bonanova 84 Posted March 7, 2018 Report Share Posted March 7, 2018 Alternate approach to the ant-checkerboard problem asks: What is the probability of flipping 8 heads before 5 tails? Quote Link to post Share on other sites

0 Solution plainglazed 67 Posted March 12, 2018 Solution Report Share Posted March 12, 2018 Spoiler Flipping a coin eleven times results in _{11}C_{8}+_{11}C_{9}+_{11}C_{10}+_{11}C_{11 }or 232 trials that have 8 or more heads. There are _{11}C_{7} or 330 trials that would have exactly seven heads, half of which would have eight heads with one more flip: (_{11}C_{7}/2+_{11}C_{8}+_{11}C_{9}+_{11}C_{10}+_{11}C_{11})/2^{11}=.193848 Quote Link to post Share on other sites

0 CaptainEd 26 Posted March 11, 2018 Report Share Posted March 11, 2018 Naive again 12C8/4096 = .12084961 Quote Link to post Share on other sites

0 bonanova 84 Posted March 11, 2018 Author Report Share Posted March 11, 2018 8 hours ago, CaptainEd said: Naive again Hide contents 12C8/4096 = .12084961 That was the first-cut answer in the ant-chessboard problem. Spoiler The value is too low. It counts only once the (multiple) cases where 8 heads is reached before the "full complement" of 4 tails occurs. The solution is easier to find if we (reasonably) assume that Spoiler the final toss will be Heads. Quote Link to post Share on other sites

0 CaptainEd 26 Posted March 11, 2018 Report Share Posted March 11, 2018 Oops! Good point! Ok, 11C7/2048 = 0.1611328125 Quote Link to post Share on other sites

0 bonanova 84 Posted March 12, 2018 Author Report Share Posted March 12, 2018 17 hours ago, plainglazed said: Hide contents Flipping a coin eleven times results in _{11}C_{8}+_{11}C_{9}+_{11}C_{10}+_{11}C_{11 }or 232 trials that have 8 or more heads. There are _{11}C_{7} or 330 trials that would have exactly seven heads, half of which would have eight heads with one more flip: (_{11}C_{7}/2+_{11}C_{8}+_{11}C_{9}+_{11}C_{10}+_{11}C_{11})/2^{11}=.193848 Spoiler Equates with the 3x speed version of ants on a checkerboard:probability of reaching either of the ends of the five meeting points = 794 / 4096 Quote Link to post Share on other sites

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Alternate approach to the ant-checkerboard problem asks: What is the probability of flipping 8 heads before 5 tails?

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