Help me with the two-kids probability problem

Factual Questions
21

In the Monty Hall problem, the benefit of switching is that you got to pick the BEST OF two choices, the best of the two other doors. The host as eliminated a goat when you switch doors. The probably that the host was choosing between two goats is only 1/3, which is the odds of you losing when you switch, so you chance of winning by switching is 2/3.

There is no host switching the other child. Its the same child. There is no information gained by interactions with the first child. Names aren’t telling us anything about gender.

No, if a girl answers the door, (B,B), and (B,G) are rules out, order matters.
also you can tell by symmetry that 2/3 is wrong. What if we are told a girl does NOT answer the door, what is the gender of the person answering the door ? Answering the door doesn’t relate to gender, so there is no bias… So how could the flipping of the situation (considering who opens the door vs who does not open the door … both a 50 ,50 split… give a different result ? We often do this flipping, its far easier to flip to considering losing and decide that winning is 100% minus chance of losing… etc

In probability we do distinguish between combinations and permutations, one of which order matters, the other doesn’t. combinations, order matter. You have a combination lock, you need 123… 321 won’t work. Lotto eg pick 6 numbers out of 1 - 50 , its permutations, they don’t know the order you picked. The results they tell you are given in ascending order…

22

I assumed you learned the first child’s gender. That changes the probability from 1/4 or 1/3 to 1/2.

There are three relevant pieces of information.
The family has two binary-gendered kids. Probability of both being girls: 1/4
At least one kid is female. Probability of both being girls: 1/3
A specific kid is female. Probability of the other kid also being a girl: 1/2

~Max

23

To expand on this: there’s a difference between picking pairs and picking individuals.

Take these pairs,
A: younger boy, older boy
B: younger boy, older girl
C: younger girl, older boy
D:younger girl, older girl

Picking a pair at random, then discovering one of the pair is a girl, gives you a 1/3 chance that the other member of the pair is a girl. That is, there is only one pair out of three that has two girls.

Picking an individual at random, then discovering she is a girl, gives you a 2/4 chance that the other member of the pair is a girl. That is, there are two girls who are part of a pair with two girls.

24

it’s all about information.

As I said, there are 4 possible arrangements of families - BB, BG, GB, GG
All are equally likely. So basically, one Boy, one Girl, is twice as likely as two girls.

here, what you do and don’t know affects the probability.
Once you know a girl has answered the door, you have eliminated only one possibility.

You don’t know if she’s the older or the younger. If you knew that, you’d eleiminate two possibilities.

I.e.your first child is a girl… what are the odds the next one will also be a girl?
50-50 because the first does not affect the second. But you know the order.

Bob says he has two children, one is a daughter - what are the odds the other is also?
1/3 because the odds of having GG is 1/2 x 1/2 = 1/4, or you have one child (G or B) and the odds the other is different is 1/2, but the case BB (also 1/4) has been eliminated when the door opens. Partial information - at least 1 girl, but first or last?

The cards thing is different because there are only 3 sets: RB, BB, RR (as opposed to BG abd GB)
You picked a card at random - there are 6 cards, you pick one. 50-50 R or B.
(not really different from “I randomly choose a pair, and then I randomly choose a card.”)
You picked R. So your odds are either RB or RR. 50-50 R or B.

Note it’s different if order matters. If you must pick the left-hand card of the pair, your choices are R,B,R so 2/3 chance of picking R. then, it’s still 50-50 if you picked R what the next card is. If you picked B, the only pair with a left-hand B is BB.

The rules of the game matter. What you know matters. If order is part of the information and the possibilities, order matters.

25

And given the typically poor wording of the set-up, and the typically poor detailed reading comprehension of the audience, there’s lots of room for people to argue about the math when they really should be arguing about the English.

So far we have had a bunch of posters, but far short of all, confidently asserting the wrong answer for the question they seem to be answering. At least as best I can tell.

26

No its 1/2

When a girl answers the door, you dont just discount the 2-boy possibilty. You also have to discount the “ 1 of each, and the boy answered the door” universes as well.

27

Age is irrelevant. The only purpose it serves in this problem is as a way to label the two children. But meeting one of them at the door also accomplishes that: It lets us label the children as “the one you met at the door” and “the one you didn’t meet”.

To illustrate this, let me describe one of the contrived situations where the answer is 2/3:

You meet a stranger, and ask them "How many children do you have?, and they say “Two”. You then ask them “Do you have at least one daughter?”, and they say “Yes”. At this point, there is a 2/3 chance that the family is a boy and a girl, and a 1/3 chance that it’s two girls.

You then ask them “Is your other child a boy?”, and they say “Which child do you mean by ‘your other child’?”. You say “The one other than your daughter”, to which the person gets confused and says “What do you mean, my child other than my daughter? Both of my children are daughters.”

You can have “The older child, and the other child”. You can have “The taller child, and the other child”. You can have “The first child you thought of just now, and the other child”. You can have “The child I’ve met, and the other child”. But it’s not meaningful to say “At least one child is a girl, and the other child”.

28

I flip two coins but keep them covered.
I reveal one coin - it is heads.
What is the chance the other coin is also heads?

29

That depends on your procedure for revealing a coin, and what you mean by “keep them hidden”. If you keep them hidden from yourself, or you see them but choose a coin at random to reveal, then the other coin has a 50% chance of being heads. If you look at the coins yourself first and then choose which one to reveal based on what you see, then it depends on how you’re basing your choice on that information.

30

Not if I have already outlined exactly what I’m doing. Which I did, and is also the OP’s question.

I’m going to flip two coins. I will reveal one. What is the chance the other coin has the same orientation?

That’s it.

ETA. People are trying to shoe-horn Monty Hall into a question that isn’t the Monty Hall problem. This appears to be the root of the OP’s frustration.

31

No, you haven’t outlined exactly what you’re doing, and without that information, we cannot calculate a probability, because the details you’re not telling us actually matter.

32

So two heads or two tails will meet the requirement? If you reveal a heads then the second head completes it, if you reveal a tails then the second tail completes it? So with 4 possible scenarios (HH, TT, HT, TH) your likelihood is 50%.

If you ask the question after you reveal the first coin then it becomes 2/3 because you’ve eliminated one of the four and it’s a different question.

33

In what sense is it a different question? Moreover why have I only eliminated one of the four?

34

Before you reveal the first coin, either HH or TT counts as fulfilling the requirements; TH and HT do not. After you reveal the first coin, either HH or TT is eliminated. Revealing the coin provides relevant information. How you frame the question and when you ask the probabilities matters greatly.

35

Which details? Are you assuming I might change my mind about the question I ask? I told you ahead of time what the question is. Are you assuming that I can sway the result by choosing which coin to uncover? In what manner can this affect the outcome? Are you assuming it isn’t a fair toss of the coins?
What if this is done by a machine?

Like I said, this isn’t the Monty Hall problem. There isn’t the opportunity for me to do anything that transmits information. Three coins, yes, we can do Monty Hall. Two coins, no.

36

One of us isn’t following the hypothetical here.

“I’m going to flip two coins. I will reveal one. What is the chance the other coin has the same orientation?”
The word “will” is key here. Does it mean, “I will reveal one, and then ask you if the other coin has the same orientation”?

If so, If I reveal H, I’ve eliminated TT. What remains is HH and HT (or HH and TH, depending one which one I reveal). So it’s 50%

If it means, “I will reveal one, but before I do, What is the chance the other has the same orientation?” Then nothing is eliminated.

37

I think there is a confusion in evaluating probability versus looking the odds as a counting problem.

Counting involves calculating the number of individual ways a system can appear. That is nuanced because you must be careful about what information you use to break down the possibilities.

Or you can assign probabilities to the outcomes. But when someone action occurs you need to update your probabilities. It isn’t just a matter of evenly spreading out the remaining non-zero ones. You must update them all.

The question with the children exemplifies this. There are three possibilities.
Two boys, 0.25
Two girls, 0.25
One of each. 0.5.

You need to revaluate the probability given the extra information provided when the door opens.

Chance of two boys given a girl answers: 0
Chance of one of each given a girl answers, and thus the second child is a boy : 0.5
Chance of two girls given a girl answers 0.5

Same thing for a boy answering.
It is 50:50.

Or you can count the possibile outcomes. Here is where there is a nuance. One of each can occur in two ways. But you can’t reason about the outcomes without identifying how these two ways can affect your interaction. The manner in which you do must leave you with identical probabilities for all the classes you end up with.
There are an arbitrary number of ways you could try to distinguish between the two, but only one which involves what happens when you knock on the door. Which child answers.

You have an equal chance of:
Two boys and a boy answers.
Two girls and a girl answers.
Boy girl and boy answers.
Girl boy and girl answers.

Two of these are eliminated when the door opens, not one.

38

I agree with all the people saying that this only happens in very contrived examples.

You need to somehow select “all the houses that have at least one girl”. That’s not something that happens very often in real life.

In the example with the girl opening the door, you need to add a rule like “if there is a girl in the household, she will always be the one to open the door”. In the case you will have the equivalent situation.

If you try walking in your own neighborhood and knocking on doors, you won’t recreate the situation at all, and the probability will be 50/50. The difference between the situation above is that sometimes the boy in the boy/girl households will open the door, and you won’t have the selection of households you need.

The probability calculations are straightforward, but the exact wording is very important, to know what kind of situation you are in.

39

Yes, that’s exactly what I’m saying. If, for instance, you look at both coins, and decide that you’re going to reveal a “Heads” if possible, and only reveal a “Tails” if there are no heads, then if you reveal H, then there’s a 2/3 chance that the other coin is T, and if you reveal T, there’s a 1 chance that the other coin is T. Which, yes, is the Monty Hall problem, and that’s one of the myriad possibilities that matches your description.

40

The point is - did you choose to reveal “heads” (on seeing both coins, or one) or did you simply reveal one of the coins at random wthoout knowing which or what it was?

If you had a choice of two coins - result known to you - and picked one, then the 2/3 rule applies. You had 2 coins, you decided to pick one that was H. Therefore, you could choose one from either HT, TH, or HH. there was a 25% chance you’d have tossed TT and this question would be impossible.

If you revealed that the first toss was Heads, then a second toss would be 50-50. But since you have 2 independent coin tosses, but get to pick only one to reveal, you could be revealing either coin A or coin B. So 50-50 which coin you picked to reveal - what if it had been tails? That’s an additional probability that has to figure into this question…

If you randomly picked one, then it’s the 50-50. “My first child was a girl, what was my second child?” or “my first child was a boy, what’s my second child?”