Hi everyone,

This is not a homework question and it has been quite some time since I tried to work anything like this out:

I roll three 6-sided dice three times. Two of the dice are white and the third is black. When the two white dice show a total result of 9 or higher then the result of the black die is checked.

I want to calculate the odds of rolling 9 or higher on the white dice AND getting the same result on the black die 3 time in a row EXCEPT a 6 (eg getting three 1's in a row).

So I can calculate getting a result of 9 or more on two dice three times in a row and I can calculate the odds of getting the same result three times on the black die, excluding a 6. Where my brain melts is combining the two sets of probabilities.

If I understand the game correctly, to win you must roll 9 or greater three times in a row with the white dice AND have the 2nd and 3rd toss of the black die equal the first toss of the black die. Since these are independent events you can simply multiply the probabilities of each:

1. Probability of rolling 9 or greater with two dice: there are 4+3+2+1 = 10 ways to roll a 9 or greater out of 36 total possible combinations of rolling two dice. So the odds of getting 9 or greater on one roll of two dice is 10/36. The odds of rolling 9 or greater three times in a row is (10/36)^3.

2. Probability of the 2nd and 3rd roll of the black die equaling the first roll of the black die is (1/6)^2.

The total probability of winning this game is therefore (10/36)^3 x (1/6)^2 = 1000/1679616 = 0.00059537, or approximately 1 in 1680 attempts.