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/F...ML/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.