Player 1 has m checkers on the bar and a free points in the opponent's home board to enter.
Player 2 has n checkers on the bar  and b free points to enter.
P[m,a,n,b] is the probability that player 1 enter all her checkers before player 2, when she has first roll.
P[n,b,m,a] is the probability that player 2 enter all her checkers before player 1, when she has first roll.
1<=a<=5 and 1<=b<=5
Constraints: 1<=m<=4 and 1<=n<=4

P[m,a,n,b] = ? for any values of m,a,n,b

There are the following possibilities:
a) player 1 roll a doublet and enter all her checkers, with probability = a/36
b) player 1 roll a non-doublet and enter 2 of her checkers, with probability = 2*C(a,2)/36 = A(a,2)/36,
       where C(a,2) is the number of combinations of 2 taken from a set of a objects and A(a,2) is the number of permutations of 2 taken from a set of a objects, A(a,2) = a*(a-1)
    Now the probability that player 1 enter all her checkers first is 1 - P[n,b,m-2,a].
c) player 1 roll a non-doublet and enter 1 checker, with probability = ((36-(6-a)²)-a-A(a,2))/36
= (a(12-a)-a-A(a,2))/36
    Now the probability that player 1 enter all her checkers first is 1 - P[n,b,m-1,a].
d) player 1 dance this roll, with probability (6-a)²/36
    Now the probability that player 1 enter all her checkers first is 1 - P[n,b,m,a].

P[m,a,n,b] = 0 if n<1.

It results the first equation:
   a + A(a,2)*(1-P[n,b,m-2,a]) + (a*(12-a)-a-A(a,2))*(1-P[n,b,m-1,a]) + (6-a)²*(1 - P[n,b,m,a])
 = 36*P[m,a,n,b]

When player 2 roll first, it results the second equation:
    b + A(b,2)*(1-P[m,a,n-2,b]) + (b*(12-b)-b-A(b,2))*(1-P[m,a,n-1,b]) + (6-b)²*(1 - P[m,a,n,b])
 = 36*P[n,b,m,a]

The unknown values for this linear system are P[m,a,n,b] and P[n,b,m,a], if we calculate first positions with fewer checkers on the bar, that is we calculate first for max(m,n)=1, then for max(m,n)=2 and so on 

I attached the files probabilities.java and results.html (with extensions changed)