bonanova Posted February 25, 2014 Report Share Posted February 25, 2014 (edited) Joe and Moe have nothing to do, so they take turns flipping a fair coin. Moe's mother interrupts their play by calling them inside for lunch. Given that Joe was the first to flip, what is the probability that Joe flipped more heads than Moe did when they stopped? When Moe's mother interrupted their play by calling them inside for lunch, one of them had flipped more heads than the other. Given that Joe was the first to flip, what is the probability that it was Joe who flipped more heads? w075 Edited February 27, 2014 by bonanova Clarified a point Quote Link to comment Share on other sites More sharing options...

0 Rainman Posted February 28, 2014 Report Share Posted February 28, 2014 I think my point still holds, only now the conditions are "magically reversed". Now if Moe was last to flip, the probability is 50% by symmetry. If Joe was last to flip, the probability depends on the number of flips. Quote Link to comment Share on other sites More sharing options...

0 Rainman Posted February 27, 2014 Report Share Posted February 27, 2014 If Joe was also last to flip, the probability is 50%. If Moe was last to flip, the probability depends on how many flips they got to. Which should mean the answer depends on the probabilistic distribution of N = number of flips. For example, is it equally likely that they flipped 10 times each as it is that they flipped a googol times each? Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted February 27, 2014 Author Report Share Posted February 27, 2014 If Joe was also last to flip, the probability is 50%. If Moe was last to flip, the probability depends on how many flips they got to. Which should mean the answer depends on the probabilistic distribution of N = number of flips. For example, is it equally likely that they flipped 10 times each as it is that they flipped a googol times each? Yes, I see your point. I think the OP was not well formed. I edited it. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted February 28, 2014 Author Report Share Posted February 28, 2014 I think my point still holds, only now the conditions are "magically reversed". Now if Moe was last to flip, the probability is 50% by symmetry. If Joe was last to flip, the probability depends on the number of flips. If ties are eliminated does not symmetry trump number of flips? That is, if "tails" was used instead of "heads" would not the answer have to be the same? Quote Link to comment Share on other sites More sharing options...

0 Rainman Posted February 28, 2014 Report Share Posted February 28, 2014 I think my point still holds, only now the conditions are "magically reversed". Now if Moe was last to flip, the probability is 50% by symmetry. If Joe was last to flip, the probability depends on the number of flips. If ties are eliminated does not symmetry trump number of flips? That is, if "tails" was used instead of "heads" would not the answer have to be the same? For symmetry to hold, you need to change "heads" to "tails" in the given condition as well. There is no guarantee that P(Joe flipped more heads|Someone flipped more heads) + P(Joe flipped more tails|Someone flipped more tails) = 1, because the given part has changed. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted February 28, 2014 Author Report Share Posted February 28, 2014 I think my point still holds, only now the conditions are "magically reversed". Now if Moe was last to flip, the probability is 50% by symmetry. If Joe was last to flip, the probability depends on the number of flips. If ties are eliminated does not symmetry trump number of flips? That is, if "tails" was used instead of "heads" would not the answer have to be the same? For symmetry to hold, you need to change "heads" to "tails" in the given condition as well. There is no guarantee that P(Joe flipped more heads|Someone flipped more heads) + P(Joe flipped more tails|Someone flipped more tails) = 1, because the given part has changed. I envisioned that removing ties made the given parts equivalent. That is, if "someone has flipped more heads" then necessarily "someone has flipped more tails." But all it means for sure is that "someone else flipped fewer heads." The counter example is if no tails have been flipped at all and Joe has flipped one more time than Moe. Then it's not true that someone has flipped more tails. I missed that point. I was contending, nevertheless, that the answer is always 50% that it was Joe. But the above distribution (e.g. HHHHH / HHHH)cannot occur for Moe [Moe never has more flips than Joe.] So the symmetry is broken, and proximity to a 50% result depends on the total number of flips being large, just as you say. The best wording would have been that after 2n+1 flips someone has more heads. What is the probability that it is Joe? Analysis: the first 2n flips expect to give J&M equal numbers of heads. Joe's last flip has a 50% chance of giving him another one. Thanks for thinking this through with me, patiently. Quote Link to comment Share on other sites More sharing options...

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## bonanova

Joe and Moe have nothing to do, so they take turns flipping a fair coin.

~~Moe's mother interrupts their play by calling them inside for lunch.~~~~Given that Joe was the first to flip, what is the probability that~~~~Joe flipped more heads than Moe did when they stopped?~~When Moe's mother interrupted their play by calling them inside for lunch, one of them had flipped more heads than the other.

Given that Joe was the first to flip, what is the probability that it was Joe who flipped more heads?

w075Edited by bonanovaClarified a point

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