Now that I’ve had a little time to sleep on it, I understand. I spent an hour or two switching my mind back and forth between the 2/3 and 50/50 answers last night, and I picked the wrong one to write about. I apologize!
If Monty had randomly picked one of his two doors (and it was empty,) then the chance would truly be 50/50. Since it wasn’t a random pick, he simply revealed to me what I already knew - that at least one of his 2 doors was empty. Therefore the odds are 2/3 to switching. Thanks!
Gotcha. I messed up pretty bad on my original post. Still, better to ask and be proven wrong, then never ask at all and remain ignorant!
No, it doesn’t work. You have information about one specific child, the one who’s in your daughter’s ballet class (and it’s not really plausible that they both are, because if they were, you wouldn’t use that wording).
Just because the second daughter is not in your friend’s daughter’s class does not mean it is not a girl, or that she is not in ballet. She just is in a different age group.
Similarly, just because the child is in ballet does not mean it is a daughter. Boys can take ballet. They aren’t nearly as common, but they do happen.
But the wording does suggest the Smiths only have one daughter.
That’s the opposite of what I’m saying. 4/5 is the correct answer, but I can’t get that answer using your approach. Evidently either do you, which means there must be something wrong with that approach.
Ah, miscommunication here. Monty opens all the doors except one, when there are more than 3 doors to begin with. That’s the intuitive way to show the 2/3 probability for 3 doors, or the 99/100 probability for 100 doors.
Bingo, only that was Ms Savant’s initial explanation, so it’s not “different”. Plus it was mentioned several times upthread.
But we don’t know whether that specific child is the first or second child, so do we really know anything specific?
I’ll have to give this some thought. It’s just this kind of thing that kept me from getting an A in prob/stat all those years ago.
What if it was “She mentioned she has a daughter, but from the way she put it I couldn’t tell if she had only one. Later I found out she has two children.”
If she mentioned she has a daughter, was that specific enough to select one of the two children?
Here’s a puzzle for you (easy). I have two coins, totalling to 55 cents. One of them is not a nickel. What are my two coins? 
Never mind. I understand now.
My solutions to 3 probability problems:
The Let’s Make a Deal problem is amazing since it seems to confuse so many (including myself) while eliciting such vehement responses!
I think I’ve finally convinced myself of the right answer (FLW)*. 3 doors, 2 goats, 1 car. The rules of the game I will adhere to are (discussions of what the correct rules are notwithstanding): You choose 1 door originally, Monte opens another door which always has a goat behind it, Monte asks if you want to switch. As has been stated several times in this forum (and others), not switching will result in losing (not getting the car) 1/3 of the time, no matter what Monte does. So does this mean that switching wins 2/3 of the time? Maybe. The question is, does switching change the probabiities? Note that, if Monte doesn’t do anything, then switching will also fail 2/3 of the time (wins 1/3 of the time). The key is in the fact that your choice affects Monte’s choice, so in a handwavy way, yes, the probabilites do change. So consider the possibilites. If you initially choose the door with the car, which happens 1/3 of the time, Monte has 2 choices for the other doors. Switching clearly loses. If you choose a goat-door first (happens 2/3 of the time), Monte has only 1 choice (the remaining goat-door). In this 2/3-of-the-time case, the remaining door (i.e., the one not originally selected by you or Monte) has the car, so switching wins. So 2/3 of the time switching wins and 1/3 of the time it loses.
Chances that 2nd kid is a boy: A person says they have 2 kids (odds of having a boy or girl are 50/50). They then show you that one of them is a girl. Given the odds of boy vs. girl in any independent birth, most people would say the chances of the 2nd kid being a boy is 1/2. But this doesn’t take into account that the order of the births comes into play since the parent reveals them in order. Thus, girl-boy is a separate outcome from boy-girl. They count as separate outcomes even though they have the same “mix”. As is usual in probability calculations, the sample (aka possible outcome) space is constructed and then the proportion of cases which fit the final outcome are tallied. In this case the sample space before any kid is revealed is girl-girl, girl-boy, boy-girl and boy-boy. There is a 1/4 chance of any one of these distinct outcomes to occur. Now when the girl is revealed, the sample space reduces to girl-girl, girl-boy and boy-girl. The problem then becomes one of a conditional probability. Note that for the second and third outcomes, the parent could have chosen the girl first to present, so by doing so, they haven’t revealed that we are now dealing only with the 3 outcomes the girl belongs to. Now the chances of the 2nd kid being a boy is 2/3 based on the new sample space.
Chances of the card pulled out of the hat was the original Ace = 1/3: A card is put into a hat that has an equal probabilty of being an Ace or a King. An new Ace is then put into the hat (2 cards are now in the hat). One card is now withdrawn. What are the chances it is the original Ace?
Let A0 = original Ace, An = new Ace and K = King
Again, thinking of sample spaces helps. The samples after An is put in the hat and shuffled are
The notation indicates that the order of the cards counts since, just to fix the procedure, we will always select the 2nd (unknown) card in the possible pairs (it could always be the first, it doesn’t matter).
If we pick a card out of the hat and find it is an Ace (we now have a conditional probability), then there are only 3 samples that could have produced this outcome (the 2nd card) and only one outcome of these is the original Ace, so picking it has a chance of 1/3.
Note: In second problem, the parent HAS revealed the smaller sample space. Sorry
Not switching results in losing 2/3 of the time.
Good point! Nice catch.
You bungled a couple statements, but arrived at the right conclusion. I think they might have just been misstatements of what you meant.
Not switching will result in losing 2/3 of the time.
Yes, if Monte simply offers you a chance to swap without revealing anything, your odds remain at 1/3 for each door.
Correct.
Child order is important because it equates the sample sizes for each group. The probability calculation could be carried out by having unequal sample sizes, and instead adding a “sample size scaling factor” to the calculation. That is unnecessarily over-complicated. But this is the key element to the calculation. If you are told the oldest is a boy, then the probability for the second child is 50:50.
Specifying that there is a daughter eliminates one of the possible outcomes, the case where there were 2 boys. Specifying which child is the daughter eliminates a second case - the counter birth order case. It’s difficult to get a real life situation where you have revealed only enough to eliminate 1 case without enough information to eliminate 2 cases.
Thanks for your comments. Another mea culpa: I said “birth order” but I didn’t really mean when they were born but rather how their presence was revealed to the person pondering the situation. Clumsy choice of words on my part. I now think introducing “order” into the explanation is confusing; just counting the number of outcomes suffices.
How their presence was revealed makes all the difference. You originally wrote “when the girl is revealed”, which implies the sex of one person is given. It doesn’t matter if it’s the oldest, if it’s the tallest, if it’s the first person who happens to come in the room, or if it’s the child in the same grade as your son. If you’re given the sex of a specific child, the odds of the other one is 50/50. It’s only if you know that, of the two, at least one is a girl, that you get a 2/3rds chance.
I may be a little late, but I wanted to join into this discussion before i reached the end of the thread and forgot to. I too am getting tired of the debate on the 3 doors and would like to comment on the male/female problem.
Before I go into this I want to say that I only took the first level of statistics in college and it wasn’t my best area of math (actually it was prob my worst). So I think I am probably misunderstanding something and would like it explained.
I’m not sure how to interpret this. if a person says he has 2 children and one is a girl, you can draw a chart like this:
child a - girl child b - boy
child a - girl child b - girl
child a - boy child b - girl
child a - boy child b - boy
in the realm of combinations, i can see a 2/3 chance of child b being a girl if child a is a girl, and vice versa a 2/3 chance of child a being a girl if child b is a girl. On the other hand, if this is a combinations problem, isn’t
child a - girl & child b - boy
and
child b - boy & child a - girl
the same thing so you would elimate one of those and be left with a 50/50 chance of the second child being a girl or a boy.
As far as permutations go, if the girl is the first to walk in the room and the second can be a boy or a girl, i think it would still be a 50/50 chance that the second child is a boy or girl. with a 25% chance for each possibility over all and both the scenarios with child a being a male eliminated.
So as I said before, I don’t think I am getting something that everyone else is. When people are listing combinations with
child a - d(m) and child b - c(f)
I read that the d is definitely and the c is probably. please elaborate on this notation and what it means.
the only way i can see a 2/3 chance is possibly if you are asked to guess the birth order along with the gender, you might have a better chance of being correct if you guess that the other child is a girl. i’m not sure what the odds are I’ll have to do the math. but I wanted to add that before someone responded.
scratch that, i crunched some numbers on that theory and unless i did something wrong, it worked out to 50/50. now you can guess the birth order (not specifically whether the girl i mentioned is first or second) and you would have a 1/3 chance of guessing the correct answer. but i don’t think that’s relevant.
That’s not necessarily the case, although it’s true if you’re equally likely to meet a girl first as a boy if the parent has 1 of each.
If I toss 2 coins behind a curtain. examine them both and if (at least) 1 lands heads announce “One of the coins is heads”. I then show you the coin that landed heads (if they both did I pick one at random) and ask “What is the probability that the other coin behind the curtain (the one you can’t see) landed heads?” The answer is 1/3.
Similarly if the parent will you always introduce a daughter 1st, the probability he has 2 daughters is 1/3 - it doesn’t matter that you’ve actually met her.