A friend of mine asked a Paul-Epstein-like problem that I didn't
know the answer to. How can you arrange 15 checkers in your
home board so as to minimize the chances that you'll leave a
blot during the bearoff? (The initial position is not allowed
to contain a blot.)

If it were 14 checkers instead of 15, then you could just stack
them all on the 1pt, and the probability of leaving a blot would
be zero. But with 15 checkers on the 1pt, you're likely to leave
a blot. In fact, if you have an odd number of checkers that are
all on the 1pt, then the only way to avoid leaving a blot is to
get down to 3 checkers and then roll doublets. So the overall
chances of not leaving a blot are less than 1/6 (since you're
not guaranteed to get down to 3 checkers).

With 5 checkers, it's easy to show that having 3 checkers on
the 1pt and 2 checkers on the 2pt is best. If you roll a single
ace (but not double aces) on your first roll, then you won't
leave a blot. Otherwise, you need to roll either 2 distinct
non-aces or double aces on your first roll, and then doublets
on your second roll. So your chances of not leaving a blot are

10/36 + (21/36)*(1/6) = 3/8.

The worst arrangement, of course, is having 2 checkers on the
1pt and 3 checkers on the 2pt. Then the only way you can avoid
blotting is if you roll snake eyes on your first shake.

---
Tim Chow

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