peteb said:
Your new mathematics-obsessed friend says to you, "I have two children.
One is a boy born on Tuesday. What is the probability I have two boys?"
It's 1/2. And no, I am not missing a subtle point about probability. You are.
davee123 said:
Tuesday does indeed have nothing to do with it, assuming he's truly anal, or without further information. Hence, the probability, assuming his statements are true, is 1/3.
And the answer to the problem you reduced it to, by omitting "Tuesday" from peteb's question, is also 1/2.
D H said:
Everything indeed. There is a big difference between
- "I have two children. One is a boy."
- "I have two children. The older of the two is a boy."
- "I have two children. One is a boy born on Tuesday."
Most of us are pretty lousy at dealing with conditional probabilities. Many people, even many otherwise very intelligent people get the Monty Hall problem wrong, and many apparently got this one wrong as well.
You are right in your assessment at the end, but you need to add more to your list.
- A: "I have two children." B:"Is one of them is a boy?" A: "Yes."
- A: "I have two children." B:"Is one of them is a boy born on a Tuesday?" A: "Yes."
The answer, when these two are the condition, is 1/3 and 13/27, respectively. The answer when any of yours is the condition is 1/2. The reason they are different, is that my condition #1 is different than your condition #1, and my #2 is different than your #3.
There could be some people who would respond "Yes" to my #1, but who, for whatever unspecified reason they are telling you what they told you in your list, would not say "One is a boy." They would say "One is a girl." Since you provided no reason for why they would choose one statement over the other, we can only assume both are equally likely when both are possible. So, while it is true that there are twice as many parents of two who have a boy and a girl, as those who have two boys; the same number in each group would tell you "One is a boy." The same arguments apply to the Tuesday version.
And the really ironic thing, is that the people who get the wrong answer for the Monty Hall Problem you mentioned are taking the same approach to it, as you do to this one. As Fox and Levav write, they are "subjectively partitioning the sample space into n interchangeable events, editing out events that can be eliminated on the basis of conditioning information, counting remaining events, then reporting probabilities as a ratio of the number of focal to total events."
So, for your first condition, you partition the sample space into the four interchangeable events BB, BG, GB, and GG; edit out the GG event which is the only one that is impossible if you have a boy; and report the probability as the ratio of focal event left (BB) to total events left (BB, BG, GB). Just like with Monty Hall they partition the sample space into the three interchangeable events where the prize is behind D1, D2, or D3; edit out the D3 (or whatever) event which is the only one that is impossible after you see what door the host opens; and report the probability as the ratio of focal event left (D1) to total events left (D1, D2).
The technically-correct answer is not the ratio of the number of cases, but the ratio of the sum of the probabilities that each of the appropriate cases would produce the observed result. In the Two Child Problem, if P is the probability a parent of a boy and a girl would say "One is a boy," the answer is 1/(1+2P). In the Monty Hall Problem, if P is the probability Monty Hall would open Door #3 when he could also open Door #2, the answer is 1/(1+P). And in the Tuesday Boy problem, the answer is (1+12P)/(1+26P) for a similar P. If P=1 in any of these, you get the "wrong" answer that Fox and Levav talk about. If P=1/2, you get the right answer.
The difference in my list of statements, is that P=1. In yours, it is really ambiguous what P is, but we can't assume anything except 1/2.
peteb said:
Thus the intuitive component is found in recognizing that it is the presence of any conditional of this type that alters the result, not just the one specific conditional mentioned in this example.
No. It is
requiring the presence of the conditional information that alters the probability. If a parent is selected at random from a group of parents who have two children and at least one boy, the answer is 1/3. If a parent is selected at random from a group of parents who have two children, and
one gender is observed, the answer is 1/2 regardless of what that gender is.
davee123 said:
"The boy in question was born on a particular day of the week. I haven't told you which day that is, and it goes without saying that of course he was born on a particular day. But in just a second, I'm going to tell you which day that was, but I haven't yet."
Probability of two boys is still 1/3.
"The boy was born on a Tuesday".
Probability of two boys is now 13/27.
If this were true, then the answer would be 13/27 regardless of what day you mention. And if it is 13/27 regardless of what day you mention, it is also 13/27 for the first question. This is called the Law of Total Probability. If {X1, X2, ..., XN} are independent events that represent all possibilities, then for any event Y, P(Y)=P(Y|X1)*P(X1)+P(Y|X2)*P(X1)+...+P(Y|XN)*P(XN). If all the conditional probabilities are equal to the same P, then P(Y)=P because P(X1)+P(X2)+...+P(XN)=1.