Puzzling Probabilities Part 3: The Two Child Problem
Updated: Apr 1, 2021
Hello everybody! I have one more neat probability problem to show you.
Let's say you meet three mothers: Mrs. A, Mrs. B and Mrs. C. Each has two children.
Mrs. A tells you her oldest child is a girl. What's the probability she has two girls?
Mrs. B tells you one of her children is a girl. What's the probability she has two girls?
Mrs. C tells you one of her children is a girl born on Tuesday. What's the probability she has two girls?
Go ahead! Try all three and see what kind of answers you get, then read on.
Ready? Let's continue.
At first glance, it seems like the answer to all three questions is 50%.
It seems that the random comments Mrs. A and Mrs. C made about their daughter (being older or being born on Tuesday) have no effect on the overall probability. After all, each is talking about only one of their children and making no references to the other, so we assume their second child has a 50% chance of being a girl and a 50% chance of a boy.
It turns out that's not true. The probabilities are, in turn, 1/2 for Mrs. A, 1/3 for Mrs. B and 13/27 for Mrs. C. Why? Let me show you.
Let's work through the probabilities! (Here we're listing the oldest child first and the youngest child second.)
For Mrs. A's children, there are two potential possibilities for her children's genders that meet the condition that her oldest is a girl: Girl-Girl and Girl-Boy. There are 2 total possibilities and 1 Girl-Girl combo, so the probability is 1/2.
For Mrs. B's children, there are three that meet the condition she has at least one girl: Girl-Girl, Girl-Boy, and Boy-Girl. This gives us a probability of 1/3 that her children are both girls.
For Mrs. C's children, we have a lot more possibilities because we're now taking the day of the week the child is born on into account:
There are a total of 28 combinations that include a girl born on Tuesday, but notice we're overcounting one of them. We have the possibility of two girls born on Tuesday listed twice, so there are only 27 combinations. 13 of them have two girls, so our overall probability is 13/27.
Realize that the answers to our probabilities came together quickly as soon as we were able to come up with a list of all of the equally likely possibilities that meet the conditions Mrs. A, Mrs. B or Mrs. C gave us, making sure we don't overcount anything and don't miss anything.
Like I mentioned earlier, this set is called the sample space. The event space is a second term that describes all the possibilities that match what we want (here, two girls). For example, Mrs. A's event space is one possibility (Girl-Girl) and her sample space is two possibilities (Girl-Girl and Girl-Boy). For Mrs. C, all 27 possibilities (minus the red one we overcounted) are her sample space and all 13 combinations colored in green are her event space.
This is the basis of probability - collecting your sample space and your event space carefully - and it's fairly revealing here.
Here's a question. What's different about Mrs. A's sample space versus Mrs. B and Mrs. C?
Mrs. B and Mrs. C could be talking about either child, while Mrs. A is only talking about one!
The essence of this problem lies in the fact that we know which child Mrs. A is referring to (the older one!) but not which child Mrs. A and Mrs. C are mentioning. After all, it's equally likely that Mrs. B's girl is younger or older, so we have a lot more wiggle room. Once again, our intuition fools us, and since Mrs. B and Mrs. C's sample space are extended, their probabilities aren't quite 1/2 like we expected.
It's a little bizarre that the way we word a problem or what information we include can change overall probabilities so fundamentally. In fact, it lies a little bit in the way English works that these "paradoxes" even emerge.
Here's my best attempt at a short explanation of that claim: If, for example, we had Mrs. B and Mrs. C flip a coin to decide which child to talk about, and when we ask them the question, "Is your child a girl?" they answer "yes" or "no" according to the results of that coin flip, the probabilities will all reduce to 1/2 because the other child is no longer in the picture at all. However, in English, when we ask Mrs. B if she has a girl, she's always compelled to mention the fact that she has a girl if she has one at all, which means we reach our sample space of three (GG, GB, BB). People mention that they have a girl if they have at least one, they don't mention that they have a girl if they have only one or that they have a girl only if she happens to be standing in front of them. It turns out that this dynamic makes the problem a lot more ambiguous, which is where we can run into problems.
If you want to see a longer explanation, feel free to check out this source: https://gdaymath.com/lessons/powerarea/4-6-more-practice/
(It's the best source I've found on this topic!)
This problem is often called the Boy-Boy Paradox (when it's stated in terms of boys) or the Two Child Problem, and though it's not really a paradox, it is a really fascinating probability problem.
Here's a neat application. Because probabilities often depend so much on how they are worded and lots of these intricacies can emerge, there are often mistakes. This YouTuber used a version of this problem to explain why TED-Ed made a mistake on their frog riddle.
Here's the original frog riddle: https://ed.ted.com/lessons/can-you-solve-the-frog-riddle-derek-abbott
(Not to discount TED-Ed, but everybody makes mistakes!)
Guess what else? This problem is also called Bertrand's Box. That's because it was originally stated not in terms of genders but in terms of gold and silver, and it was posed by the French mathematician Joseph Bertrand:
Imagine there are three identical treasure chests in front of you.
One contains two gold coins.
One contains two silver coins.
One contains one gold and one silver coin.
You pick a random box and take out a gold coin. What's the probability the other coin in the box is also gold?
I'll give you the chance to think about this one and to investigate which of the above problems it's roughly equivalent to! (Hint, the answer is the same as the answer to the Monty Hall Problem) When you're ready, go to this link and make sure to also read the comment by JeffJo at the bottom which connects it to the Two Child Problem: http://blog.zymergi.com/2013/06/bertrands-box-paradox.html
I'll leave you with some other great explanations of the problem and resources:
I hope you enjoyed the post and I apologize for the delay this week! This is probably going to conclude my Puzzling Probabilities series for the time being, but feel free to let me know if you find another neat probability problem (or any problem) that you'd like me to explore and I just may write a post about it! Comment down below if you have any questions and as always, use the forum to discuss anything and everything you want to!