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May 14, 2013 at 20:58 comment added user112109 @Douglas Zare: What you say is true: you get paid B-G. When playing ten rounds, always stopping if boy/black shows up, then according to your prediction you win in 52 % and lose in 48 % of all cases. Or would you like to revise your prediction?
May 14, 2013 at 17:04 comment added Douglas Zare @Rhett Butler: Here is a difference. When you bet on roulette, you don't get paid $B/(B+G)$. If you make bet $1$ at each step, you get paid $B-G$, say. Your payoff is a martingale. You do not get paid $B/(B+G)$. $B/(B+G)$ is not a martingale. Stop pretending that recognizing the fact that $B/(B+G)$ does not have expected value $1/2$ under some stopping rules means that we have a roulette strategy which wins on average. Your statement that "my" strategy (which is not mine) must win at roulette is mathematically wrong. It does not become right by spamming it or being increasingly rude.
May 14, 2013 at 15:56 comment added user112109 @Douglas Zare: What is the difference between sequences of bits in form of boys and girls, interrupted at "boy" and started new at an arbitrary point, and bits of black and red interrupted at "black" and started new at an arbitrary point? If you can find a reasonable explanation, I will not hesitate to apologize.
May 13, 2013 at 22:11 comment added Douglas Zare @Rhett Butler: That is clearly false and insulting. I hope that when you realize your mistake, you post apologies with the enthusiasm you are using to spam your errors and lies.
May 13, 2013 at 17:33 comment added user112109 @:Qiaochu: Douglas' receipe could be used to play roulette, always betting on black and, after stopping (for a while), starting a new sequence. If the chance of red for the current single sequence is less than 31 % in the average, even the sporadic appearance of the zero could not hinder you to get a rich man.
May 13, 2013 at 17:28 comment added user112109 @rgrig: But it lets us hope.
Jul 11, 2010 at 18:06 comment added T.. @DZ: $k=1$ is no counterexample unless one defines population (the set of births for which G/(G+B) is calculated) to be a set of families that have reached the STOP state. It's quite clear that the one-completed-family (B,G) distributions can be asymmetrical under a stopping rule, that this asymmetry can be reflected in the proportion of girls, and that it is dampened (the girl proportion tends to 1/2) when convolving many such distributions. What is not clear is whether any bias exists in models not of that form.
Jul 11, 2010 at 11:34 comment added Douglas Zare @T: You intuition says something about k families, but your unsupported conclusion is incorrect for k=1, and for other finite values of k, just less obviously. Your intuition is wrong. You call k=1 a trivial calculation, but it's a counterexample to your statements. Your argument for the expected value of the weighted average is wrong, since the weighting depends on the variables. Again, only the magnitude of the bias depends on the precise formalization, not its existence, and you can compute the bias if you correctly analyze even very simple models.
Jul 11, 2010 at 6:27 comment added T.. @Doug: you did not calculate E[G/(G+B)] for the population (where it is 1/2) but for one family that completes its reproduction (where it is 3/8). The population is not a union of $k$ completed families and is not reasonably approximated by such. The argument is not about how to perform trivial expected value calculations, as I think was clear.
Jul 11, 2010 at 4:40 comment added Douglas Zare @T: You are wrong. 1/2 B: 0% girls. 1/4 GB: 50% girls. 1/4 GG: 100% girls. 1/2 * 0 + 1/4 * 1/2 + 1/4 * 1 = 3/8. I thought that there wouldn't be any argument about the expected value of G/(G+B) in that example, which illustrates what is going on in the more complicated actual example of a finite number of families.
Jul 10, 2010 at 5:22 comment added T.. The expected value of G/(G+B) is 1/2, not 3/8, under that stopping rule for a population in the ordinary sense of that term, i.e., one where the set of families is changing over time and without an artificial exclusion of incomplete families. By symmetry, E[G/(G+B)]=1/2 for any set of births that is defined without reference to concepts that (given the stopping rule) break gender symmetry, such as boys, girls, completed families, or siblinghood. This includes "all people born since 1920" but not "eldest children" or your definition of population "all children from completed families".
Jul 10, 2010 at 4:29 comment added Douglas Zare @T: G/(G+B) is not a martingale, and for some stopping rules where the population size is not fixed, the expected value is not 1/2. Here is a much simplified example: Have 1 or 2 children, and only have a second child if the first child is a girl. 1/2 chance of B, 1/4 GB, and 1/4 GG. E[G/(G+B)] = 3/8.
Jul 9, 2010 at 17:22 comment added T.. That's not a problem, because for the same reason that $(B - G)$ is a martingale, $G/(G+B)$ has expectation 1/2, exactly and for any population size.
Mar 25, 2010 at 21:20 comment added Qiaochu Yuan Douglas has already pointed out in a comment to vonjd's answer that the proportion of girls in the population is not a martingale.
Mar 23, 2010 at 21:36 comment added rgrig It's amazing how many people post wrong solutions after the correct one has been up-voted so many times that it's hard to miss.
Mar 23, 2010 at 21:25 history answered Jay CC BY-SA 2.5