought.

[GreyWanderer]

I'm not trying to convince anyone of anything regarding the Monty Hall problem.
Just play the game for money and you will see that Marilyn was right.

I want to offer another version of the somewhat simliar sibling problem that has a
slight twist which makes clear how important subtle concrete details are
in stating the problem.

Suppose you meet a girl on the street and learn she has one sibling. What is
the probability that the sibling is a boy? 50-50. Why? Here are the possible
scenarios.

1) boy1, boy2
2) boy-young, girl-old
3) boy-old, girl-young
4) girl1, girl2

On the street, you met either girl-old, girl-young, girl1, or girl2. If you met
girl-young or girl-old, the sibling is a boy. If you met girl1 or girl2, the sibling is
a boy.

The reason is that choice 4 becomes more likely, because you can meet either one of the girls.

It becomes:

1: 0%
2: 25%
3: 25%
4: 50%

In the original problem it's:

1: 0%
2: 33%
3: 33%
4: 33%

[GreyWanderer]

I'll do an experiment on the Monty Hall problem. Feel free to try this yourselves if you don't trust me.

I'll do the experiment 20 times. First I'll choose by dice roll which doors are correct.

3, 1, 2, 3, 3, 1, 2, 3, 2, 2, 3, 1, 2, 1, 2, 3, 3, 3, 2, 3

Which door you choose at first doesn't matter, so I'll choose it at random by dice roll.

2, 1, 1, 2, 3, 2, 3, 2, 3, 1, 2, 1, 3, 1, 3, 3, 1, 3, 3, 2

Then I'll start playing. I'll switch doors every time.

1:
I choose 2, Monty opens 1, I switch to 3. Win

2:
I choose 1, Monty opens 2, I switch to 3. Lose

3:
I choose 2, Monty opens 3, I switch to 2. Win

4:
I choose 2, Monty opens 1, I switch to 3. Win

5:
I choose 3, Monty opens 1, I switch to 2. Lose

6:
I choose 2, Monty opens 3, I switch to 1. Win

7:
I choose 3, Monty opens 1, I switch to 2. Win

8:
I choose 2, Monty opens 1, I switch to 3. Win

9:
I choose 3, Monty opens 1, I switch to 2. Win

10:
I choose 1, Monty opens 3, I switch to 2. Win

11:
I choose 2, Monty opens 1, I switch to 3. Win

12:
I choose 1, Monty opens 2, I switch to 3. Lose

13:
I choose 3, Monty opens 1, I switch to 2. Win

14:
I choose 1, Monty opens 2, I switch to 3. Lose

15:
I choose 3, Monty opens 1, I switch to 2. Win

16:
I choose 3, Monty opens 1, I switch to 2. Lose

17:
I choose 1, Monty opens 2, I switch to 3. Win

18:
I choose 3, Monty opens 1, I switch to 2. Lose

19:
I choose 3, Monty opens 1, I switch to 2. Win

20:
I choose 2, Monty opens 1, I switch to 3. Win

14 out of 20 wins. That's more than 2/3.

You'll see that the only times I lost was when the door I chose was the winning door. That was 6 out of 20 times. If you don't trust my dice rolls, then try it yourself.

[GreyWanderer]

Another way to look at it is like this. You can choose door 1 OR you can choose whichever one is correct of door 1 or door 2. You do this by first standing in front of door 3 and let Monty open the false one of door 1&2. Then you switch back to the one that Monty didn't open. That way you'll win if the prize was originally behind either door 1 or door 2.

[don't ask]

The easiest way to look at it is that you have the choice of picking one door for yourself and leaving Monty 2 to play with or you choose one door for Monty and you open the other 2. Which strategy sounds like it would win 2/3 of the time?

[Chastain86]

Ahhh, the yearly drudging out of the Monty Hall question. Is it 2004 already?

[Nightime]

The easiest way to look at it is that you have the choice of picking one door for yourself and leaving Monty 2 to play with or you choose one door for Monty and you open the other 2. Which strategy sounds like it would win 2/3 of the time?


That is the easiest way, but you have to be careful if you look at it that way.


Remember, it depends on whether or not Monty knew what was behind the door.

If Monty is choosing randomly between the doors you didn't pick, then it is very different.



In that case, if you choose door number 3, and Monty opens door number 2 showing it to be empty, you have a 2/3 chance of winning if you stay with your original door.




The concept also applies to the lottery problem. Say there are 1000 lottery tickets, one and only one winner, and you take one ticket and leave.

All the rest of the lottery tickets are thrown in a big pile on the floor, and people begin picking them up randomly and seeing if they are winners.


By the time there is only one ticket left on the floor, should you switch to take that ticket?

After all, there was only a 1/1000 chance that your original ticket was right.


But no... you should not switch. In fact, the odds are overwhelmingly in your favor that the ticket you already have is the winner.

[RM Mentock]

Nightime, as near as I can tell, that's backwards. If the choices are random, then there is no advantage to staying or switching.

[Nightime]

Thank you for answering. You are right, I had not thought it through enough.


If the door opening is random, then there is no advantage to staying or switching.

[Malacandra]

No, it's now a true 50-50 chance. Call the door that you open #1, the door that Monty opens #2 and the third door #3. The car has an equal chance of being behind any of them. One-third of the time Monty will reveal it and you have already lost. The rest of the time he will not reveal it and the car is equally likely to be behind door #1 or door #3.

This problem is a hardy annual, isn't it?

[mad_spoon]

I really can't see why anyone has a problem with this. THe answer IS 2/3. There isn't any doubt about it, and anyone that says its 50-50 is wrong.

Think of it like a flow chart. You have 3 possible choices at the start- car, goat or goat, although you don't know what one you picked. It DOES matter which one you pick. If you pick a goat, then the presenter will open the door to the other goat. If you change you win. If you pick the car first off, if you change, you lose.

So, if you pick a goat at the start; changing later means you win. No exception. If you pick the car at the start, changing = lose (duh).
There are 2 goats. There is one car. The chance of a randomly selected door being a goat is 2/3. Therefore, because if you pick a goat, changing= winning, the chance of winning if you change doors is 2/3.

Its not hard. OK?

[Chronos]

according to the law of averages, if the first child is female, then the other must be male. if you keep flipping heads, you're bound to flip tails soon, right?I'm surprised that nobody else has commented on this. This is absolutely wrong. The law of averages does not say that any given sample must contain equal numbers of boys and girls (assuming that either is equally probable). What it does say is that as your samples get larger, you're likely to get closer to 50% of each. The ratio of boys to girls approaches 1.

A pitfall here, though: The expected difference between the number of boys and the number of girls does not approach zero. In fact, as you make your sample size larger, the difference between the number of boys and girls is expected to get larger. It just doesn't get larger as fast as your sample gets larger, so relative to your sample size, the difference is still decreasing.

And if you keep flipping heads, you're still just as likely to flip another head on the next flip as you are to flip tails (assuming a fair coin). Each coin flip is completely independent of all previous flips. This is not only consistent with the Law of Averages; indeed, the Law of Averages depends on this property.

[Stratocaster]

This is the toughest part of the problem. Why does it matter if Monty knows what is behind the door? <snip> I don't think this has been brought up in previous threads, and I hope someone can confirm or disprove this last line of thought.I can't confirm either way. I can only point out that this is a new record for a response to one of my questions: almost 4 years from when I posted it. Where the hell were you? One more day and I was going to stop checking. ;)

[Nightime]

So I just made it, then! I figured you hadn't been able to sleep easy these 4 years, the question gnawing at your mind.


Sadly, despite taking 4 years to compose an answer, I still got it wrong. :(

[Manduck]

We are given that MH opens an empty door. It doesn't matter how he decided to open it, whether he knew it was empty or just picked it at random. You still have the situation where there is a 1/3 probability that your original guess was correct. So if you stick with your first guess you have only a 1/3 chance to win; therefore if you switch you must have a 2/3 chance to win, no matter how Monty picked his door.

[Nightime]

Ah ha! Now I can use my newfound knowledge.


It doesn't matter how he decided to open it, whether he knew it was empty or just picked it at random.



It *does* matter how he decided to open it. Here is why:


You pick the wrong door 2/3 of the time, true, but half of those times Monty will open the winning door if he is opening randomly among the two remaining doors!


That means that only 1/3 of the time will you pick the wrong door, AND have the option to switch to the winning door.


You also have a 1/3 chance of picking correctly from the start.


Therefore, if Monty is opening randomly, and he opens a wrong door, your chances are exactly the same whether you stay or switch.

[Manduck]

Ah ha! Now I can use my newfound knowledge.






It *does* matter how he decided to open it. Here is why:


You pick the wrong door 2/3 of the time, true, but half of those times Monty will open the winning door if he is opening randomly among the two remaining doors!


That means that only 1/3 of the time will you pick the wrong door, AND have the option to switch to the winning door.


You also have a 1/3 chance of picking correctly from the start.


Therefore, if Monty is opening randomly, and he opens a wrong door, your chances are exactly the same whether you stay or switch.

It's given that he opens an empty door. You have to make your decision whether to switch after seeing that he has opened an empty door. That is part of the puzzle. Given all that, you should switch. The possibility that he might open the prize door is not part of the problem. So given that he opened an empty door, his method for picking that door is irrelevant. In fact you would have no way of knowing whether he picked it at random anyhow, so the correct answer for the random scenario must be the same as the correct answer for the non-random one.

[RM Mentock]

I can't confirm either way. I can only point out that this is a new record for a response to one of my questions: almost 4 years from when I posted it. Where the hell were you? One more day and I was going to stop checking.
Thank prokopowicz
It's given that he opens an empty door. You have to make your decision whether to switch after seeing that he has opened an empty door. That is part of the puzzle. Given all that, you should switch. The possibility that he might open the prize door is not part of the problem. So given that he opened an empty door, his method for picking that door is irrelevant. In fact you would have no way of knowing whether he picked it at random anyhow, so the correct answer for the random scenario must be the same as the correct answer for the non-random one.
No way of knowing? Why not? Because it doesn't say so in the problem?

Depends upon which version of the problem you are reading. Generally, if it does not say so explicitly, it is assumed that he does know what is behind the doors, and it is not random.

Of course his method is relevant. Nightime's example shows why.

[Stratocaster]

I'm still missing why his randomness (or lack thereof) has any bearing at all on the decision, given the fact that Monty opens an empty door. To me it's the equivalent of Monty showing you one of the sides of a tossed die (but none of the other sides) and asking you the odds that the upper side matches what you guessed before the toss (with me?). Whether he randomly selected to show you a side 4, or whether he randomly selected a side and it turned out to be a 4, you have exactly the same probability calculation in front of you. Why would it be different?

Same in this problem, right?

[RM Mentock]

I'm still missing why his randomness (or lack thereof) has any bearing at all on the decision, given the fact that Monty opens an empty door. To me it's the equivalent of Monty showing you one of the sides of a tossed die (but none of the other sides) and asking you the odds that the upper side matches what you guessed before the toss (with me?). Whether he randomly selected to show you a side 4, or whether he randomly selected a side and it turned out to be a 4, you have exactly the same probability calculation in front of you. Why would it be different?

Same in this problem, right?
But you're saying that in both cases, the choice is random in your two examples. Would he show you the side if it had your number on it? That's the crux.

What is interesting is that this "random" line of thinking is what got all those mathematicians into trouble in the first place. They were so used to all the "non-intuitive" solutions to randomized choices, that they objected strongly, and in writing, to Marilyn's solution. Marilyn, of course, showed them wrong. Now, people have a hard time seeing how the opposing example can be correct, because of the Monte Hall problem.

[Stratocaster]

But you're saying that in both cases, the choice is random in your two examples. Would he show you the side if it had your number on it? That's the crux.Maybe we're talking about two different points. In any single instance where Monty shows you a losing choice (or a die face that isn't your choice), for that single iteration it matters not whether it was deliberate or not on Monty's part, you still have the same probability caclulation in making a decision. Period. How could it be otherwise?

Let me put it another way. Say Monty did billion iterations where he deliberately showed you a losing door; let's collectively call those events Set A. Let's say he also did a billion+ iterations where he randomly selected a door, and the subset of those where he randomly selected a losing door amounted to a billion iterations (call it Set B). I posit that in either Set A or Set B, it is the same advantage to change your choice on any given iteration (all of them where Monty showed you a loser), and the overall results of those two billion iteration would show this.

Do you agree with this? I'm really just trying to understand if we're discussing different points, I guess. And, just as I did 4 years ago, I fully allow I may just be completely missing the point.

[RM Mentock]

Maybe we're talking about two different points.
We don't appear to be, to me.
In any single instance where Monty shows you a losing choice (or a die face that isn't your choice), for that single iteration it matters not whether it was deliberate or not on Monty's part, you still have the same probability caclulation in making a decision. Period.
I disagree.
How could it be otherwise?
Rather than using your example with billions of instances, let's break it down into counting number of possible outcomes.

With three doors, you can choose any of the three, and they are equally probable. Call them A, B, C, with C being the one with the prize. Each choice has 1/3 chance of occurring. So, Ab would represent you choosing A, and Monty showing B. Notice Ac is not an option, if Monty is not choosing randomly. So, the possibilities are Ab, Ba, Ca, and Cb, but the probability of Ca is only 1/6th, since Ca and Cb can be considered equally probable. Just because there are four possibilities does not mean that they are equally probable. The sum of all the probabilities is 1, of course. In the case of Ab and Ba, you would win if you switched, so the probability of winning is 2/3 if you switch. You don't win by switching with Ca or Cb. Of course, you should switch.

If Monty is choosing randomly, then all six of Ab, Ac, Ba, Bc, Ca, and Cb are possible, and they all have probability 1/6. Which ones satisfy the terms of our problem? Only Ab, Ba, Ca, and Cb, since the other two would involve revealing the prize, and the problem states that that does not occur. These are the same options as before, but their probabilities are different! In this case, the probability of winning is only 1/2 if you switch. In other words, it doesn't matter if you switch or not.

The assumptions about how you got to a particular situation do indeed affect the probabilities. Too many mathematicians were treating the problem in the second fashion--but everyone knows that game shows are rigged, right? Well, at least, that Monty Hall knew what he was doing.
I'm really just trying to understand if we're discussing different points, I guess. And, just as I did 4 years ago, I fully allow I may just be completely missing the point.
Wow, you did ask four years ago, and there wasn't a real response. I'll be d*rned.

[Manduck]

If Monty is choosing randomly, then all six of Ab, Ac, Ba, Bc, Ca, and Cb are possible, and they all have probability 1/6. Which ones satisfy the terms of our problem? Only Ab, Ba, Ca, and Cb, since the other two would involve revealing the prize, and the problem states that that does not occur. These are the same options as before, but their probabilities are different! In this case, the probability of winning is only 1/2 if you switch. In other words, it doesn't matter if you switch or not.


Remember, the probabilities of all outcomes must add up to one. Since we are given that Ac and Bc did not happen, their probabilities become zero. What then are the probabilities of Ab, Ba, Ca, and Cb?

Here's the way I look at it: The probability that my first guess was wrong is 2/3. That means that the sum of the probabilities of Ab, Ac, Ba, and Bc must be 2/3. It doesn't matter how that 2/3 is distributed among those four because I know that if I chose wrong on the first guess I will win if I switch. I also know that there is a 2/3 probability that I was wrong on the first guess. Therefore I should switch.

[RM Mentock]

Remember, the probabilities of all outcomes must add up to one. Since we are given that Ac and Bc did not happen, their probabilities become zero. What then are the probabilities of Ab, Ba, Ca, and Cb?
That depends on how you define them, of course. I defined them two different ways.

Here's the way I look at it: The probability that my first guess was wrong is 2/3. That means that the sum of the probabilities of Ab, Ac, Ba, and Bc must be 2/3. It doesn't matter how that 2/3 is distributed among those four
[quote]
In my previous post, I showed that it does make a difference.
[quote]
because I know that if I chose wrong on the first guess I will win if I switch. I also know that there is a 2/3 probability that I was wrong on the first guess.
Not if Monty's choice is random AND you have thrown out those times where he did choose the prize door--which would be 1/3 of the time.
Therefore I should switch.
If the probability is 1/2, it won't matter if you switch. But there is a situation where you definitely should not switch--Evil Monty. Evil Monty is the guy who knows all about this problem and how everybody has convinced themselves that they should switch. So, what does he do? If you choose the wrong door, he *always* shows the right door. If you choose the right door, you'll switch. So you never get the prize.

Good Monty would open all the doors that don't have prizes--so you know exactly which one has the prize.

Monty's behavior and motivation definitely changes the odds.

[Manduck]

RM, I think you are getting lost in the details and missing the big picture. The problem is really simple. There are only two facts you need to consider:

1. If you chose the wrong door initially, you will win if you switch. This is an inescapable fact because we are given the MH will eliminate a non-prize door, leaving only one other choice.

2. There is a 2/3 chance you will pick the wrong door at the start, because there are 3 doors and you have no information about them when you make your choice.

The only probability that matters is that 2/3. No matter what Monty's motivations are, he can't change that 2/3 because you make that choice before Monty does anything. The only way Monty can affect you chance of winning is by opening the prize door, but we are given that he doesn't do that, so that is not an event that we should even consider.

[Nightime]

It seems to me that back when this problem first came up, a lot of the answers basically said "you have a 2/3 chance of picking wrong from the start, so you should always switch".

Unfortunately, this is a poor way to answer the question, and leads to difficulty later on when thinking about variations of the problem.

The real reason you should switch in the original problem is *not* merely because you had a 2/3 chance to be wrong, but rather because 2/3 of the time the winning door will still be available to switch to.



If Monty is opening randomly, then:


1/3 of the time you pick right, and Monty opens an empty door.

1/3 of the time you pick wrong, and Monty opens an empty door.

1/3 of the time you pick wrong, and Monty reveals the winning door.



Since we are told Monty opened an empty door, we know we are in one of the first two scenarios. But as you can see, these scenarios are equally likely. So there is no advantage to switching.



Another illustration of why the chances are equal:

What if TWO people pick doors, and the third door is chosen randomly and turns out to be empty?

By the reasoning that "I had a 2/3 chance to be wrong and should switch", both people would have a better chance if they switched to the other's door, but clearly that is impossible.

[GreyWanderer]

Nighttime is right. That last post showed it quite clearly.

What Monty does is basically stopping the game (by opening the winning door) half of the times that it would be best to switch doors. So instead of it being 1/3 chance that you will lose and 2/3 chance that you will win it will be 1/3 chance that you will lose, 1/3 chance that you will win and 1/3 chance that Monty will ruin the game by opening to winning door.

[Stratocaster]

Let the doors be a, b, and c. A capital letter signifies that the player selected this door. Underlining means that it's the door Monty opens randomly. The color red indicates the winning door. Assuming each of these events is random, the following depicts the 27 possible outcomes:

1. Abc
2. Abc
3. Abc
4. Abc
5. Abc
6. Abc
7. Abc
8. Abc
9. Abc
10. aBc
11. aBc
12. aBc
13. aBc
14. aBc
15. aBc
16. aBc
17. aBc
18. aBc
19. abC
20. abC
21. abC
22. abC
23. abC
24. abC
25. abC
26. abC
27. abC

In 9 of the outcomes, Monty reveals the winning door (1, 5, 9, 10, 14, 18, 19, 23, 27). Not counting those already noted, Monty reveals an additional 6 doors that the player selected (2, 3, 13, 15, 25, 26). These 15 are not relevant to the question at hand.

In each of the remaining 12 possibilities, Monty reveals a losing door. In 6 of the 12, the player selected the winning door initially (4, 7, 11, 17, 21, 24). Switching doors would mean the player would lose.

In the other 6, the player initially picked a loser, so switching would win for them (6, 8, 12, 16, 20, 22).

So, in any random scenario where the player is shown a losing door, switching or not has no bearing on winning.

Does this make sense? Did I mess any of this up (I had a couple of beers at dinner, so check me)? If this is right, I still can't get my head around it. Outcomes 4, 6, 7, 8, 11, 12, 16, 17, 20, 21, 22, and 24 are the 12 possible combinations where Monty could show an empty door, whether he did so randomly or deliberately.

Did somebody already try this? Man, I'm confused. I'm getting another beer.

[Manduck]

Again I think people are confusing themselves with irrelevant details. If the probability of picking the wrong door is 2/3 (which I think we all agree on) and it's a fact that you will win if you pick the wrong door and then switch (which is obvious) then it follows that you have a 2/3 chance of winning if you switch.

A mistake some people are making is to say that the probability of MH opening the winning door is 1/3; but, the probability of him opening the winning door is actually 0, because we are told that he does not open the winning door.

Try to imagine yourself actually playing this game. You have three doors, all alike. You pick one. Then some guy opens one of the others and you see that there is no prize behind it. Maybe he knew that door was empty and opened it deliberately; maybe he picked it at random and it's a fluke that it was empty - you don't know. So, do you switch? You know that you should switch if you picked the wrong door at first, right? And how likely was that?

The decision to switch is correct whenever you first choice was wrong. The rightness or wrongness of your first choice is not influenced by whether Monty Hall knows where the prize is, so the decision whether to switch should likewise not depend on what MH knows. You know that your first choice was probably wong,so you know that if you switch you will probably win.

[GreyWanderer]

A mistake some people are making is to say that the probability of MH opening the winning door is 1/3; but, the probability of him opening the winning door is actually 0, because we are told that he does not open the winning door.

I think everyone agrees on this. The only thing people are saying is that if Monty _does_ open winning doors too (which would be pretty stupid, and ruin the game) it wouldn't matter if you switched or not.

[Manduck]

I think everyone agrees on this. The only thing people are saying is that if Monty _does_ open winning doors too (which would be pretty stupid, and ruin the game) it wouldn't matter if you switched or not.

If he opens a winning door, you already lost the game, so sure, in that case it doesn't matter what you do. But if he is allowed to open a winning door, but happens to open a losing door, then you still improve your chances by switching (because your first guess was probably wrong).

[Nightime]

But if he is allowed to open a winning door, but happens to open a losing door, then you still improve your chances by switching (because your first guess was probably wrong).


Nope. The reason for this is that opening an empty door randomly adds another element of chance, while opening an empty door that you already know is empty adds no element of chance.



Chances of guess right: 1/3

Chances of randomly opening an empty door after guessing right: 1

(this is a key point... since you already guessed right, Monty will *always* randomly open an empty door)


1/3 x 1 = 1/3



Chances of guessing wrong: 2/3

Chances of randomly opening an empty door after guessing wrong: 1/2

(since you guessed wrong, there is only a 1/2 chance that Monty will open an empty door)


2/3 x 1/2 = 1/3



So the chances are exactly equal that you will guess right and see an empty door, and that you will guess wrong and see an empty door.


So yes, *after* seeing an empty door opened randomly, there is no advantage to staying or switching.

[Manduck]

You are presented with the option to switch after seeing an empty door opened, so the probability of that empty door being opened isn't 1/2; it is 1 (because it has actually happened - anything that you know has actually happened has probability 1).

[Nightime]

Right.


And since you know the empty door was opened, you know you are in one of two scenarios:


1. You guessed wrong, and Monty opened an empty door.

2. You guessed right, and Monty opened an empty door.



Since these scenarios are equally likely, there is no advantage to staying or switching.

[Manduck]

Those scenarios aren'y equally likely. Since you know that Monty opened an empty door, the probability of that event is 1. So the probabilities are:

1. You guessed wrong, and Monty opened an empty door: 2/3 * 1 = 2/3
2. You guessed right, and Monty opened an empty door: 1/3 * 1 = 1/3

If you calculate the probability of scenario 1 the way you do, i.e. (2/3 * 1/2 = 1/3), you are ignoring information. You are pretending that the probability of Monty opening that empty door is 1/2, when in fact Monty has actually opened that door already, so the probability of that event is 1.

[Nightime]

I'll try a different tactic: What if 2 people were playing?


I pick door number 1.

You pick door number 2.



Monty randomly opens door 3, and it is empty.



According to your logic, both of us should think "I probably chose wrong, so I should switch to the one remaining door".


But this makes no sense, because we would just be switching with each other.

[don't ask]

Play Monty 6 times:

Let's say you always pick door 1.

1/3 of the time you're right.

first game Monty picks door 2, it is empty. You swap and lose.

second gameMonty picks door 3, it is empty. You swap and lose.

2/3 of the time you are wrong.

third game Monty picks door 2, it is empty. You swap and win.

fourth gameMonty picks door 3, it is empty. You swap and win.

fifth game Monty picks door 2, it is the prize. Game over.

sixth gameMonty picks door 3, it is the prize. Game over.

There are no other possibilities.

You are wrong with this:

1. You guessed wrong, and Monty opened an empty door: 2/3 * 1 = 2/3

As you can see it is 1/3*1 = 1/3 because if he revealed the prize you don't see the empty door.

[RM Mentock]

Those scenarios aren'y equally likely. Since you know that Monty opened an empty door, the probability of that event is 1.
If Monty is opening the doors randomly, the probability is 1/2.

So, half the time Monty ruins the game by opening the prize door--and that doesn't satisfy the original game scenario so it doesn't allow you to switch, and it reduces your chances to switch to the prize door.

[Manduck]

Nightime - your two-person game is quite a bit different from the original problem. Here's how I see it:

You say that Monty reveals an empty door after we make our picks. That means that one of us must have chosen the right door. Let's say you pick first. If your first choice is wrong, that means my first choice must be right. We know this because you are saying that the remaining door is not the winner.

So, there is a 2/3 chance that your first guess was wrong, so you should switch. Likewise, there is a 2/3 chance that my first guess was right (because my first guess is right whenever yours is wrong), so I should stay with my first choice.

RM- It's given that Monty does not open the prize door, so that event cannot have a non-zero probability. If you do assign it a non-zero probability, it would mean that Monty might not open an empty door, but we already know (because it is given) that he does open an empty door. So the probability of him opening the prize door has to be 0 and the probability of him opening a non-prize door has to be 1.

[Stratocaster]

Again I think people are confusing themselves with irrelevant details.What extraneous details are included in what I lined up?

I itemized all the possible outcomes of 3 events: The player's initial selection, the winning door, and the door Monty reveals. There are 27 possibilities, each of them equally likely.

I eliminate from this group all combinations where Monty reveals the winning door and/or the player's initial selection, since those outcomes aren't relevant. The 27 is reduced to 12, 12 combinations where Monty is revealing an empty door. If I am thinking of this clearly, there are 12, and only 12, combinations of these 3 events where Monty reveals a loser.

Within those 12, there are 6 where the player originally selected correctly and 6 where he didn't. Put another way, there are 6 where switching is the right move and 6 where it isn't.

Again, what is wrong with what I outlined? (There may well be something, but I can't see what.)

[Nightime]

So, there is a 2/3 chance that your first guess was wrong, so you should switch. Likewise, there is a 2/3 chance that my first guess was right (because my first guess is right whenever yours is wrong), so I should stay with my first choice.


Neither of us knows the other is playing, and we choose at precisely the same time.


Yet there is a 2/3 chance that your original guess is right?


Maybe if you are psychic. Otherwise our chances are equal.


Just as the chances are always equal for the two remaining doors, if Monty randomly opens an empty door.





1/3 of the time you guess right, and Monty opens an empty door.

1/3 of the time you guess wrong, and Monty opens an empty door.

1/3 of the time you guess wrong, and Monty opens the winning door.



The fact that Monty opened an empty door tells us that we are in one of the first 2 scenarios. They are equally likely.

[RM Mentock]

RM- It's given that Monty does not open the prize door, so that event cannot have a non-zero probability. If you do assign it a non-zero probability,
It's not a matter of assigning probabilities. It's a matter of computing probabilities. You have to make sure that your assumptions are consistent.

If you do assign it a non-zero probability, it would mean that Monty might not open an empty door, but we already know (because it is given) that he does open an empty door. So the probability of him opening the prize door has to be 0 and the probability of him opening a non-prize door has to be 1.
Let me get this straight. Your assumption is that Monty is faced with two doors, one of which is the prize door, and he is choosing randomly between them, and you say the probability of him choosing the prize door is zero?

Try playing this game six times, with Monty picking randomly, with all six possible outcomes. What do you do with the situations where Monty picks the prize door? (If he is picking randomly, he *will* sometimes pick the prize door.) How do you propose to incorporate them into the calculations? Do you think it affects the relative probabilities of the other events?

[Nightime]

Again, what is wrong with what I outlined? (There may well be something, but I can't see what.)


Your list looks good to me, Stratocaster. It shows why the doors are equally likely to be the winner if Monty chooses randomly.



If this is right, I still can't get my head around it. Outcomes 4, 6, 7, 8, 11, 12, 16, 17, 20, 21, 22, and 24 are the 12 possible combinations where Monty could show an empty door, whether he did so randomly or deliberately


True. But if Monty knows what is behind the doors, he can change the probability of each of these outcomes.

Look at outcomes 6 and 9 on your list. If Monty knows what is behind the doors, outcome 9 will be replaced by another instance of outcome 6, thus making outcome 6 twice as likely as if Monty is picking randomly.

The same happens to all the outcomes in which you guess wrong. Each will occur twice as often, thus giving you a 2/3 chance if you switch.

[Nightime]

That last line should say:

"The same happens to all the outcomes in which you guess wrong AND Monty opens an empty door. Each will occur twice as often, thus giving you a 2/3 chance if you switch."

[GreyWanderer]

Listen here:

The chance of you picking the wrong door at first is 2/3. But half of the times you start by picking the wrong door and thus would win, Monty ruins the game for you instead. So that outcome that happens 2/3 of the time is split into a 1/3 where you will win and a 1/3 where Monty will ruin the game.

[Manduck]

You guys are wearing me out {:^P

I'll try to get to everybody over the next few posts.

First, Stratocaster:

The mistake your are making is eliminating several of your 27 outcomes and assuming that the probabilities of the others remain equal. In fact, some outcomes are more likely than others.

Let's say that the prize is behind door A. The possible sequences of my pick/Monty's pick are:

AB
AC
BA
BC
CA
CB

(Actually BA and CA aren't possible, but bear with me for a moment.)

So, what is the probability of AB? The probability of A is 1/3 (because I'm choosing among 3 doors). Once I have chosen A, the probabilty of Monty choosing B is 1/2 (because Monty could choose either B or C). So the probability of AB is:

AB: 1/3 * 1/2 = 1/6

By the same argument, the probability of AC is

AC: 1/3 * 1/2 = 1/6

Now we come to BA. The probability that I pick B is 1/3; so, given that I have picked B, what is the probability that Monty picks A? The prize is behind door A, and we are given that Monty does not reveal the prize, so the probability that Monty picks A must be 0. I can't stress that enough. That is the key to the problem. We know that Monty doesn't pick A, so we have to assign it a 0 probability. So the probability of BA is:

BA: 1/3 * 0 = 0

So, if I pick B, and we know that Monty doesn't pick A, then he must pick C because that is the only remaining choice. That means that the probability that he picks C is 1. So

BC: 1/3 * 1 = 1/3.

The same logic that applies to BA and CA also applies to CA and CB, so their probabilities are

CA: 1/3 * 0 = 0
CB: 1/3 * 1 = 1/3

So the possible outcomes and their probabilties are

AB: 1/6
AC: 1/6
BA: 0
BC: 1/3
CA: 0
CB: 1/3

The cases where I will win if I switch are BC and CB. Their probabilities add up to 2/3. so I should switch.

Where your analysis goes wrong is that you say that each outcome has 1/27 probability, then you eliminate 15 of them as being impossible, but not reexamining the probabilities of remaining 12 outcomes. So the probabilities of all possible outcomes in your analysis was adding up to 12/27, which can't be right. The total probabilities of all possible outcomes has to be 1. Also, when you eliminate a combination, you can't assume that that doesn't affect the distribution of probabilities among the other combinations. You can see that in my analysis above, where eliminating BA made BC more likely, but didn't afftect the probabilities of AB or BC.

[Manduck]

Neither of us knows the other is playing, and we choose at precisely the same time.


Yet there is a 2/3 chance that your original guess is right?


Maybe if you are psychic. Otherwise our chances are equal.



In your description of the game, you said that you pick a door, I pick a door, and the remaining door is opened by Monty. From that I inferred that the game didn't allow us to both open the same door. If we are allowed to choose the same door, that changes everything. But that still leaves the situation where we both pick the two empty doors, leaving only the prize door for Monty. You description of the game says that Monty opens an empty door, so that situation can't arise, so IF we pick different doors, one of them must be the winner. IF we aren't allowed to pick the same door, then the player who goes first should switch and the other player shouldn't.

If I was correct, that we can't pick the same door, then presumably the game would have some mechanism to enforce this. I don't know, you made up the rules.


Just as the chances are always equal for the two remaining doors, if Monty randomly opens an empty door.


Not true, because it is given that Monty doesn't open the prize door. If the prize door is one of the two remaining, then the probability of him opening that one is 0 and the probability of him opening the other is 1. Anything else would violate the premises of the problem. See my response to Stratocaster above for the mind-numbing details.



1/3 of the time you guess right, and Monty opens an empty door.

1/3 of the time you guess wrong, and Monty opens an empty door.

1/3 of the time you guess wrong, and Monty opens the winning door.

The fact that Monty opened an empty door tells us that we are in one of the first 2 scenarios. They are equally likely.

They are not equally likely withing the problem domain. The problem says that scenario three doesn't happen. Therefore its probability is 0, and the 1/3 that was being assigned to it must be distributed among the other two scenarios. Clearly it belongs with scenario two, because that is the only other one that has me picking the wrong door.

[Manduck]

It's not a matter of assigning probabilities. It's a matter of computing probabilities. You have to make sure that your assumptions are consistent.


Whether you call it assigning or computing, every event has to have a probability. An event that you know does not happen has to have 0 as its probability. An event that you know does happen has to have 1 as its probability. We are given that Monty opens an empty door, so that event has to have probability 1, and the event that he opens the prize door has to have probability 0. We are given this in the statement of the problem; we have no choice about those probabilities.



Let me get this straight. Your assumption is that Monty is faced with two doors, one of which is the prize door, and he is choosing randomly between them, and you say the probability of him choosing the prize door is zero?


That's not my assumption; that is given. We are told that Monty opened and empty door. Given that, he can't open a prize door, because that would take us into a situation that this problem isn't about.


Try playing this game six times, with Monty picking randomly, with all six possible outcomes. What do you do with the situations where Monty picks the prize door? (If he is picking randomly, he *will* sometimes pick the prize door.) How do you propose to incorporate them into the calculations? Do you think it affects the relative probabilities of the other events?

If you play this game and Monty picks the prize door, then that particular iteration of the game isn't one that we are being asked about in this problem. The problem asks "What should you do, given that Monty opens an empty door and gives you the option to switch".

Again, imagine you are playing this game. There are 3 doors. A guy tells you there is a prize behind one of them, and invites you to pick one. So you pick one. The guy then opens one of the other doors, and you see that there's nothing behind it. The guy asks if you want to switch. Let's stop here for a second.

Maybe if you played the game another time, Monty would open the prize door and say "ha ha you lose. sucker!" But should that affect your answer in the game you're playing right now, in which you know that he opened an empty door? The prize is either behind the door you picked first, or the other unopened door. There is a 1/3 chance you picked the correct door first. That is all you need to figure out your answer. It doesn't matter what might have happened in some other iteration of the game, because you know that in this iteration, it didn't happen.

[Manduck]

Listen here:

The chance of you picking the wrong door at first is 2/3. But half of the times you start by picking the wrong door and thus would win, Monty ruins the game for you instead. So that outcome that happens 2/3 of the time is split into a 1/3 where you will win and a 1/3 where Monty will ruin the game.


Not really, because the statement of the problem says that Monty doesn't pick the prize, so that event is out of play. The last two paragraphs of my previous post explain it (I hope).

[Manduck]

Whew! I always hated essay tests :D

[Stratocaster]

You guys are wearing me out {:^P

I'll try to get to everybody over the next few posts.

First, Stratocaster:

The mistake your are making is eliminating several of your 27 outcomes and assuming that the probabilities of the others remain equal. In fact, some outcomes are more likely than others.If you haven't been worn out, let me try again. I feel like I'm close to a breakthrough.

The combinations that I laid out are equally likely, right? The 12 combos that remain, whatever their probabilities, are equally likely, correct? The combos that aren't relevant don't matter.

Here's an anology. I'm trying to determine how often a 6 or a 4 will come up, relative to each other, if I roll a die. All faces of the die have the same probability: 1/6. But every roll where a non-4 or a non-6 comes up is simply ignored, and I record how often a 4 or a 6 is rolled.

Those outcomes, a 4 or a 6, are still equally likely. The probability just has a different algebra behind it, and I'm certain that over a long haul I will conclude that 4's come up about 50% of the time, as do 6's (in rolls that "count").

So, how is that different from what I've laid out? Each of the 27 combos is equally likely, if all events are randomly chosen. Whenever one of 15 comes up, though, I simply ignore it--I only care about outcomes where Monty picks an empty door. I posit that each of the 12 has a 1/12 probability. What's wrong with that logic?

And when are we going to discuss the great philosophical question, what, exactly, does the phrase "empty door" mean?

[Manduck]

Yeah I cringe a bit every time I type "empty door", but I don't want to type "door with nothing behind it" five times a paragraph, so I use that sloppy shorthand :p.


The combinations that I laid out are equally likely, right? The 12 combos that remain, whatever their probabilities, are equally likely, correct? The combos that aren't relevant don't matter.


Nope - the combos aren't equally likely. I think I explained that pretty thoroughly in my previous response to you. If you choose a prize door, there are two possible picks Monty can make, but if you choose an empty door [heh] Monty can only make one choice. That is because we know that Monty picks an empty door [sic;)]. So that makes a sequence like CB twice as likely as a sequence like AB (if the prize is behind door A).



Here's an anology. I'm trying to determine how often a 6 or a 4 will come up, relative to each other, if I roll a die. All faces of the die have the same probability: 1/6. But every roll where a non-4 or a non-6 comes up is simply ignored, and I record how often a 4 or a 6 is rolled.

Those outcomes, a 4 or a 6, are still equally likely. The probability just has a different algebra behind it, and I'm certain that over a long haul I will conclude that 4's come up about 50% of the time, as do 6's (in rolls that "count").



Your dice example is not like the Monty Hall problem because it doesn't have the "x happens, then y happens" setup, so you don't have to deal with contingent probabilities.

But anyway - the probability of a 4 coming up on a roll is 1/6, also the probability of a 6 coming up is 1/6. If you now say that any other number must be rerolled until a 4 or 6 comes up, the probability of eventually getting a 4 becomes 1/2, and the probability of a 6 also becomes 1/2. Those probabilities are equal, but they are no longer 1/6. The are equal because there is nothing in the re-roll rule that biases toward one outcome or the other.

Now, in the Monty Hall problem, the analog to the re-roll rule is to reject (or replay) any game where Monty opens the prize door. So, that leaves us with the situation where you have an option to switch to the other unopened door. You only have two unopened doors, so one of them must have the prize. When will the prize be behind the door that you didn't pick? It will be there whenever you picked a door without a prize with your first pick. You know that that will happen 2/3 of the time. So: switch. The game is already biased toward switching because of that 2/3 probability of an initial wrong guess.


So, how is that different from what I've laid out? Each of the 27 combos is equally likely, if all events are randomly chosen. Whenever one of 15 comes up, though, I simply ignore it--I only care about outcomes where Monty picks an empty door. I posit that each of the 12 has a 1/12 probability. What's wrong with that logic?


You have to work out the probabilities taking into account what you know about the game. You should ask "what can happen first" and "how likely is it". and then, for each of those, "what can happen next" and "howl likely is that".

So, there a two things that can happen first. Either you pick the prize door (probability 1/3) or you can pick one of the loser doors (probability 2/3).

Given you picked the right door, Monty can open one loser door (probability 1/2) or the other loser door (probability 1/2).

Given you picked a loser door, Monty can only open the other loser door (probability 1).

Sooo we have

Probability of I-pick-winner, Monty-picks-loser1 = 1/3*1/2 = 1/6
Probability of I-pick-winner, Monty-picks-loser2 = 1/3*1/2 = 1/6
Probability of I-pick-loser, Monty-picks-other-loser = 2/3*1 = 2/3

[Asteroide]

Manduck

if you state from the outset that monty has a 0 probability of selecting the prize door, then you are stating that his choice is non-random by definition, and we're back to the initial scenario where Monty knows what he's doing, and a switch is advantageous.

If we're describing a given instance where Monty has selected a non-prize door through a random process (non-prize probability 2/3), then the two remaining doors have an equal chance of being the prize-door.

The key is that when Monty's selection is random, he's not giving you any extra information about the remaining door - he doesn't have any to give.

[Manduck]

Asteroide

Nope - If Monty happens to pick a non-prize door, then the prize must either be behind the door you picked initially, or the other unopened door. You already know that there is a 1/3 chance that the prize is behind your first pick, so the probability that it's behind the only other alternative has to be 2/3. It doesn't matter how Monty arrived at his choice, once you know that he did in fact pick a non-prize door.

[don't ask]

Asteroide

Nope - If Monty happens to pick a non-prize door, then the prize must either be behind the door you picked initially, or the other unopened door. You already know that there is a 1/3 chance that the prize is behind your first pick, so the probability that it's behind the only other alternative has to be 2/3. It doesn't matter how Monty arrived at his choice, once you know that he did in fact pick a non-prize door.

By that logic if you choose door A and I choose door B and Monty opens door C to reveal nothing each of us has 1/3 of a chance of winning. What happened to the other third?

[Nightime]

Manduck, I am beginning to despair of even explaining this to you.


However, you can prove to yourself that the chances are equal by performing a simple test.



Take 3 coins, and put one heads up. That is the "winning coin".


Now you need some method of choosing randomly. Once you have it, choose a random coin. That is your chosen coin.


Now choose a random coin from the remaining 2 coins. That is the coin "Monty" reveals.




Do this experiment over and over, and keep track of the times that the coin "Monty" reveals is a losing coin.



What you will discover is that half of the times Monty reveals a losing coin, you have already chosen correctly.

[Manduck]

By that logic if you choose door A and I choose door B and Monty opens door C to reveal nothing each of us has 1/3 of a chance of winning. What happened to the other third?

I think what you're asking is, what if I choose a door at random, and you choose one of the other two doors, and Monty reveals that the last door is empty, what are our respective probabilities of winning the prize. If I have that right, here's how it breaks down.

Lets call the doors A, B, and C and let's say that the prize is behind door A. The possible outcomes are (first letter is my door, second letter is your door, third letter is Monty's empty door):

ABC
ACB
BAC
CAB

Note that BCA and CBA don't appear on the list. That's because those are the cases that have Monty's door containing the prize. You have stated that Monty opens an empty door, so BCA and CBA can't happen.; their probabilities are 0.

So, the probability that I pick A with my guess is 1/3, because I am picking from all three doors. If I pick A, the probability that you will pick B is 1/2, because you have two doors to choose from. Similarly, the probability that you will pick C is also 1/2. In either case Monty has to open the remaining door, so the probability of him doing that is 1. So the ABC and ACB sequences have these probabilities:

ABC: 1/3 * 1/2 * 1 = 1/6
ACB: 1/3 * 1/2 * 1 = 1/6

There is a 1/3 probability that my first pick will be B, If I pick B, the probability that you pick C has to be 0. This is because it's given that Monty will open an empty door. He can't do that if you pick the empty door, so you have to leave it for him to pick. Your choice has to be A (probability 1). So:

BAC: 1/3 * 1 * 1 = 1/3

The same logic applies if I pick C:

CAB: 1/3 * 1 * 1 = 1/3

So I win in the cases ABC and ACB, which each have a probability of 1/6; the probability that I will win is 1/3.

You win in cases BAC and CAB; their total probability is 2/3.

So you have an advantage in this game. If that seems counterintuitive, it's because the restriction that Monty's door will be empty creates an artificial situation that forces you to pick the winning door if I happen to pick a loser. If you were going to implement a game like this in real life, you would have a do-over if the prize turned out to be behind Monty's door. You would keep replaying until Monty's door came up empty.

[Nightime]

If that seems counterintuitive, it's because the restriction that Monty's door will be empty creates an artificial situation that forces you to pick the winning door if I happen to pick a loser.


:eek:


Surely you are not suggesting that the only way for Monty to open a losing door is an "artificial situation"?


Consider that he may have opened it by chance. Nothing artificial about that.

And indeed, that is precisely what the problem says happened.


It wasn't destiny, or preordained. It just happens that, by chance, he opened a losing door. Thus the remaining doors are equally likely to be the winner.



If angry poltergeist were whispering hints to Monty to guarantee he opens a losing door, then yes, you should switch.

[Manduck]

:eek:


Surely you are not suggesting that the only way for Monty to open a losing door is an "artificial situation"?


Consider that he may have opened it by chance. Nothing artificial about that.

And indeed, that is precisely what the problem says happened.


It wasn't destiny, or preordained. It just happens that, by chance, he opened a losing door. Thus the remaining doors are equally likely to be the winner.



If angry poltergeist were whispering hints to Monty to guarantee he opens a losing door, then yes, you should switch.

Whether by chance or not, the possibility of Monty picking the prize door is RULED OUT by the statement of the problem. There may be other iterations of the game where he does pick the prize, but the problem only concerns iterations where he does not pick the prize.

So, in don't ask's variation, I pick a door, he picks one of the other two doors, Monty picks the remaining door, and Monty's is reviealed to be empty. In that situation, the probability is 2/3 that don't ask won the prize.

Before Monty opens his door, the probabilites are indeed 1/3 that I win, 1/3 that don't ask wins, and 1/3 that neither wins. If Monty now opens the door and reveals that the prize isn't there, the probabilities become 1/3 that I win and 2/3 that don't ask wins. That is because we are able to incorporate new information into the analysis: We know that if I picked a losing door, don't ask must have picked a winning door.

It makes sense when you think about it: The first player to pick has a choice of three doors, so his probability has to be 1/3. The second player does not have a free choice of the remaining two, because he has to leave an empty door for Monty.

[Nightime]

Please try the experiment I described two posts ago.


You will be surprised to find that half the times that "Monty" opens a losing door, your original choice is correct!


Seriously. Just try it.

[Nightime]

I have to admit... when I first heard about the original question (where Monty *does* know which door is the winning door) I had to actually do the experiment before it became clear.

[Manduck]

Nightime, your coin experiment doesn't mirror the Monty Hall problem, because your experiment allows Monty to pick the winning 'coin', and the original problem does not allow this. The point of the problem is that you are supposed to evaluate the probabilities in the full light of the knowledge that Monty opened a non-winning door.

If you have to make your decision whether to switch before you know what's behind Monty's door, then, sure, it doesn't matter whether you switch or not. But as the problem is formulated, you don't make your decision until after you have seen that Monty's door was empty, and you have to incorporate that information into your calculation or you will get the wrong answer. Saying that switching doesn't help is ignoring information. Switching doesn't help as long as the possibility exists that Monty picked the prize door, but as soon as you know that he didn't, switching becomes the best strategy.

[Nightime]

You misread the experiment.


You only have to keep track of the times that "Monty" reveals a losing coin, because, as you say, the other times are not allowed by the problem.



However, even though you only consider the times where a losing coin is revealed, half of those times you will have already selected the winning coin. Just try it.

[Manduck]

Just try it.

Okay, at last I think I see what you're driving at. Yes, the coin experiment turns out as you predict. But, what that proves is that if you play this game, with the rule that Monty picks his door at random, and that the game is void if Monty picks the prize, then you will probably win about half the non-void games. But that isn't what the problem asks.

The problem asks, what is the correct strategy in the situation where Monty has opened a door without the prize. When you find yourself in that situation, you have to ask what is the probability that your first guess, in that iteration of the game, was right. It couldn't be simpler: You had a choice of 3 doors, so that probability has to be 1/3. Since there is only one other door, its probability has to be 2/3.

I think what you're saying is, that when you see Monty reveal the empty door, that you are in one of those iterations where Monty doesn't reveal the prize, so that everything must now be modified by the probability that you would go into one of those iterations. But that isn't how probability works. Once you know an event has happened, its probability is 1. Regardless whether it was something less before it happened.

It's as if you were playing poker and got dealt a full house. You might think that you had little chance to win, because it's unlikely that you would be dealt a full house. But you already have the full house, so it's now irrelevant how unlikely it was for you to get it in the first place.

[Nightime]

Here is an illustration of why your theory of probability is wrong:


Suppose a friend gives you a scratch-off lottery ticket.

Given:

1. There is a 1/2 chance that it is a fake your friend made to trick you into thinking you won.

2. There is a 1/2 chance that it is a real ticket, in which case you have a 1/1000000 chance of winning.



You scratch the ticket, and it says you are a winner.


What are the chances that it is a fake?



By your logic, once you see the "winner" on the ticket, the probability of the ticket saying "winner" is 1, and it doesn't matter how unlikely it was.

So by your logic, there is a 1/2 chance that the ticket is fake.



But I think you can see right away that your method is wrong, because the ticket is almost certainly fake.

[The PHB]

Manduck, I hope you don't mind somebody else joining in this thread. I'm impressed with your stamina, but it looks like you could do with a hand.

Nightime, let's look at the game with your friend and the lottery ticket a bit closer.

Suppose your friend has two tickets - one is genuine and one is a fake he made. He offers you the choice and you pick one. Don't scratch it off yet. Just think about it. What is the probability that you have chosen the genuine one? Answer: 1/2.

Now ... well, actually, why do we care what happens now? The probability that you have the genuine ticket is 1/2. If you scratch it off and it's a winner - or if you scratch it off and it's a loser - or if you decide not to scratch it off at all but to frame it as a memento of the game - the probability that it is genuine is still 1/2. How could it be otherwise?

Let me know if you want the version with the mathematical formulae and stuff.

[Asteroide]

Manduck,

you seem to be saying that Monty's opening of a door doesn't affect the probability of your initial selection being a winner (1/3).

This might seem plausible when Monty picks a loser, but how about when Monty selects the winning door - does your initial pick's probability of winning remain similarly unaffected ?

In 1/3 of the 'Random Monty' plays your initial pick's probability of winning drops to 0, and in 2/3 of the 'Random Monty' plays your initial pick's probability of winning increases to 1/2.

1/3*0 + 2/3*1/2 = 1/3 1/3 being your overall win probability.



Nightime

I'm not sure where you were going with the fake lottery ticket thing - you're saying that the fact that it's a winner makes it more likely a fake ? Sounds more like the murky realms of psychology than a probability question !



The crux of the question for me is that when Monty knows which is the winning door, he is in effect creating the probability subset of 2/3 that previous posters have described. The probability of your initial choice being the winner is unaffected because Monty's selection is not random - he's knowingly eliminating a losing choice from a subset which has a 2/3 probability of winning.

When he makes a random pick (whatever the outcome) He is essentially playing the same game as you, with the same probability of winning or losing. In those cases where he picks a losing door, he is simply reducing the overall field of possible doors.

[The PHB]

Originally posted by Asteroide
You seem to be saying that Monty's opening of a door doesn't affect the probability of your initial selection being a winner (1/3). This might seem plausible when Monty picks a loser, but how about when Monty selects the winning door - does your initial pick's probability of winning remain similarly unaffected ?
But as Manduck has pointed out about 100000000 times, Monty does not select the winning door. That's a given. The problem as defined states that Monty picks a losing door. So there's just no point wondering what happens if, in some alternative universe perhaps, he picks a winning door. In this game, he doesn't. (Well, OK, there may be some value in the exercise from a theoretical point of view. But as dozens of posters to this thread have demonstrated, it doesn't help answer the problem that was orginally posed!)

[don't ask]

I think all that happened was the first time I posted someone chimed in to say that if the original problem was changed and Monty chose randomly there would be no advantage in swapping. This is not the original problem, but everyone accepts the answer to the original problem. It was just an observation that you had to know the circumstances under which Monty chose the empty door to know if it altered the odds. The observation is correct.

[Asteroide]

Hi PHB, while Manduck has indeed made his point 100000000 (or so) times, he is still mistaken !

The very nature of probabilities is that they deal in hypotheticals, but are still useful as predictors in the rea

[Asteroide]

oops !

PHB

...real world.

If you try the 2 Monty experiments as set out, you will find that Nightime's probabilities are an accurate predictor and Manduck's are not.

Experiment number 1: Monty knows which is the winning door, switching improves your odds to 2/3.

Experiment number 2 : Monty selects a losing door by chance, switching does not improve your odds of winning (odds will be 1/2).

This may seem counter-intuitive to you, but experiment will bear it out. You are saying that Monty does choose a losing door - this means that either his selection was not random (experiment 1, switch away), or that you are already in the subset of cases where Monty's random selection gives a losing door (experiment number 2), and thus already deep into the world of theoretical probabilities !

As don't ask just pointed out, what makes experiment 2 different is that Monty doesn't know where the prize is ...

Test it with coins !

[Nightime]

If you scratch it off and it's a winner - or if you scratch it off and it's a loser - or if you decide not to scratch it off at all but to frame it as a memento of the game - the probability that it is genuine is still 1/2.



So... if the ticket says "winner", which it will do slightly more than half of the time (fake tickets always say winner, and real ones say it 1/1000000 of the time), you have a 1/2 chance of having won the lottery.

Which means that... according to your logic, slightly more than 1/4 of the times you play this game, you will win the lottery.



Yes, more than 1/4 of the time you play you will win the lottery, despite the fact that a real ticket only has a 1/1000000 chance of winning.


Wow.


I may have just stumbled onto a get rich quick scheme.


All I have to do is learn to make realistic replicas of lottery tickets that I know will be "winners". Then I can take a fake ticket and a real ticket, and let a probability-challenged person choose between them.



If it is a "winner", they will exclaim "There is a 1/2 chance that this ticket is worth a million dollars!"


Then I can sell it to them for fifty bucks.

[zut]

Nightime
I'm not sure where you were going with the fake lottery ticket thing - you're saying that the fact that it's a winner makes it more likely a fake ? Sounds more like the murky realms of psychology than a probability question !I think I know where Nightime was going with this. Let's rephrase the question in a different form:

I have two standard decks of cards, one with a blue back and one with a red back. With you out of the room, I pick a random card from the blue deck and place it face-up on the table. From the red deck, I pick the ace of spades and place it face-up on the table. I cover each card with a sheet of paper. Now I call you into the room and ask you to choose one of the cards. You do, and I flip over the sheet of paper to reveal...the ace of spades. (The other card is not revealed.) Now, what is the chance that the card you chose, the ace of spades, comes from the blue deck? Is it 50/50? After all, you had a 50/50 shot at choosing the blue deck card, right? This is exactly equivalent to the lottery ticket question, only with a 1/52 chance of "winning" instead of a 1/1000000 chance. And, if you don't believe that... can I interest you in a little sporting proposition?

This is somewhat related to the modified MH problem under discussion, in that the type of information you know beforehand (i.e., one of the cards is certainly an Ace of spades, and the other may be with 1/52 chance) dictates how you set the odds once the last piece of information is revealed (i.e., the card you've actually chosen turns out to be the ace of spades).
.
.
...if you play this game, with the rule that Monty picks his door at random, and that the game is void if Monty picks the prize, then you will probably win about half the non-void games. But that isn't what the problem asks.But that most certainly is what the problem asks. What else would it be? You know that you are one of hundreds of people to play "Let's Make a Deal." You know (in this variation) that Monte picks his doors randomly (i.e., stealing nomenclature from above, there are six possibilities, AB, AC, BA, BC, CA, and CB, where the prize is behind door A, and your choice and Monte's are the first and secon letter, respectively). You know that, in this case, he happens to pick an empty door, and thus you can discard the notion that you're part of a trial where MH has chosen a prize door (a BA or CA trial, in other words). Most importantly, because Monte chooses randomly, his choice and the actual location of the prize are independant, and thus the relative probability of each of the remaining possible cases remains unchanged.

How is, in your view, this modified MH problem any different from the coin trial that Nightime proposed? If you believe that there is a fundamental difference, describe how we could run an experiment, using coins, that mirrors exactly the modified MH problem, and then let's run it.

[Malacandra]

Let's try this, then:

In Season One, Monty knows where the car is. The show runs for 24 episodes. In eight of these - one-third of the time - the contestant picks the right door first time and Monty has a choice of two doors he can open without ruining the game. In the remaining sixteen, the contestant does not pick the right door first time and so Monty does not have a choice. The optimum strategy is to switch.

In Season Two, Monty does not know where the car is. The show runs for 24 episodes. In eight of these, the right door is the one the contestant first picks. In another eight, the right door is the one Monty picks. In the remaining eight, the right door is neither of these. Now, the terms of the question state that Monty has not opened the right door, so we are clearly in the position of the contestant in one of the sixteen episodes in which Monty doesn't ruin the game. In half of these the contestant had the right door in the first place and so switching doesn't improve his chances (nor worsen them).

Supposing that you are taking part in an episode of Season Three and don't know if the rules have changed... you should switch, as it wouldn't have hurt under Season Two rules and helps under Season One rules.

[The PHB]

Originally posted by Malacandra
Supposing that you are taking part in an episode of Season Three and don't know if the rules have changed... you should switch, as it wouldn't have hurt under Season Two rules and helps under Season One rules.
Yes ... just so long as you do know in advance that the Season Three rules will be the same as either the Season One rules or the Season Two rules.

But suppose Monty has (unknown to you of course) cooked up the following new rule for Season Three: if the contestant has picked the winning door, then open a losing door; but if the contestant has picked a losing door, then open the winning door. Two-thirds of the time, of course, Monty will open the winning door and you'll lose - but when you are presented with the opportunity to switch, you shouldn't.

Since my earlier posts I've read up a bit of background about the Monty Hall problem. The key point to my mind, which I'm trying to make here, is that it's critically dependent on what you know in advance about the rules of the game.

If you know in advance that Monty will open a losing door, then obviously his selection is not random and your tactic is to switch. If you know in advance that Monty will open a random door, and that he might therefore open the winning door, then it doesn't make any difference whether you switch. If you don't know anything about Monty's motivation or method, then all you can fall back on is that your initial pick had a 1/3 chance of being right. If you don't have any more information, you can't improve on those odds.

I'm off to do some more research and to find a probability-challenged person for Nightime to play with.

[Asteroide]

Zut, my understanding now of Nightime's lottery ticket example is that one of the rules was that all fake tickets are winners, and only one real ticket is a winner. This wasn't stated first time round.

In your example with the cards however, I think I must be missing something. You seem to be describing a one-off event, where all the parts are determined by human agency - more of a psychology problem than a probability question.

The player walks into the room - points at one of the cards - you uncover that card, it's the A of spades (not sure why that's relevant) - there's another card next to it, which will never be uncovered (how's that relevant ?) - assuming the player is aware that there are two decks, he now has to guess which deck you chose the A of spades from - surely this would be entirely at your discretion, and not subject to any probability calculations ?

[zut]

Asteroide: I'm not sure why you think this is a psychology experiment; it seemed clear to me. Nonetheless, let me try again.

I've set up some rules for a game. You know exactly what the rules are.:
1. I have two decks of cards, red-backed and blue-backed.
2. I pick a random card from the blue deck.
3. I obtain the ace of spades from the red deck.
4. I place these two cards on the table in such a way that you can't see either of them.
5. You walk into the room, knowing full well exactly what the rules are as listed above. You don't know which blue-deck card I chose (randomly, remember!), and you don't know which card on the table is which, but you know the process by which I chose those cards.
6. You pick one of the two cards on the table, and I reveal (this particular time) that that card is the ace of spades. You still don't know what the other card is, nor do you know which is from the red deck and which from the blue deck.
7. Now, knowing that the card you picked is the ace of spades, and knowing the rules I set up above, what do you suppose the chances are that the card you picked has a blue back? Is it 50/50?

No psychology here. We could run it once, or run it a million times; the question is: assuming you happen to pick up an ace of spades, what are the chances that it's a blue card?

Note that this card example and the lottery ticket example are analagous: blue deck = real ticket, red deck = fake ticket, ace of spades = "you're a winner!"

[Asteroide]

OK, I get it :

- The red card is necessarily the Ace, and we both know it (not a random pick)

- The blue card is a random pick - 1/52 that it's the Ace of spades

I guess what I was missing was that the Ace was not a random draw, and that the player had all this info.

Agreed that it's the same concept as the lottery, assuming all fake lottery tickets are winners.

The psychology aspect would come in if you were selecting cards at your discretion - which is apparently not the case !

[GreyWanderer]

Okay, at last I think I see what you're driving at. Yes, the coin experiment turns out as you predict. But, what that proves is that if you play this game, with the rule that Monty picks his door at random, and that the game is void if Monty picks the prize, then you will probably win about half the non-void games. But that isn't what the problem asks.

The problem asks, what is the correct strategy in the situation where Monty has opened a door without the prize. When you find yourself in that situation, you have to ask what is the probability that your first guess, in that iteration of the game, was right. It couldn't be simpler: You had a choice of 3 doors, so that probability has to be 1/3. Since there is only one other door, its probability has to be 2/3.

I think what you're saying is, that when you see Monty reveal the empty door, that you are in one of those iterations where Monty doesn't reveal the prize, so that everything must now be modified by the probability that you would go into one of those iterations. But that isn't how probability works. Once you know an event has happened, its probability is 1. Regardless whether it was something less before it happened.

It's as if you were playing poker and got dealt a full house. You might think that you had little chance to win, because it's unlikely that you would be dealt a full house. But you already have the full house, so it's now irrelevant how unlikely it was for you to get it in the first place.

Back to the experiment. Imagine if every time you tried that experience a different person was the one playing.

After the experiment you found that half of those persons playing won by switching. That is what the experiment says, right?

EVERY SINGLE of those persons were in the exact same situation as you are. What advice would you give them? To switch or not? Half the people will win if they switch. That means it's a 50% probability, by definition.

But if you did the experiment with Monty not opening a random door 2/3 would win by switching. Of course, then you would advice them to switch.

This is also the situation you describe. Without knowing if Monty chooses a door by random or not you can't know which situation you are in. Yes, he does exactly the same in both situations. But it's what he does over time that matters. When dealing with chance you HAVE TO consider what will happen over time, because that's in the definiton of probablity.

[RM Mentock]

Maybe if you played the game another time, Monty would open the prize door and say "ha ha you lose. sucker!" But should that affect your answer in the game you're playing right now, in which you know that he opened an empty door? The prize is either behind the door you picked first, or the other unopened door. There is a 1/3 chance you picked the correct door first. That is all you need to figure out your answer. It doesn't matter what might have happened in some other iteration of the game, because you know that in this iteration, it didn't happen.
Basically, you're saying that "random" doesn't mean anything, in this context. It doesn't change anything--but we know that that is not true. If Monty knows what he's doing, he can always pick the prize door--if you don't first. That would destroy any chance of you switching to it, which would mean that your chances of having the prize when Monty opens an empty door wuold increase to 100%.

[John W. Kennedy]

Here is a simple Ruby program to demonstrate the whole thing.# Define the doors.
# Their only property is that they are either open or closed
class Door
# Open the door
def open; @open = true; end
# Close the door at the beginning of the game
def close; @open = false; end
# Is this door open?
def open?; @open; end
end

# Define the player
# This is a simple player
class Poor_player
# Our first choice is random
def initial_choice; @initial = rand 3; end
# Our second choice is to stand pat
def second_choice; @initial; end
end

# Define Monty
# Monty knows where the prize is, and will tell you if you won
class Monty
# Tell Monty at the beginning which one wins
def whisper(prize); @prize = prize; end
# Ask Monty whether the door is the right one
def winner?(choice); choice == @prize; end
# Monty chooses a door to open
def open(choice)
# If the player's choice is the winner, Monty can choose one of two doors
# Otherwise, he has only one choice
which = choice == @prize ? rand(2) : 0
# Scan the doors
$doors.each_with_index do |door, i|
case i
# If this is the door the player chose, or the winning door, don't open
when choice, @prize
else
# Otherwise, possibly open this door
if (which == 0)
door.open
return i
# If we didn't open, we will have to open the next
else
which -= 1
end
end
end
end
end

# Simulate a game
def simulate
# Start by closing all the doors
$doors.each {|door| door.close}
# Tell Monty where the prize is
$monty.whisper(rand(3))
# Tell the player to make his first choice
choice = $player.initial_choice
# Tell Monty to open a door
$monty.open(choice)
# Tell the player to make his second choice
choice = $player.second_choice
# Ask Monty whether the player won
$monty.winner?(choice)
end

# Now we are ready to play
# Create Monty
$monty = Monty.new
# Create the player
$player = Poor_player.new
# Create three doors
$doors = []; 3.times {$doors << Door.new}

# Run 10000 simulations
sum = 0; 10000.times {sum += 1 if simulate}
puts "Poor player won #{sum} times."

# Give the player an improved rule for the second choice
def $player.second_choice
# Scan the doors until we find one that isn't open, and that we didn't choose
$doors.each_with_index {|door, i| return i unless i == @initial or door.open?}
end

# Run 10000 more simulations
sum = 0; 10000.times {sum += 1 if simulate}
puts "Smarter player won #{sum} times."
It shows the smarter player winning two-thirds of the time, consistently.

[RM Mentock]

It shows the smarter player winning two-thirds of the time, consistently.
And if Monty's choice is random?

[Manduck]

Hi everybody, I'm back!

I haven't posted since the weekend what with the real job and everything, but I have been continuing to think of this thread; the coins; the lottery tickets; the playing cards - and, uh, well, I changed my mind.

It was bothering me that the coin game came up 50-50, but I thought there must be some subtlety that would reconcile it with my analysis; I just have to figure it out. But I didn't come up with anything. So that planted a seed of doubt in my mind.

Then I looked at the lottery ticket argument (yes, I was lurjing when I should have been working :)). The point of the lottery ticket example, I think, is that adding information during the game can affect your assessment of the probabilities of earlier event. I.e., when you make your random pick, there is a 50% probability that it's a fake ticket, and 50% that it's genuine. But when you scratch it off, you either find that it's a loser or a winner. If it's a loser, you know it must be genuine, so the probability that it's fake is 0. If it's a winner, the probability that it's fake becomes 1999999/2000000 (if my calculations are right).

Okay, so how does that apply to the Monty Hall problem? I wasn't sure that it did, but to try out the idea, I used the old trick of considering a 100-door variant of the game. What would happen if MH opened 98 empty doors after I made my pick?

Consider for a moment a situation where there are 99 doors, and one of them may or may not conceal a prize. You don't know whether a prize is behind one of them or not. Now suppose you opened 98 of the doors at random, and none of them had the prize. What is the probability that 98 tries would come up empty if the prize was behind one of the doors? It's 1/99. Of course, if the prize wasn't there, the probability of getting 98 empty doors is 1.

So I'm playing 100-door Monty, and Monty opens 98 doors at random without finding the prize. I have to believe that there is very little chance that the prize is among those 99 doors. That gives me new information about the probability that I picked the prize in the first place. If I had picked one of the 99 losers with my first guess, there is only a 1/99 chance that he could have opened those 98 empty doors. That 1/99 is just the right amount to balance the 99/100 chance that the guess was wrong.

Now if Monty knew where the prize was and deliberately showed 98 empty rooms, that doesn't provide any new information about the 99th one, because he can always open 98 empty doors whether the prize is among the 99 or not. So, the probability of my first guess being correct is not altered by that information, so I should switch.

In the random case, Monty's door-opening does provide new, if incomplete, information about my original guess which has the effect of creating a situation where it doesn't matter whether I switch.

... which is what Nightime and others have been saying all weekend. Congratulations, boys, you won me over. And please ignore all my posts from the previous two days :D

[it was the lottery tickets that did it]

[Malacandra]

Hiya Manduck,

It was just such an approach that convinced me of the solution to the original Monty Hall problem when I first heard about it. I remember running it through my head one afternoon while I was playing cricket.

Now what? :D

[The PHB]

I can't believe how stupid I've been

I mean, I've got a degree in maths from Cambridge University, and I took all the probability and statistics courses offered, and I only just missed getting a First. I'm not boasting - I'm trying to underline the extent of my stupidity here. Incidentally I saw the light yesterday, before seeing a similar recantation from Manduck. Looks like the penny dropped for both of us around the same time.....

Here's my new take on the lottery game advanced by Nightime. I originally said:

Suppose your friend has two tickets - one is genuine and one is a fake he made. He offers you the choice and you pick one. Don't scratch it off yet. Just think about it. What is the probability that you have chosen the genuine one? Answer: 1/2.

Now ... well, actually, why do we care what happens now? The probability that you have the genuine ticket is 1/2. If you scratch it off and it's a winner - or if you scratch it off and it's a loser - or if you decide not to scratch it off at all but to frame it as a memento of the game - the probability that it is genuine is still 1/2. How could it be otherwise?
In my enthusiasm to apply intuition and "common sense" instead of proper analysis, I think I overlooked the fact that the chosen ticket is not a sort of 50/50 hybrid of a genuine ticket and a fake ticket - it is either definitely genuine or definitely fake. Before I scratch off the ticket, the only information I have to help me decide whether it's genuine is the fact that I picked it randomly from a choice of two - i.e. a 1/2 chance that it's genuine. But scratching off the ticket introduces new information. And that's what changes the probability.

If there's anybody out there who sees all this intuitively (unlike me), but doesn't know how to do the mathematical calculations (which I do, at least when I'm not being stupid), then this might be interesting. The standard approach to calculation conditional probabilities (i.e. the probability of something happening, gien that some non-independent other thing has happened) is Bayes' Theorem. This states thatP ( A | B ) = P ( A ) * P ( B | A ) / P ( B )where P ( A | B ) means the probability of event A happening, given that event B has happened.

Applying Bayes' Theorem to the lottery ticket game, let's suppose there are 1,000 genuine tickets (I could say 1,000,000 but I'm sure I'd get bored typing all the zeroes repeatedly); then let's say W denotes the event of me picking a winning ticket and G denotes the event of me picking a genuine ticket. So we want to know P ( G | W ), i.e. the probability that my ticket is genuine, given that it is a winning ticket.

We know that

P ( G ) = 0.5, because I had a 50/50 pick between the fake ticket and the genuine ticket;
P( W | G ) = 0.001, because there are 1,000 genuine tickets and only one is a winner; and
P( W ) = P ( [b]W[b] | G ) * P ( G ) + P ( [b]W[b] | notG ) * P ( notG ) = (0.001*0.5)+(1*0.5) = 0.5005, because I had a 50/50 chance of picking a fake ticket which is definitely a winner, and a 50/50 chance of picking a genuine ticket which has probability 0.001 of being a winner.

So plugging this into Bayes' Theorem gives P ( G | W ) = 0.000999. Not 0.5.

I'd like to apologise to Nightime, Asteroide, Malacandra, GreyWanderer and everybody else whose time I've wasted, and I'd like to thank you all for your patience in the face of my stupidity.

But a final comment, if you'll allow... what does this actually say about the Monty Hall problem? Not very much, is what I think. Obviously the mindset of the Bayesian approach is useful, and the value of identifying how much new information is made available at various stages of the process is clear. But at the end of the day, it comes down to what you know about the rules of the game. If you know Monty will open a losing door, you should switch. If you know he'll open a random door, switching and not switching are equally good. But how much do you know about what Monty will do? Why not ask Monty Hall himself....

The following extracts are from "Behind Monty Hall's Doors: Puzzle, Debate and Answer" in The New York Times, Sunday, July 21, 1991 which I found referenced here (http://www.wiskit.com/marilyn.gameshow.html). The entire New York Times article is available in the Internet Archive here (http://web.archive.org/web/20030404053104re_/www.dartmouth.edu/~chance/course/topics/Monty_Hall.html) .

The contestant picked Door 1. "That's too bad," Mr. Hall said, opening Door 1. "You've won a goat." "But you didn't open another door yet or give me a chance to switch." "Where does it say I have to let you switch every time? I'm the master of the show."
On the second trial, the contestant again picked Door 1. Mr. Hall opened Door 3, revealing a goat. The contestant was about to switch to Door 2 when Mr. Hall pulled out a roll of bills. "You're sure you want Door No. 2?" he asked. "Before I show you what's behind that door, I will give you $3,000 in cash not to switch to it." - "I'll switch to it." - "Three thousand dollars," Mr. Hall repeated, shifting into his famous cadence. "Cash. Cash money. It could be a car, but it could be a goat. Four thousand." - "I'll try the door." - "Forty-five hundred. Forty-seven. Forty-eight. My last offer: Five thousand dollars." - "Let's open the door." - "You just ended up with a goat," he said, opening the door.
Because of the ambiguity in the wording, it is impossible to solve the problem as stated through mathematical reasoning. "The strict argument," Dr. Diaconis said, "would be that the question cannot be answered without knowing the motivation of the host." Which means, of course, that the only person who can answer this version of the Monty Hall Problem is Monty Hall himself. Here is what should be the last word on the subject:

"If the host is required to open a door all the time and offer you a switch, then you should take the switch," he said. "But if he has the choice whether to allow a switch or not, beware. Caveat emptor. It all depends on his mood.

"My only advice is, if you can get me to offer you $5,000 not to open the door, take the money and go home."

[Manduck]

But a final comment, if you'll allow... what does this actually say about the Monty Hall problem? Not very much, is what I think. Obviously the mindset of the Bayesian approach is useful, and the value of identifying how much new information is made available at various stages of the process is clear. But at the end of the day, it comes down to what you know about the rules of the game. If you know Monty will open a losing door, you should switch. If you know he'll open a random door, switching and not switching are equally good. But how much do you know about what Monty will do?

I have to admit that this argument started when I proclaimed, many posts ago, that it doesn't matter whether Monty's choice is random or not.

[John W. Kennedy]

And if Monty's choice is random?Then, as expected, there is no profit to switching, and Monty spoils 1/3 of the games.

[CurtC]

For those who maintain that Monty's motivations don't come into play, all you need to know is that in this instance, he did show you an empty door and offer the switch, I have an example.

Let's say you're walking down a city street and come upon a game of three-card monty. The street hustler moves the three cards around very quickly, and you select which card might be the ace. The street hustler then turns over one of the cards that you didn't pick, which is not the ace, and offers to let you switch your guess.

Should you switch? You'd be a fool to switch. I can calculate that your probability of winning would approach zero if you did. Can you see why? Can you see that the only difference between this, and the Monty Hall problem is what assumption we make about the host's motivations?

[Asteroide]

This would depend, of course, on whether Monty has free will :)

[Malacandra]

CurtC is quite right, of course, which is why I never play three-card monte, but as I understood the Monty Hall problem as I originally heard it, it was in the rules that Monty always opened a door, never spoiled the game, and always offered you the switch. If those aren't the rules, it's just a question of whether you feel lucky :)

[RM Mentock]

CurtC is quite right, of course, which is why I never play three-card monte, but as I understood the Monty Hall problem as I originally heard it, it was in the rules that Monty always opened a door, never spoiled the game, and always offered you the switch. If those aren't the rules, it's just a question of whether you feel lucky
Most people's introduction to the problem (it also appeared in Games magazine some months before--in their last issue before a financial hiatus, so they didn't have a chance to publish the irate and wrong responses) was in Parade in 1991. The rules weren't explicit like that.

http://kevingong.com/Math/MontyHall.html

[CurtC]

Malacandra, in all the times that I've heard this puzzle (I first heard it as the weekly puzzler on Car Talk, before either Marilyn or Cecil took it on, in the late 80s), I have never, ever, heard it be explicitly stated that Monty is required to offer the switch. On the contrary, I have often heard it stated by the "2/3" defenders that it doesn't matter, that we're given that in this instance the switch was offered and that's all we need to know. You can see that same response earlier in this thread.

[RM Mentock]

Malacandra, in all the times that I've heard this puzzle (I first heard it as the weekly puzzler on Car Talk, before either Marilyn or Cecil took it on, in the late 80s), I have never, ever, heard it be explicitly stated that Monty is required to offer the switch. On the contrary, I have often heard it stated by the "2/3" defenders that it doesn't matter, that we're given that in this instance the switch was offered and that's all we need to know. You can see that same response earlier in this thread.
The Car Talk Puzzler archives show it in about October of 1997 (http://cartalk.com/content/puzzler/1997.html). That's long after Cecil and Marilyn (and Games), but I suppose they could have used it twice.

The puzzle itself has been traced back lots further than that. Martin Gardner had a version in the sixties or seventies.

[CurtC]

Here's a usenet thread (http://groups.google.com/groups?q=cameron+hall+group:alt.fan.cecil-adams&hl=en&lr=&ie=UTF-8&group=alt.fan.cecil-adams&selm=31AB3481.6345%40hp-usa-om24.om.hp.com&rnum=10) from 1996, where I state that I heard it several years before that on Car Talk.

[RM Mentock]

Here's a usenet thread (http://groups.google.com/groups?q=cameron+hall+group:alt.fan.cecil-adams&hl=en&lr=&ie=UTF-8&group=alt.fan.cecil-adams&selm=31AB3481.6345%40hp-usa-om24.om.hp.com&rnum=10) from 1996, where I state that I heard it several years before that on Car Talk.
TGF gooja, AFU, and afca. Car Talk apparently did use it twice. OTOH, it says that Cecil attacked it before Marilyn, but Cecil's column (http://www.straightdope.com/classics/a3_189.html) has the Marilyn text.

[Malacandra]

I heard about it in 1991 or 1992, when Marilyn had just tackled it and controversy was raging. Originally I thought Marilyn was full of it, but I pondered it further and concluded she was right... for which I'm sure she's grateful.

I should have known better, knowing enough bridge to have heard of the Principle of Restricted Choice (viz: when a player may hold one of two cards, or both, and may play either of them if he has both, the playing of either of them is suggestive that he does not have the other, since if he had both, either one of them would turn up only half the time. If that extremely terse summary makes any sense! :D )

[Asteroide]

FWIW

One of the earliest known appearances of the problem was in Joseph Bertrand's Calcul des probabilites (1889) where it was known as Bertrand's Box Paradox. It later reappeared in Martin Gardner's 1961 book, More Mathematical Puzzles and Diversions, as The Three Prisoner Problem and then resurfaced in 1975 - inspired by Monty Hall's U.S. gameshow "Let's Make a Deal" - in an article in The American Statistician by Steve Selvin entitled A Problem in Probability.
http://barryispuzzled.com/zmonty.htm

I first came across it in a long and tortuous article in a French science mag.

It seems like every time time it comes up again, new and exciting twists are added. So, hours of fun for all the family !