You say to Mary, “Hey, these may seem like random questions, but (Q1) do you just happen to have exactly two children, (Q2) with the added conditions that at least one is a boy, (Q3) who was born on a Tuesday?” Then however many answer “yes” to Q1, 75% will answer “yes” to Q2. Of those, 2/3 will also have a girl. And 13.8% of those who answer “yes” to Q1 will also answer “yes” to Q3. Of those, 14/27 will also have a girl.
Now the tricky part. What part of… “Mary has 2 children. She tells you that one is a boy born on a Tuesday” tells you that these three questions were asked? Please be specific - point out the words.
Because what I meant was that when a potential Mary tells us this kind of information without any such prompting, then all of these Marys who have BB will tell is that at least one is a boy. Half of those with BG or GB will say “boy” and the other half will say “girl.” And all with GG will say “girl.” I’m saying that we have to ignore why a potential Mary tells us this kind of information because the problem does not indicate why. And so use these proportions. The probability that such a Mary also has a girl is 50%. And the same is true if she adds “Tuesday.”
Absolutely nothing tells you that. That’s my point. We don’t know the context of why Mary gave us that information, and different context will give different answers, and so we can’t calculate the probability.
People give the right answer to the Monty Hall Problem, for the wrong reason, when they say “Your original changes can’t change” or “It’s like getting the better of two doors.” The right reason is that Monty will open door #3every time the car is behind door #2, but only half the time it is behind door #1. This makes the odds for #1 (1/2):1, or 1:2, or a probability of 1/3.
Very few people give the right answer for the Two Child Problem because far too many teachers tell them that an incorrect solution - one just like the one in the MHP that says switching can’t help - is correct in the TCP. The correct solution is that every Mary who has BB will tells about a boy, but only half of the Marys with BG or GB will. This makes the odds that she has a boy and a girl (1/2+1/2):1, or 1:1, or a probability of 50%.
THEY ARE THE SAME PROBLEM, just with a different number of cases. If you apply actual probability theory, instead of memes that sound good but eschew mathematics, this ios easy to see.
This is not necessarily true; since it limits the puzzle with an additional condition/information, cannot be part of the puzzle. If Mary had a little boy, first, last or both, she tells us she has a boy. Why would she not?
I actually do have two children. I am thinking of a gender that applies to at least one. What is the probability that I have a boy and a girl?
Since I have given you no gender information, the answer must be 1/2.
But if I were to tell you that I was thinking of “boy,” then my statement, the question, and the answer is the same as Mary’s. Call the answer P. And if I were to tell you that I was thinking of “girl,” then it is a functionally equivalent question with the same answer P. But if the answer is P regardless of what word I was thinking of, I don’t have to tell you the word. Just knowing that I was thinking of a word makes the answer P.
Do you recall that the answer is 1/2?
Probability can be thought of as a measure of our ignorance. If I roll a six-sided die out of your sight, the probability that it landed on a four is 1/6. If I tell you the result is even, that changes to 1/3. But nothing about the die changed, just your knowledge.
My point is that yes, we can calculate “a” probability with no knowledge, under certain conditions. In this case, the conditions are that we know nothing that would make a Mary with BG more, or less, likely to say she has a “boy” instead of saying she has a girl. So within our knowledge, there is a 50% chance of either. This is true even if there is an underlying reason that Mary knows but we don’t.
The issue is, you’re letting Mary pick the gender to start the problem, so 50-50; it’s a combination of the “I have a boy” and “I have a girl” problems, so yes, 50-50.
If we fix that Mary said “I have a boy” then the question of whether she was going to speak up, and what gender she picks, is not part of the problem. Just as saying “that boy was born on a Tuesday” doe not give her 7 days to pick from, little Bobby himself picked the day.
Here’s my take on the full Monty: P is pick, O is open C is correct, doesn’t matter what numbers I say here, permute them to all combinations of 123 and the answers are the same.
1,2,or 3 have the same 1/3 chance of being correct. Monty will never early-open the correct door.
1PC 2 3O - is 1/6 likely
1PC 2O 3 is 1/6 likely - if 1 is correct, 50-50 Monty opens 2 or 3.
1P 2C 3O is 1/3 likely - if 2 is correct, Monty must open 3
1P 2O 3C is 1/3 likely - if 3 is correct, Monty must open 2.
You pick door number 1 (1P), Monty opens door 2, so 1C=1/3, 3C=2/3
You pick door number 1 (1P) Monty opens door 3, so 1C=1/3, 2C=2/3
Which leads me to the contrary conclusion your odds tell you to switch - where did I go wrong?