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Probability of hitting a checker from a long running start

Posted By: ah_clem
Date: Sunday, 6 July 2025, at 3:51 p.m.

In Response To: Probability of hitting a checker from a long running start (Jason Lee)

I spent a few minutes playing with this approach in google sheets, going up to n==12. Here's what I get

n p(n)
1 0.3055555556
2 0.3333333333
3 0.3888888889
4 0.4336419753
5 0.4606481481
6 0.5648148148
7 0.2893518519
8 0.3808727709
9 0.4301268861
10 0.4363854595
11 0.4444492074
12 0.507958962

I'm guessing it settles in around 45% or thereabouts. If I'm going to spend any more time on it I'll code it up is Python instead of making spreadsheet formulas by hand. Should be fairly straightforward to compute the first hundred thousand or so values of p(n).

OTOH, it might not converge - looks like there may be "spikes" at multiples of 6, but this probably smooths out for sufficiently large n.

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