Heads up

[bonanova 83]

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

[plainglazed 67]

Spoiler

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

 

[CaptainEd 23]

Naive again

12C8/4096 = .12084961

[bonanova 83]

8 hours ago, CaptainEd said:

Naive again

 

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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.

 

[CaptainEd 23]

Oops! Good point!

Ok, 11C7/2048 = 0.1611328125

[bonanova 83]

17 hours ago, plainglazed said:

  Hide 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

 

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