# Must Comment on Monty Hall sub riddle/logic trick

this is a sub riddle on the the monty hall page which has so far not met any opposition, ill paraphrase:

“Two children are coming over your house for some reason. The first child walks in the door, she is a female. What is the probability that the second child is male?”

the answer they give is 2/3 and here is the evidence they provide:

"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 kept running this through my head until I eventually convinced myself. However this explanation is just a trick of logic. There are only two choices now that we find the first child to be a girl. The Male/Female, Female/Male separation is pointless and shouldn’t be considered as two separate options. To make things simpler just define child A as the first child to walk in the door, otherwise you are adding a new dimension to the puzzle.

Please, correct that foolish confusing logical trick and make me feel special.

Link to column: On Let’s Make a Deal, you pick Door #1. Monty opens Door #2–no prize. Do you stay with Door #1 or switch to #3?. The “two children” problem is way down the page. – CKDH

It seems to me that if the first child - Child A - is female, you’ve eliminated possibilities 3 and 4 right away. The only possibilities from that point are 1 and 2, each with equal probability - 50% chance male or 50% female. I think that’s what you’re saying, so I agree with you.

You’ve misread the problem. It doesn’t say that we know child A is a girl, we know that one child is a girl. The first child to walk through the door might well be child B.

I think this is the column to generate the most discussion, by far.

The second child is either male or female, with a 50% probability. The gender of the first child is irrelevant.

Actually, 50% may not be correct since the ratio of males to females in humans is not exactly 50%. Sex ratio - Wikipedia gives a ratio of approximately 105 males to 100 females. If this is correct then the probability that a random child is male would be approximately 51.2%

Obviously, I was responding to the question as stated in the OP. The answer to the problem described by Priceguy would be different, but the actual real world ratio would still need to be taken into account.

Because of the way the OP is written, there is no difference between Child A and Child B in the four possibilities given. Child A could have been the first or second child thorough the door and Child B could have been the first or second child through the door. So Child A and Child B are equivalent. This means that “(1) Child A is female and Child B is male” and “(3) Child A is male and Child B is female” are actually two different ways of saying the same thing and one of them is superfluous. Therefore there are only actually three possibilities:

(1) One child is female and the other child is female.
(2) One child is female and the other child is male.
(3) One child is male and the other child is male.

We know (3) is not true because the first child through the door was a girl. Therefore, either (1) or (2) is true and the odds are 50/50 (discounting the real world disparity between genders).

Another possibility is to define Child A and Child B in distinct terms. The obvious way would be to define Child A as the first child seen and Child B as the second child seen. In this case, the four possibilities are accurate because Child A and Child B are no longer interchangeable terms.

(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.

But in this case, once we see the first child is a girl, we know that Child A is female and both (3) and (4) are not true. Once again, we are left with (1) and (2) as the two possible answers and 50/50 odds.

The 2/3 answer is correct. Look at it this way: there is a 1/4 probability that both are girls, 1/4 that they are both boys, and 1/2 that it’s one of each. If you don’t believe that, flip two coins at the same time and note how often they come up the same and how often they come up different.

The problem removes the “both boys” case from the discussion, so you’re left with 1 case of both girls and 2 cases of one of each. Since now you only have 3 cases, that makes 1/3 both girls and 2/3 a girl and a boy. QED.

The problem with that thinking is that the “A” has some meaning to it, yet it is trivial. THe only situation it would matter in was if the “A” was replaced with some type of new dimension. For example?

Two children are coming over to your house. One is going to explode. A girl walks in the door. What are the odds that a girl is going to explode.

In this case the answer is 2/3s

Atomic Moose, welcome to the Straight Dope Message Boards, we’re glad to have you with us. When you start a thread, it’s helpful to other readers if you provide a link to the column in question; keeps us all on the same page and saves search time.

I agree with your comment that A and B need meaning; in the original, probably “older” and “younger” although “exploding” has a nice ring to it.

Remember that “probability” is not a thing that exists in reality and can be counted or weighed. The exact wording of the situation can make a huge difference. As noted in the column:

• 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? (Answer: 2/3.) Situation: FF, FM, MF, MM based on oldest, youngest, and eliminate MM.

• There is a family with two children, at least one of which is a girl. It’s a matter of chance who arrives first. The first child enters–a girl. The second knocks. What are the odds it’s a boy? (Answer: 1/2.) Situation: FF, FM, MF, MM based on who comes to door first, and eliminate MF and MM.

To clearly distinguish between the two subtle-but-significant conceptions of the problem (the usually unspoken source of the controversy, in my opinion), it may be helpful to consider how the coin-flipping analogy changes depending on how you think of it: with two identical quarters, vs with two easily differentiated coins, like a penny and a quarter. It might change the way you think about how you’re identifying the components of the problem, and illuminate why the basic conception of the scenario matters.

The problem actually has 8 states, the four stated above combined with child a walks in first, child B walks in first. So ourt total states are:

(1) Child A is female and Child B is male, Child A Walks in first

(2) Child A is female and Child B is female, Child A walks in first

(3) Child A is male and Child B is female, child A walks in first

(4) Child A is male and Child B is male, child A walks in first

(5) Child A is female and Child B is male, child B walks in first

(6) Child A is female and Child B is female, child B walks in first

(7) Child A is male and Child B is female, child B walks in first

(8) Child A is male and Child B is male, child B walks in first

The probability of each state is 1/8 or 12.5%. When a girl walks in first half these scenarios are eliminated: 3,4,5,8. But the other four are possible giving us 50%.

Nope. In the OP, specifying the gender of the first child through the door is the equivalent of flipping one of the coins - it eliminates two of the four possible outcomes and you only have two left. The problem doesn’t just remove the “both boys” case, it also removes the “boy then girl” case. The cases you have left are “both girls” and “girl then boy”.

The problem is that the OP, in his paraphrasing has altered the question. In the original OP (i.e. Jordan Drachman of Stanford, California in Cecil’s column), the statement is:

The point here (along with the Monty Hall problem) is that there is aditional information. Therefore the 50/50 chance of being right has been changed. Neither process is entirely random anymore.

2/3 is the correct answer for the riddle in Cecil’s column; 1/2 is the correct answer for the riddle in the OP.

Ignoring the mistatement of the problem in the OP and using the problem as stated in Cecil’s column and assuming (apparently incorrectly, but for the sake of the puzzle…) that the biological odds of having either a boy or a girl are equal, the answer is 50%.

Here is the stated explanation for the 2/3 answer:

The problem with this explanation is that cases (1) and (3) are the same case, a girl and a boy. It doesn’t matter what order you state them in, it’s still a girl and a boy (or a boy and a girl). Therefore the choices really are

(1) Both children are female

(2) Both children are male

(3) There is one of each.

Once we’ve seen that one of them is a female we can eliminate (2). This leaves us with

(1) Both children are female

and

(3) There is one of each.

Giving us a 50% chance that her sibling will be male.

This fits with the real world intuitive notion that a random child will have a 50% chance of being male and that knowledge of the gender of another child doesn’t change that.

This is incorrect logic. It presupposes that each of the choices (1, 2 or 3) has the same possibility of being true. But that is not true; choice 3 has a twice-greater chance of being true than choices 1 and 2. Therefore, if you eliminate 2, you are stuck with 2:1 odds.

Are you saying that, among all families with two children, one third have a boy and a girl, one third have two girls, and one third have two boys? That’s demonstrably false. The odds are not the same.

By analogy, if you flip a coin twice, would you say there are three cases: HH, HT, and TT ? You think that HT is the same as TH?

But the point is that you have extra information in this case. You’re not just told, “There’s a child, guess it’s gender.” You’re told there are two children and one is a girl. Please note: there is a difference between the following two questions:
(a) I flipped a coin and got Heads. What is the probability of getting heads on the next flip?
(b) I flipped a coin twice and got heads on one of the flips. What is the probability the other toss also was heads? (answer: 1/3).

I thoght of a good way to understand this while walking to school today. The 2/3rds answer is found by thinking of the problem this way:

There are about an equal number of sperm with both the x and y chromosomes in the man’s body. For the sake of simplicity lets say there is one of each. One boy sperm, one girl sperm. Now in this problem it states that there are two children… so then lets give the man four sperm, two boys and two girls… This problem assumes that once the girl sperm successfully gains the right to impregnate the egg that it is now gone and the man is left with two male sperm and one female sperm for the second egg. However this is not the case. Past events cannot be brought up in cases of logic like this unless specified as such. If it stated that the man only had four sperm (two of each) then it would work, but then it would be quite easy to solve and not make much sense.

oh i said “x and y chromosomes in the man’s body.” and that sounds stupid… I meant boy/girl producing sperm.

On a somewhat related issue, Cecil mentions the fact that Monty always knows what’s behind the doors and raises the possibility of him affecting the odds by being able to choose whether or not to offer the option of switching doors.

In the canonical game, there are three doors and Monty always offers the switch. We’ve established that 2/3 of the time, switching will lead to a prize and it’s always a good idea to switch doors.

But suppose Monty was a bastard and was trying to keep people from collecting the prize. And suppose he had control of whether or not to offer a switch (although he still has to play the game by the same rules if he chooses to offer a switch - he can’t intentionally open the door with the prize before offering a switch).

In such a case in the two out of three times the player picked the wrong door, Monty would not offer to switch - the player had already lost and he wouldn’t want to give them a chance to win. In the one out of three times the player picked the right door, Monty would offer the switch and hope to fool some people into unknowingly giving away the prize.

But as Cecil pointed out, in a rigged game like this players would quickly figure out Monty’s pattern and realize that anytime they were offered a chance to switch it was because Monty was trying to screw them, so everyone would refuse to switch. The only way Monty could entice people to switch would be to occasionally bluff and make a offer to switch when they hadn’t picked the prize.

So my questions are (1) What percentage of the time should Monty make a bluff in order to make the decision to switch or not a 50/50 choice? (2) Is there some percentage of bluffing that would lower the amount of winners below the 33% I described above? (3) Is there any benefit to Monty in occasionally making a reverse bluff and not offering to switch when the player had initially picked the right door?

If the probability that Monty opens a door when you’ve picked the prize is [symbol]a[/symbol] and the probability that Monty opens a door when you’ve picked the wrong door is [symbol]b[/symbol], then the probability that you’ve picked the right door given that Monty opens a door is [symbol]a[/symbol]/([symbol]a[/symbol] + 2[symbol]b[/symbol]). If [symbol]a[/symbol] = 2[symbol]b[/symbol], you’ve got even odds.