I don’t think your analysis is right; For one, if you know that the random variable D is Monday, your problem reduces to mine; In the simulation, you have set D incorrectly ( D=Y1in the case both are boys. This makes no sense. You are ignoring the case where the first boy is not born on Monday, but the second one is.), and it is not a simulation of the probability you have written.
The second example; I’m not sure what your conclusion on it is. It seems like the Monty Hall problem to me, i.e., Alice is still having a chance of 1⁄3, but Charlie now has 2⁄3 chance to die. Because:
Yes, that’s kind of my point. There’s two wildly different problems that looks the same on the surface, but they are not. One gives the answer of your post, the other is 1⁄3. I suspect that your initial confusion is your brain trying to interpret the first problem as an instance of the second. My brain sure did, initially.
On the first one, you go and interview 1000 fathers having two children. You ask them the question “Do you have at least one boy born on a Monday ?”. If they answer yes, you then ask then “Do you have two boys ?”. You ask the probability that the second answer is yes, conditioning on the event that the first one is yes. The answer is the one of your post.
On the second one, you send one survey to 1000 fathers having two children. It reads something like that. “1. Do you have at least one boy ? 2. Give the weekday of birth of the boy. If you have two, pick any one. 3. Do you have two boys ?”. Now the question is, conditioning on the event that the first answer is yes, and on the random variable given by the second answer, what is the probability that the third answer is yes ? The answer is 1⁄3.
My main point is that none of the answers are counter-intuitive. In the first problem, your conditioning on Monday is like always selecting a specific child, like always picking the youngest one (in the sentence “I have two children, and the youngest one is a boy”, which gives then a probability of 1⁄2 for two boys). With low n, the specificity is low and you’re close to the problem without selecting a specific child and get 1⁄3. With large n, the specificity is high and you’re close to the problem of selecting a specific child (eg the youngest one) and get 1⁄2. In the second problem, the “born on the monday” piece of information is indeed irrelevant and get factored out.
You’re correct, but I still don’t find the scenario I am describing intuitive by my system 1. If I think about it (especially now that I have analyzed the problem rigorously), yes, I’ll feel that the probability should increase (though of course, with nowhere near the precision of the Bayes rule), but if you just asked me this problem yesterday, and I wasn’t watching for trap questions, I’d give you a wrong answer.
I just found Wikipedia has a whole page on this, named the girl and boy paradox. They cover lots of details there.
I don’t think your analysis is right; For one, if you know that the random variable D is Monday, your problem reduces to mine; In the simulation, you have set D incorrectly ( D=Y1in the case both are boys. This makes no sense. You are ignoring the case where the first boy is not born on Monday, but the second one is.), and it is not a simulation of the probability you have written.
The second example; I’m not sure what your conclusion on it is. It seems like the Monty Hall problem to me, i.e., Alice is still having a chance of 1⁄3, but Charlie now has 2⁄3 chance to die. Because:
P(Bob | Alice) = 1⁄2
P(Bob | Charlie) = 1
Yes, that’s kind of my point. There’s two wildly different problems that looks the same on the surface, but they are not. One gives the answer of your post, the other is 1⁄3. I suspect that your initial confusion is your brain trying to interpret the first problem as an instance of the second. My brain sure did, initially.
On the first one, you go and interview 1000 fathers having two children. You ask them the question “Do you have at least one boy born on a Monday ?”. If they answer yes, you then ask then “Do you have two boys ?”. You ask the probability that the second answer is yes, conditioning on the event that the first one is yes. The answer is the one of your post.
On the second one, you send one survey to 1000 fathers having two children. It reads something like that. “1. Do you have at least one boy ? 2. Give the weekday of birth of the boy. If you have two, pick any one. 3. Do you have two boys ?”. Now the question is, conditioning on the event that the first answer is yes, and on the random variable given by the second answer, what is the probability that the third answer is yes ? The answer is 1⁄3.
My main point is that none of the answers are counter-intuitive. In the first problem, your conditioning on Monday is like always selecting a specific child, like always picking the youngest one (in the sentence “I have two children, and the youngest one is a boy”, which gives then a probability of 1⁄2 for two boys). With low n, the specificity is low and you’re close to the problem without selecting a specific child and get 1⁄3. With large n, the specificity is high and you’re close to the problem of selecting a specific child (eg the youngest one) and get 1⁄2. In the second problem, the “born on the monday” piece of information is indeed irrelevant and get factored out.
You’re correct, but I still don’t find the scenario I am describing intuitive by my system 1. If I think about it (especially now that I have analyzed the problem rigorously), yes, I’ll feel that the probability should increase (though of course, with nowhere near the precision of the Bayes rule), but if you just asked me this problem yesterday, and I wasn’t watching for trap questions, I’d give you a wrong answer.
I just found Wikipedia has a whole page on this, named the girl and boy paradox. They cover lots of details there.