p(2s\T)=p(2s & t)/p(T)
p(2s & T)=1/4(13/49)
p(T)=1/4(13/49)+1/2*(1/7)
p(2s\T)=(13/49)/(13/49+14/49)
p(2s\T)=13/27
Think you understand Monty Hall? Try the Tuesday boy problem.
11–20 of 152 posts
Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#12> Now suppose we know simply that a man has two children and that one of them is a son. [snip] It follows that the sexes of his two children, ordered from oldest to youngest, are either BB, BG or GB. Since these cases are equally likely, and since only one of them involves having two boys, we would say the probability that the man has two boys is 1/3.
Errr. except these cases aren't equally likely - because the boy we know about could be either of the boys in the BB scenario, but only one either of the other two. So the probability is still 1/2.
Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#13Doesn't this rest on the simple ambiguity in the phrasing? > I have two children and one is a son born on a Tuesday. If by that is meant: > I have two children. Here is some information about one of them: son, born on Tuesday. Then the probability of the other child being a son is 1/2. If on the other hand we mean: > I have two children. One or more is a son. Exactly one of them was born on a Tuesday. Then we get the…
if exactly one of the children is born on Tuesday, would't the probability of having a son be 5/13 and not 13/27.
Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#14Let's try a simpler problem. Suppose we know that a certain man has two children and we also know that the older one is a boy. In this case we would say that the probability that the other child is a boy is 1/2. After all, the sex of one child is independent of the sex of the other child. That the older child is a boy has no bearing on the sex of the younger child. Now suppose we know simply that a man has two childr…
For a randomly selected family with two children, there are four possible boy/girl combinations:
B B,
B G,
G B,
G G
In the first case we are told that the older child is a boy. This leaves only two cases: B B, B G Therefore, there is a 50% chance the second child is a boy.
In the second case, we are told only that [at least] one child is a boy. This leaves three possibilities: B B, B G, G B Therefore, the probability that both children are boys is 1/3.
Enumerating possible states of the world like this is the fundamental insight you need to have to be able to understand these types of problems - but it does take a while to get used to!
Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#15I hate this one because while it says: "I have two children and one is a son born on a Tuesday." It actually means: "I have two children and only one of them is a son born on a Tuesday." You are supposed to just assume this modification.
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Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#16> Now suppose we know simply that a man has two children and that one of them is a son. [snip] It follows that the sexes of his two children, ordered from oldest to youngest, are either BB, BG or GB. Since these cases are equally likely, and since only one of them involves having two boys, we would say the probability that the man has two boys is 1/3. Errr. except these cases aren't equally likely - because the boy w…
Since the constraint simply says that at least one child is a boy, we don't need to distinguish between two types of BB.
This is a curious misconception, by the way. The typical failure mode I see on these questions is people having difficulty accepting BG and GB as different possibilities.
Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#17I hate this one because while it says: "I have two children and one is a son born on a Tuesday." It actually means: "I have two children and only one of them is a son born on a Tuesday." You are supposed to just assume this modification.
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Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#18Let's try a simpler problem. Suppose we know that a certain man has two children and we also know that the older one is a boy. In this case we would say that the probability that the other child is a boy is 1/2. After all, the sex of one child is independent of the sex of the other child. That the older child is a boy has no bearing on the sex of the younger child. Now suppose we know simply that a man has two childr…
For a randomly selected family with two children, there are four possible boy/girl combinations: B B, B G, G B, G G In the first case we are told that the older child is a boy. This leaves only two cases: B B, B G Therefore, there is a 50% chance the second child is a boy. In the second case, we are told only that [at least] one child is a boy. This leaves three possibilities: B B, B G, G B Therefore, the probability…
But you are assuming that each of those three possibilities has equal probability; can you explain the rationale for that? (it is clear why BB, BG, GB, and GG have equal probability in the unrestricted case, but less clear why BB, BG, and GB have equal probability in this restricted case)
Besides, this is just a rephrasing of the original article's argument, and doesn't counter mine at all. I am open to the possibility that there is a flaw in my argument, but where is it?
Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#19> Now suppose we know simply that a man has two children and that one of them is a son. [snip] It follows that the sexes of his two children, ordered from oldest to youngest, are either BB, BG or GB. Since these cases are equally likely, and since only one of them involves having two boys, we would say the probability that the man has two boys is 1/3. Errr. except these cases aren't equally likely - because the boy w…
Since the constraint simply says that at least one child is a boy, we don't need to distinguish between two types of BB. This is a curious misconception, by the way. The typical failure mode I see on these questions is people having difficulty accepting BG and GB as different possibilities.
If we are determined to look at order of birth (ie. have a BG and GB) then we should consider all cases by order of birth, so where 'B' represents the boy we know and 'b' or 'g' represents a child we dont, and the first character represents the first child, and the second the second childe - we have:
Bb, bB, Bg, gB
50%.
Re: Think you understand Monty Hall? Try the Tuesday boy problem.
#20Doesn't this rest on the simple ambiguity in the phrasing? > I have two children and one is a son born on a Tuesday. If by that is meant: > I have two children. Here is some information about one of them: son, born on Tuesday. Then the probability of the other child being a son is 1/2. If on the other hand we mean: > I have two children. One or more is a son. Exactly one of them was born on a Tuesday. Then we get the…
The ambiguity isn't in the phrasing of the father, we understand what he says about his family, it
is in the probability universe that surrounds his statement. The scenario doesn't give one; we don't have a model for the probability of people saying anything at all unprompted. We have a simplified model for the probability distribution of genders, and one for how prompts and replies modify an already known probability model, but here there is nothing to anchor any reasoning of this kind.