Help me with the two-kids probability problem

[Left_Hand_of_Dorkness]

First, there’s a closely-related problem whose solution I understand. Then, there’s the two-kids problem which makes me furious at math in general. Can you help me?

CLOSELY RELATED PROBLEM:
I make three pairs of cards. One pair has a black card and a red card. One pair has two black cards. One pair has two red cards. I randomly choose a pair, and then I randomly choose a card. It’s red. What are the chances that the other card in that pair is red?

INFURIATING PROBLEM (adapted from here):
I tell you that my sister has two kids. You go to her house and knock on the door, and a girl answer the door. “Are you Dorkness’s sister’s kid?” You ask; and the girl says yes. “Is your sibling nonbinary?” you ask, and the girl says no. What are the chances that the other kid is a girl?

I believe these problems are mathematically identical, and that the answer to both problems is 2/3. But while the first problem makes sense to me, the second problem makes me hate math. And in reading the link earlier, it suggests that the answer is 1/3.

What’s more, if I’m told names, it might change the probability. Let’s say I’m told the names are Jo and Terri (but I can’t hear the spelling). As I understand it, the possibilities are:

Jo: boy Terri: boy
Jo: boy Terri: girl
Jo: girl Terri: boy
Jo: girl Terri: girl

If the kid tells me their name is Jo, then there are only two possibilities to consider, and the probability that Terri is a girl is 1/2. Similarly, if the kid tells me her name is Terri, than there are only two possibilities to consider, and the possibility that Jo is a girl is 1/2.

It seems that as long as I know that the child is EITHER Jo OR Terri, there’s a 2/3 probability that the other kid is a girl; but if I know for sure which kid she is, the probability changes.

Is the website I linked to earlier wrong? What’s the problem with it? Or am I misunderstanding something?

Keep in mind that I teach gifted fifth graders math, but I’m far from a mathematician, so if you’re explaining the probability to me, please keep the vocabulary at the level that a non-mathematician can understand.

[colinfred]

I don’t really see the problem. If I go up to your sister and say “Please can you send Jo(e) out to me” and a girl appears I have eliminated 2 out of 4 possibilities and so it’s 50/50 whether Terri(y) is a girl.

The J(B) T(G) has been removed from the matrix as well as the J(B) T(B) you would get from before I knew her name.

[Left_Hand_of_Dorkness]

Here’s a Wikipedia article, but honestly it leaves me even more confused.

[colinfred]

Actually that’s wrong please ignore.

[zimaane]

The Wikipedia article is a bit of a mess, so I understand your confusion.

If a family has two children, and both are binary, then there are four possibilities:
(B,B) ; (B,G); (G,B); (G,G)

If a girl answers the door, then you can throw out (B,B). So the probability that the other child is a girl is 1/3.

[muldoonthief]

The reason the 2 cases aren’t the same is that there are 3 sets of cards, each equally likely. BB, RB, and RR. You choose the set before you continue.

But there are actually 4 possibilities for 2 children - BB, BG, GB, GG. Boy then girl is a different case than girl then boy, but you’re combining them into 1 set. So the odds actually are 25% BB, 50% mixed, 25% GG. You can prove this to yourself by flipping a coin twice 100 times. You’ll wind up with 25% HH, 25% HT, 25% TH, 25% TT.

So when you choose a child set randomly, it’s different odds than the card case - in cards, there’s only a 33% chance you picked a mixed set, but in children, there’s a 50% chance of a mixed set.

[Left_Hand_of_Dorkness]

@muldoonthief

The reason the 2 cases aren’t the same is that there are 3 sets of cards, each equally likely. BB, RB, and RR. You choose the set before you continue.

But there are actually 4 possibilities for 2 children - BB, BG, GB, GG.

That’s incredibly helpful, and what I was missing. Thank you!

[Danger_Man]

Another way to think about it is if you go and count all the two-kid families in the world, there will be twice as many boy/girl (in any order) families than girl/girl families.

[LSLGuy]

@Left_Hand_of_Dorkness

@muldoonthief

The reason the 2 cases aren’t the same is that there are 3 sets of cards, each equally likely. BB, RB, and RR. You choose the set before you continue.

But there are actually 4 possibilities for 2 children - BB, BG, GB, GG.

That’s incredibly helpful, and what I was missing. Thank you!

Said another way, to have the cards accurately model the kids, you’d need two sets of mixed card pairs: a red+black pair and a black+red pair.

[md-2000]

@LSLGuy

Said another way, to have the cards accurately model the kids, you’d need two sets of mixed card pairs: a red+black pair and a black+red pair.

Yes. If you dealt a pair, and the first was black, 50-50 (approx) that the next is black or red.
Similarly, if you dealt a pair, and the first was red, 50-50 the next was black or red.
(Assuming an infinitely large deck so we don’t say 1 in 26 opposite, 1 in 25 same colour).

So to create yourr sets of pairs you need to account for all possible outcomes - 4 pairs - when laying out the pairs.

Also note you have partial information in the selection - 1 child is a girl, but not whether this is the first child or second child. If you know “this is the oldest child” (or, this is the card dealt first) then there is only a 50-50 chance of same or opposite.

(If you picked a first-one-dealt card and it’s red, you didn’t pick the BB pair or the BR pair; you either picked the RB or RR pair).

In certain probability, order matters. For example with a lottery, order does not matter if it’s numbers from 1 to 49 (or 63). winning ping-pong balls 1,2,3… same as 1,3,2… and 3,2,1 etc.

In sports, order matters- who wins the semifinals determines who plays the finals vs. the bronze medal match. W-L silver is better than L-W bronze. Same with the “pulling socks from a drawer” problem. Red and black socks - if you pull BB you have a pair in 2, if you pull BRB it takes 3.

If you really want to go crazy, there’s the Monty Hall door problem.

[hibernicus]

The statement of the Monty Hall problem on the linked web page fails to explain the significance of whether there is a goat or a car behind the door. Specifically, it doesn’t say that the chooser gets to take away and own the car (or goat) behind the chosen door. Without this piece of information, the question “Should you switch doors or not?” boils down to whether you would prefer to see a goat or a car.

[Left_Hand_of_Dorkness]

@md-2000

If you really want to go crazy, there’s the Monty Hall door problem.

I’m having a blast teaching probability. My fifth graders have simulated the Monty Hall problem using playing cards. I’ve offered a big bag of candy to anyone who can guess the order in which I’ll randomly draw six colored dots, but made them figure out the probability of guessing correctly first (Factorials FTW!) We’ve calculated the odds of winning “Not My Job” on Wait Wait, Don’t Tell Me if you just guess randomly. And we’ve solved a couple of other problems. It’s so fun.

[RTFirefly]

@Left_Hand_of_Dorkness

If the kid tells me their name is Jo, then there are only two possibilities to consider, and the probability that Terri is a girl is 1/2. Similarly, if the kid tells me her name is Terri, than there are only two possibilities to consider, and the possibility that Jo is a girl is 1/2.

It seems that as long as I know that the child is EITHER Jo OR Terri, there’s a 2/3 probability that the other kid is a girl; but if I know for sure which kid she is, the probability changes.

Yeesh. I must disagree with the others about the probability that the other child is a girl. The thing is, ANY way of identifying/distinguishing them will do.

In the context of this problem, “the child I am talking to right now” (hereinafter ‘am’) and “the child I’m not talking to right now” (hereinafter ‘not’) will do just fine.

P(‘not’ is a girl, given that ‘am’ is a girl) is 50-50. Here’s why:

Assuming no nonbinary/genderqueer/whatever kids (I’m still working on getting the lingo down), there were four possibilities before you opened the door, assuming one of the children would answer it:

am=boy, not=boy
am=boy, not=girl
am=girl, not=boy
am=girl, not=girl

They were all equally likely until ‘am’ opened the door and turned out to be a girl. Now just the last two are possible, and they’re equally likely. So P(not=girl) is 1/2.

OTOH, I second what @muldoonthief said about why the cards are different than the kids.

[md-2000]

@RTFirefly

Assuming no nonbinary/genderqueer/whatever kids (I’m still working on getting the lingo down), there were four possibilities before you opened the door, assuming one of the children would answer it:

am=boy, not=boy
am=boy, not=girl
am=girl, not=boy
am=girl, not=girl

They were all equally likely until ‘am’ opened the door and turned out to be a girl. Now just the last two are possible, and they’re equally likely. So P(not=girl) is 1/2.

There are 4 equally possible 2-child families - (Simplifying, ignoring twins, but even so one twin is older…)
A: younger boy, older boy
B: younger boy, older girl
C: younger girl, older boy
D:younger girl, older girl

When a girl answers the door, you have eliminated only possibility A.
Among B,C,D the possibility is 2/3 boy, 1/3 girl.

If you know the younger child opened the door (or the older, essentially if you know the order of the child) then you eliminate 2 cases and it’s 50-50; if a younger girl answers the door, it’s either C or D.

[Max_S]

@zimaane

If a family has two children, and both are binary, then there are four possibilities:
(B,B) ; (B,G); (G,B); (G,G)

If a girl answers the door, then you can throw out (B,B). So the probability that the other child is a girl is 1/3.

I see it as eight possibilities, not four, since there’s the additional element of which child opens the door. When you know a girl opens the door, that knowledge lets you throw out half of the (B,G) and (G,B) possibilities.

\text{Probability that the child who remains inside is a girl: } \frac{4}{8} = \frac{1}{2}

Jo (girl) answers door, Terri (girl) remains inside
Jo (girl) answers door, Terri (boy) remains inside
Jo (boy) answers door, Terri (girl) remains inside
Jo (boy) answers door, Terri (boy) remains inside
Terri (girl) answers door, Jo (girl) remains inside
Terri (girl) answers door, Jo (boy) remains inside
Terri (boy) answers door, Jo (girl) remains inside
Terri (boy) answers door, Jo (boy) remains inside

If you know a girl answers the door,

\text{Probability that the child who remains inside is a girl: } \frac{2}{4} = \frac{1}{2}

Jo (girl) answers door, Terri (girl) remains inside
Jo (girl) answers door, Terri (boy) remains inside
Jo (boy) answers door, Terri (girl) remains inside
Jo (boy) answers door, Terri (boy) remains inside
Terri (girl) answers door, Jo (girl) remains inside
Terri (girl) answers door, Jo (boy) remains inside
Terri (boy) answers door, Jo (girl) remains inside
Terri (boy) answers door, Jo (boy) remains inside

If you know a girl named Jo answers the door,

\text{Probability that the child who remains inside is a girl: } \frac{1}{2}

Jo (girl) answers door, Terri (girl) remains inside
Jo (girl) answers door, Terri (boy) remains inside
Jo (boy) answers door, Terri (girl) remains inside
Jo (boy) answers door, Terri (boy) remains inside
Terri (girl) answers door, Jo (girl) remains inside
Terri (girl) answers door, Jo (boy) remains inside
Terri (boy) answers door, Jo (girl) remains inside
Terri (boy) answers door, Jo (boy) remains inside

~Max

[Dead_Cat]

@Left_Hand_of_Dorkness

I make three pairs of cards. One pair has a black card and a red card. One pair has two black cards. One pair has two red cards. I randomly choose a pair, and then I randomly choose a card. It’s red. What are the chances that the other card in that pair is red?

Is the answer to this not 1/2? You know that you must have either the RR pair or the RB pair. It’s 50:50 which.

[Max_S]

These are the probabilities if you don’t know the color of the card you draw:

\text{Probability of the other card being Red: } \frac{3}{6} = \frac{1}{2}

Choose Black, other card is Red
Choose Red, other card is Black

Choose Black, other card is Black
Choose Black, other card is Black

Choose Red, other card is Red
Choose Red, other card is Red

But if the card you draw is Red:

\text{Probability of the other card being Red: } \frac{2}{3}

Choose Black, other card is Red
Choose Red, other card is Black

Choose Black, other card is Black
Choose Black, other card is Black

Choose Red, other card is Red
Choose Red, other card is Red

~Max

[Dead_Cat]

Many thanks, that’s really clear. My intuition has always sucked when it comes to probability.

[Chronos]

The 2/3 answer only comes up in extremely contrived situations that are almost never going to come up in the real world. If you meet one kid at the door, then the other kid is 50-50 to be a boy or girl. In fact, in any situation where it’s even meaningful to refer to “the other kid”, they’re 50-50.

(given all the standard assumptions of this sort of problem, that all children are binary and that boys and girls are overall equally likely).

[Max_S]

@Left_Hand_of_Dorkness

What’s more, if I’m told names, it might change the probability.

It’s not giving the kids names that changes probability, it’s you learning more information about a specific kid. That additional element changes the space a probability function has to operate on.

Two binary-gendered kids come in four permutations:

\text{Probability of two binary-gendered kids both being boys: } \frac{1}{4}

Boy, Boy
Boy, Girl
Girl, Boy
Girl, Girl

In the original boy or girl problem with the \frac{1}{3} solution, you know “[at least] one of them is a boy”. That leaves three of the four possible permutations:

\text{Probability of two binary-gendered kids both being boys: } \frac{1}{3}

Boy, Boy
Boy, Girl
Girl, Boy
Girl, Girl

There are four permutations of two binary-gendered kids, as expected:

\text{Probability of two binary-gendered kids named Jo and Terri both being boys: } \frac{1}{4}

Jo (boy), Terri (boy)
Jo (boy), Terri (girl)
Jo (girl), Terri (boy)
Jo (girl), Terri (girl)

If it is given that one of them is a boy, but not which one, that leaves three of the four possible permutations, as expected:

\text{Probability of two binary-gendered kids named Jo and Terri both being boys,}
\text{where at least one is a boy: } \frac{1}{3}

Jo (boy), Terri (boy)
Jo (boy), Terri (girl)
Jo (girl), Terri (boy)
Jo (girl), Terri (girl)

If you know the gender of one of the children specifically (very dependent on how the problem is presented) there are now eight potential permutations:

\text{Probability of two binary-gendered kids named Jo and Terri both being boys,}
\text{where the gender of a specific child is known: } \frac{1}{4}
Jo (boy), Terri (boy), knowledge that Jo is a boy
Jo (boy), Terri (boy), knowledge that Terri is a boy

Jo (boy), Terri (girl), knowledge that Jo is a boy
Jo (boy), Terri (girl), knowledge that Terri is a girl

Jo (girl), Terri (boy), knowledge that Jo is a girl
Jo (girl), Terri (boy), knowledge that Terri is a boy

Jo (girl), Terri (girl), knowledge that Jo is a girl
Jo (girl), Terri (girl), knowledge that Terri is a girl

That very fine distinction leads to a different probability than the earlier \frac{1}{3} if the gender is also specified,

\text{Probability of two binary-gendered kids named Jo and Terri both being boys,}
\text{where a specific child is known to be a boy: } \frac{2}{4} = \frac{1}{2}
Jo (boy), Terri (boy), knowledge that Jo is a boy
Jo (boy), Terri (boy), knowledge that Terri is a boy

Jo (boy), Terri (girl), knowledge that Jo is a boy
Jo (boy), Terri (girl), knowledge that Terri is a girl

Jo (girl), Terri (boy), knowledge that Jo is a girl
Jo (girl), Terri (boy), knowledge that Terri is a boy

Jo (girl), Terri (girl), knowledge that Jo is a girl
Jo (girl), Terri (girl), knowledge that Terri is a girl

\text{Probability of two binary-gendered kids named Jo and Terri both being boys,}
\text{where Jo is known to be a boy: } \frac{1}{2}
Jo (boy), Terri (boy), knowledge that Jo is a boy
Jo (boy), Terri (boy), knowledge that Terri is a boy

Jo (boy), Terri (girl), knowledge that Jo is a boy
Jo (boy), Terri (girl), knowledge that Terri is a girl

Jo (girl), Terri (boy), knowledge that Jo is a girl
Jo (girl), Terri (boy), knowledge that Terri is a boy

Jo (girl), Terri (girl), knowledge that Jo is a girl
Jo (girl), Terri (girl), knowledge that Terri is a girl

It’s confusing because the question can be written somewhat ambiguous. “What is the probability of two binary-gendered kids named Jo and Terri both being boys, if at least one is known to be a boy?” Does the sample space include your knowledge of a specific kid’s gender or not?

If you knock on the door and learn new information, like asking the kid’s names or assuming gender, I’d say your knowledge is part of the sample space. If it’s all known beforehand, for example in a more abstract problem, I’d say probably not. A kid working through a problem should construct the sample space once, and eliminate possibilities from there.

~Max