Hi Paul,
wins with all six stones still on board is not so seldom.
The following argument shows that it happens in average at least once in 1,555 games. A plays vs B and wins in 5 moves. A moves only with his central stone in the front diagonal. (So he needs this number all three times.) B's first turn should be not just the opponent of A's lonely fighter (happens with prob 5/6). B's second roll has to be with any of the at lerast two stones who can not capture the front runner.
Total probability for this is 1/6 * 5/6 * 1/6 * 2/6 * 1/6 = 5/7,776 ca= 1/1,555
There are of course other (good) paths which saves all 6 stones of the winning player. In total I think that the probability is larger than 1 promille.
Ingo.
