Since all the checkers are at 1-point, the players have only one legal move per dice throw - if it is a double they can bear-off four checkers; otherwise they bear-off two. We start with the simpler version of a single player by defining f(n) to be the probability of the last throw being a double when we start with n checkers.
Clearly: f(1)=f(2)=1/6; f(3)=f(4)=11/36 and f(n)=1/6*f(n-4)+5/6*f(n-2).
The solution for this recursion formula is f(2n-1)=f(2n)=(2+5*(-1/6)^n)/7.
The recursion formula for two players in the game is a little more complicated:
f(n,m)=1/6*f(m,n-4)+5/6*f(m,n-2) (with similar boundary conditions)
and its solution is f(15,15)=~0.3417.
Note that for the full game, the answer would be even larger, since there are, for example, situations where one may bear off four checkers with a 6-6 throw and none with 3-4 throw.
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