Monty Hall Problem

Cecil's Columns/Staff Reports
1

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.

2

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.

3

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.

4

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.

5

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.

6

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.

7

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.

8

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

9

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.

10

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%

  5. and 2) are in the same subset. Total=50%

  6. 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.

11

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

12

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.

13

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 ¦¬)

14

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.

15

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

16

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…

17

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?

18

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

19

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 ¦¬)

20

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 ¦¬)

“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