The trick of the so-called "paradox" is turning the question into the Monty Hall but with an ambitious enough formulation that you might be confused it’s not.
p(both are girls | you're told at least one is a girl)
= p(you're told at least one is a girl | both are girls) * p(both are girls) / (
p(you're told at least one is a girl | both are girls) * p(both are girls)
+
p(you're told at least one is a girl | they aren't both girls) * p(they aren't both girls)
)
This is the crux of the thing. Different readings of the setup imply different answers to p(what you're told | the unknowns).
It's also a great case of where bayes rule shorthands can be slippery. You'll usually abbreviate it out (hell, it was tedious to write this way even with copy-paste). But if you abbreviate "you're told there's at least one girl" to "there's at least one girl", then you've stopped modeling a crucial part of the setup. p(there's at least one girl | they aren't both girls) has an unambiguous answer.
What is the probability that a family with 2 children has exactly 2 daughters *if you know that the family does not have 2 boys*?
Because most people don't talk formal probabilities, your explanations will be so vague that the other person will not realize your different understanding. You will discuss forever, you will both be right, and you will part ways with the strange feeling that maybe the other person was right, when all along you were talking about different problems. This is why this problem is so notorious.
I think this is true of the "children" question, but I actually disagree that this is what makes the Monty Hall question so confusing.
For one thing, I vaguely recall this being asked directly, and even after people agree on all the definitions explicitly, they still consider the answer wrong. (See e.g. some mathematicians like Erdos refusing to believe the correct answer without actually running simulations on computers... by that point you clearly have a real definition.)
For another, when I personally talk to people about Monty Hall, even after I explain the correct answer, and explain all the nuances, people tend to still have a hard time accepting the correct solution and claim to find it counterintuitive (as did I!).
(Also, this problem has an additional layer of ambiguity where “birth order is irrelevant,” but MF and FM are treated as distinct items in the probability set. Is the order irrelevant to the probability, or is it impossible to distinguish the age of the two children? It would be clearer to simply say “each birth is an independent event.” One of the comments on the blog explains this better than I can.)
Almost everybody understands the same problem, AND STILL GET DIFFERENT ANSWERS. If they don't understand it, they make pedantic arguments about Monty's motivations. All of which make the puzzle impossible to answer.
What they don't understand is probability. Probability is a measure of the information you lack about what causes a certain result to occur. That includes the physical details (where the prize is, what the genders are) but also the choices made for hidden reasons.
In the Monty Hall Problem, to reduce complexity, label the doors C (the contestant's original door), R (the door to its right, wrapping around if necessary), and L (the door to its left). What leads up to the game state at the time the decision to switch is made are (A) Where the prize was placed and (B) How Monty Hall chooses a door to open if the prize is behind C.
The naive answer is based on only (A). The two unopened doors (C and R, or C and L) started with the same probability. So they must now have the same probability, 1/2, right? No, wrong, because we need to take (B) into account. If the prize is behind R then the host had to open L. If the prize is behind L then the host had to open R. But if the prize is behind C then the host had to choose. Since we don't know how, we have to assume there was a 50% chance that he would choose R, and 50% for L. Once we see him open, say, R? This 50:50 reduces the probability that the prize is behind C, so switching becomes twice as likely to win.
The Two Child Problem works exactly the same way. What leads up to the point where we are asked for a probability is (A) the gender makeup of the family and (B) how the information came to us if there is a boy and a girl.
The naive answer is based on only (A). A mixed family is twice as likely as either two-of-a-kind family. So the probability of two-of-a-kind is 1/3, right? Wrong, unless we know WITH CERTAINTY that we could not have learned about the other gender. If we do not have that certainty, then just like with Monty Hall we have to assume that half of the time in a mixed we would have learned the other gender. This makes a mixed family half as likely as (A) alone would suggest; in other words, the same as two-of-a-kind.The answer is 1/2.
Joseph Bertrand pointed out, in 1889, why we need to take (B) into account. Martin Gardner, who originated the Two Child Problem, repeated it in 1959. In the same article where he introduced the predecessor to Monty Hall (called the Three Prisoners Problem), and explained why (B) is important. It should be embarrassing to anyone who thinks that the "Tuesday" variation's answer is 13/27. Because it was first mentioned at a puzzle convention named in honor of Martin Gardner and forgot his warning. Adding irrelevant information can't change the answer, and if you take (B) into account the answer doesn't change.
Exactly. As the first reply to the first comment explains “The problem is that we don't know p(you're told at least one is a girl | they aren't both girls).”
It’s funny that the same commenter who writes that “to get the right answer you must be careful about conditional probabilities” finds that doing so is “splitting hairs about something boring and irritating.”
vs
"The question writer came across a girl from a two child family, then they asked the exact question above". This is 1/2 chance - select gg from [gg, gg, bg, gb] with gg listed twice since there's two ways to select a girl from that set; ie. coming across a girl is twice as likely to occur from the gg case than it is either gb or bg.
I think that's the clearest wording to get the message across. Either way it's the exact same question but it reasonably has a completely different answer. There's no way to resolve this ambiguity with the question as written.
If you look at one random child, see it's a boy and exclude the family, even though the other child may be a girl, then you get the 1/2 probability. If however in that case you also look at the second child, see that's a girl and consider the family anyway, then you get the 1/3
I think this is a reasonable interpretation:
You meet a family at a party. They say "We have two children". You ask "Do you have any girls"? They say "yes!"
This will give you 1/3 probability that the other child is also a girl.
I think this interpretation is more intuitive because it doesn't make any assumptions about how you get your information. Usually in probability questions you assume any information you have is given to you from on high. For example, you just "know" that the family has two children, you don't somehow deduce it. Therefore I assume the same for "one child is a girl" information.
Do you mean “interpretation” or “alternative problem”.
Because if it’s an “interpretation” of the original problem you’re indeed making assumptions to fill the unspecified information.
If you mean that it’s an alternative problem which has a definite solution I agree. It’s a different problem and its relevance to the original one is to illustrate that additional assumptions were required.
The original problem cannot be answered without making additional assumptions about how you get your information. Different interpretations may reach different answers by making different assumptions.
ambiguous?