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--- boyorgirl [2025/12/29 15:53] – admin
+++ boyorgirl [2025/12/30 13:49] (current) – admin
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-What is the probability of the other child being a girl?
+=== What is the probability of the other child being a girl? ===
 Wikipedia: [[https://en.wikipedia.org/wiki/Boy_or_girl_paradox|Boy or girl paradox]]
@@ Line 46: / Line 46: @@
 P=7/13 = .538
-Here is where Presh Talwaker makes his mistake. He takes the boy's statement to mean $B_2B_2$ is not allowed, and drops it from his grid. 14 still have a girl and 2 remain. He says P=14/26=.519
+Here is where Presh Talwaker makes his mistake. The grid isn't helpful if the not all cases have equal probability.  Presh says "there are 27 check marks". No, $B_2B_2$ should have two check marks, just as in the classic "I am a  Catholic" statement, $BB$ would have two check marks, otherwise you would not get P=.5. In Presh's grid, 14 still have a girl. So he should have said P=14/28=.5. "and that's the answer!". Presh doesn't realize he is doing the "I am A Tuesday Boy" case. If he wants to analyze the "I am THE Tuesday Boy" result, he needs to drop two check marks, and get P=14/26=7/13=.538.
-If Mary makes a statement about one of her children, //chosen at random //, and the statement reveals the gender, and the child is a boy, I say P=1/2. But if the culture is that no public statements are allowed about girls, then BB BG and GB are equally probable.  I say P=2/3.
+== another musing ==
+Consider a gruesome culture where if the second child is a boy, born on the same weekday as a first boy, the second child is sacrificed and the couple tries again for a second child. Girls will be more prevalent in this culture. (A corollary is no B_2B_2 allowed.) If the first child is a boy, the probability of the second child being a girl is:
+P = 1/2 + (1/2)(1/7) P
+So P = 7/13
+Thus the probabilities for the two child families:
+BB 6/26
+BG 7/26
+GB 1/4
+GG 1/4
+The usual questions can be asked about these families, such as:
+"What is the probability of at least one girl?"
+"Given a boy exists, what is the probability of the other child being a girl?"
+"Given the first child is a boy, what is the probability of the second child being a girl?"
+But after 4 days of musing, I feel my gruesome scenatio doesn't really help solve the dilemma.  The statement "Given that a boy exists..." mandates that GG should be removed, but does the statement mandate that BB should be doubled?  In the classic question:
+BB 1/4
+BG 1/4
+GB 1/4
+GG 1/4
+Dropping GG requires multiplying the remaining probabilities by 4/3 ... or doubling BB. The probabilities must sum to 1. That depends on the reason for dropping GG.
+In my gruesome scenario, doubling BB isn't enough. We also need to multiply the remaining probabilities by 52/51 to make the sum 1.  What is the justification for such tidiness?  I think the choice between doubling BB or multiplying all probabilities by a constant doesn't address that a blend of such adjustments is possible, even in the classic scenario.
+== Conclusion ==
+I really should have read [[https://en.wikipedia.org/wiki/Boy_or_girl_paradox|Boy or girl paradox]] more carefully.  My epiphany is nothing original.  But thanks to the [[https://youtu.be/2Cidn_Z4Vrc?si=4l3nQZwgOB282vIF | Feynman inspiration ]], or I might not have persisted.
+Betty knows that Mary has two children.  Betty tells you Mary has at least one boy.  What is the probability that Mary has a girl?  That depends on how Betty acquired the knowledge.  Did Mary tell Betty: "I have at least one boy".  Or did Betty see Mary enrolling a boy in youth soccer, during the few minutes that Betty was doing the same? What if you can only assign probabilities to how Betty acquired the knowledge? You will need to blend the scenarios and get $\frac{1}{2} \le P \le \frac{2}{3}$.