There are two basic forms that the dealer could throw: a, a or a, b. His chance of getting a, a is 1/6 and of getting a, b is 5/6.

Consider a, a first. Let's say he threw two 6's. So the player could throw two sixes first (1/36) or one six then another six (10/36*1/6) or no sixes then two sixes (25/36*1/36) Adding these possibilities gives 121/1296. The dealer had a 1/6 chance of getting the a, a form, so the total probability would be 121/7776.

Now a, b. Say the dealer threw 6, 5. The player could throw 6, 5 or 5, 6 first (2*1/36) or 6 first then 5 or 5 first then 6 (2*9/36*1/6) or throw neither first and both second (16/36*2*1/36). Adding these gives 53/324, but the dealer had a 5/6 chance of getting this so the total chance would be 265/1944.

The chance of winning the whole game is therefore 121/7776+265/1944, which is 1181/7776. I think...