I don’t think you’re modeling your problem correctly, unless I misunderstood the question you’re trying to answer. You have those following random variables :
X_1 is bernoulli, first child is a boy
X_2 is bernouilli, second child is a boy
Y_1 is uniform, weekday of birth of the first child
Y_2 is uniform, weekday of birth of the second child
D is a random variable which corresponds to the weekday in the sentence “one of them is a boy, born a (D)”. There is many ways to construct one like this, but we only require that if X_1=1 or X_2=1, then D=Y_1 or D=Y_2, and that D=Y_i implies X_i=1.
Then what you’re looking for is not P(X_1=1,X_2=1 | (X_1=1,Y_1=monday) or (X_2=1,Y_2=monday)) (which, indeed, is not 1⁄3), but P(X_1=1,X_2=1 | ((X_1=1,D_1=D) or (X_2=1,D_2=D)) and D=monday). This is still 1⁄3, as illustrated by this Python snippet (I’m too lazy to properly demonstrate this formally) : https://gist.github.com/sloonz/faf3565c3ddf059960807ac0e2223200
There wass a similar paradox presented on old lesswrong. If someone can manage to find it (a quick google search returned nothing, but i may have misremembered the exact terms of the problem…), the solution would be way better presented there :
Alice, Bob and Charlie are accused of treason. To make an example, one of them, chosen randomly, will be executed tomorrow. Alice ask for a guard, and give him a letter with those instructions : “At least Bob or Charlie will not be executed. Please give him this letter. If I am to be executed and both live, give the letter to any one of them”. The guard leaves, returns and tell Alice : “I gave the letter to Bob”.
Alice is unable to sleep the following night : “Before doing this, I had a 1⁄3 chance of being executed. Now that it’s either me or Charlie, I have a 1⁄2 chance of being executed. I shouldn’t have written that letter”.
I don’t think you’re modeling your problem correctly, unless I misunderstood the question you’re trying to answer. You have those following random variables :
X_1 is bernoulli, first child is a boy
X_2 is bernouilli, second child is a boy
Y_1 is uniform, weekday of birth of the first child
Y_2 is uniform, weekday of birth of the second child
D is a random variable which corresponds to the weekday in the sentence “one of them is a boy, born a (D)”. There is many ways to construct one like this, but we only require that if X_1=1 or X_2=1, then D=Y_1 or D=Y_2, and that D=Y_i implies X_i=1.
Then what you’re looking for is not P(X_1=1,X_2=1 | (X_1=1,Y_1=monday) or (X_2=1,Y_2=monday)) (which, indeed, is not 1⁄3), but P(X_1=1,X_2=1 | ((X_1=1,D_1=D) or (X_2=1,D_2=D)) and D=monday). This is still 1⁄3, as illustrated by this Python snippet (I’m too lazy to properly demonstrate this formally) : https://gist.github.com/sloonz/faf3565c3ddf059960807ac0e2223200
There wass a similar paradox presented on old lesswrong. If someone can manage to find it (a quick google search returned nothing, but i may have misremembered the exact terms of the problem…), the solution would be way better presented there :
Alice, Bob and Charlie are accused of treason. To make an example, one of them, chosen randomly, will be executed tomorrow. Alice ask for a guard, and give him a letter with those instructions : “At least Bob or Charlie will not be executed. Please give him this letter. If I am to be executed and both live, give the letter to any one of them”. The guard leaves, returns and tell Alice : “I gave the letter to Bob”.
Alice is unable to sleep the following night : “Before doing this, I had a 1⁄3 chance of being executed. Now that it’s either me or Charlie, I have a 1⁄2 chance of being executed. I shouldn’t have written that letter”.