4 ms·
Isn't the probability still 2/3 even in the proposed alternative? The fact that we "arbitrarily announce the gender of one of the children" provides new informa
by cduan 18y ago
Isn't the probability still 2/3 even in the proposed alternative? The fact that we "arbitrarily announce the gender of one of the children" provides new information for calculating the probability, which is ignored if you assume that the original 50/50 distribution is unchanged by the announcement.
In an extreme example, consider this algorithm, analogous to the one in the article:
1. Choose a random parent that has exactly two children
2. Announce the gender of _both_ children
3. Ask about the odds that the parent has both a boy and a girl
In this situation, it would be ridiculous to say that the odds calculated at step 3 are still 50/50. Because of step 2, it is either certainly true (100%) or not true (0%) that the parent had both a boy and a girl. Granted, on average with repeated trials, half the time the step-3 value will be 100% and the other half 0%. But still, the step 2 information changes the value at step 3.
- trominos 18y agoNo. You've gained no useful information. The person who announces the gender of one child can say one of two things: "this child is a girl" or "this child is a boy." However, these two statements are symmetric in their effect on P(one child is a girl and one is a boy). There's a huge difference between announcing the gender of one child and announcing the gender of both children. If the announcer announces both children's genders, he can say that "both are boys," that "both are girls," or that "one is a boy and one is a girl." But the last statement is not symmetric to the first two in how it affects P(one child is a boy and one is a girl).