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--- boyorgirl [2025/12/28 16:04] – created 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]]
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 This is what inspired me to stick with it:
 Youtube: [[https://youtu.be/2Cidn_Z4Vrc?si=4l3nQZwgOB282vIF | Feynman inspiration ]]
+Let P be the probability of at least one of two children being a girl.
 Mary has two children, if that is all you know, these families are equally probable:
-BB BG GB GG
+BB BG GB GG. Thus P = 3/4.
-The probability of having at least one girl is 3/4.
 If Mary tells you she has at least one boy, these are equally likely:
-BB BG GB
+BB BG GB, so
-The probability of having at least one girl is 2/3.
+P = 2/3.
 If Mary tells you her first child is boy, these are equally likely:
-BB BG
+BB BG, so
-The probability of having at least one girl is 1/2.
+P is 1/2.
-If Mary doesn't say anything, but you see Mary walking with one of her children, chosen at random,
+Suppose the culture is that, at a certain hour, mothers flip a coin to choose a child to take out for a walk. You will see half the children are boys, half are girls. You see Mary with a boy. So Mary's family must be one of:
-her family must be one of:
+BB BG GB.
-BB BG GB
 But the probability of choosing a B to walk with is TWICE if BB.  BB BG GB are not equally probable, BB is twice as probable, if all you observe is the walk.
-The probability of having at least on girl is 1/2.
+P=1/2.
+But if Mary told you, prior to you observing the walk, that she has at least one boy, P=2/3. What if you subsequently observe a walk with a girl? P=1, obviously.
+What if you subsequently observe Mary walking a boy? P=1/2 or P=2/3?  I say P=1/2. Mary's prior
+statement is a corollary of this new observation.
+Consider the culture that a boy born on a Tuesday is considered special. These boys brag about being a "Tuesday Boy".
+You observe Mary walking a boy. P=1/2. The boy then says to you one of:
+  * "the Pope is Catholic"
+  * "I am Catholic"
+  * "I am a Tuesday Boy"
+P does not change, P=1/2.
+But what if the boy says "I am THE Tuesday Boy"?
+This information excludes the possibility of the sibling being another Tuesday Boy. Instead of 14 possibilities for the sibling, there are now 13 equally probable siblings, and 7 are girls.
+P=7/13 = .538
+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.
+== 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.
-But is Mary tells you that she has at least one boy, that could happen if she is walking either a girl or boy. That is more information than simply observing one child she is walking with.
+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 $\frac{1}{2} \le P \le \frac{2}{3}$.