bonanova 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 comment Share on other sites More sharing options...
0 CaptainEd Posted March 11, 2018 Report Share Posted March 11, 2018 Naive again Reveal hidden contents 12C8/4096 = .12084961 Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 11, 2018 Author Report Share Posted March 11, 2018 On 3/11/2018 at 2:57 AM, CaptainEd said: Naive again Reveal hidden contents 12C8/4096 = .12084961 Expand That was the first-cut answer in the ant-chessboard problem. Reveal hidden contents 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 Reveal hidden contents the final toss will be Heads. Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted March 11, 2018 Report Share Posted March 11, 2018 Oops! Good point! Reveal hidden contents Ok, 11C7/2048 = 0.1611328125 Quote Link to comment Share on other sites More sharing options...
0 plainglazed Posted March 12, 2018 Report Share Posted March 12, 2018 Reveal hidden contents Flipping a coin eleven times results in 11C8+11C9+11C10+11C11 or 232 trials that have 8 or more heads. There are 11C7 or 330 trials that would have exactly seven heads, half of which would have eight heads with one more flip: (11C7/2+11C8+11C9+11C10+11C11)/211=.193848 Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 12, 2018 Author Report Share Posted March 12, 2018 On 3/12/2018 at 1:29 AM, plainglazed said: Reveal hidden contents Flipping a coin eleven times results in 11C8+11C9+11C10+11C11 or 232 trials that have 8 or more heads. There are 11C7 or 330 trials that would have exactly seven heads, half of which would have eight heads with one more flip: (11C7/2+11C8+11C9+11C10+11C11)/211=.193848 Expand Reveal hidden contents 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 comment Share on other sites More sharing options...
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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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