Trying help my daughter with a problem.
The problem: Two dice. Each numbered 1-6. Both dice are rolled at the same time to get one of 11 values in the range 2-12. The 11 values have a different probability of occurring: Number 2 has 1/36 probability of being rolled each time (1 + 1). Number 3 has 2/36 probability of being rolled each time (1 + 2 and 2 + 1). and so on.
Keep rolling the dice until one of the values has been rolled N times. The first value to be rolled N times is the winner. For each of the 11 values work out the probability of that value being the winner.
Independent event probability means the multiplication of probabilities.
For N = 2, the probability of getting 3 twice in two rolls is (2/36 * 2/36) = 4/1296 = 1/324. Which means the probability of not getting 3 twice in two rolls is (2/36 * 34/36 + 34/36 * 34/36 + 34/36 * 2/36) = (68/1296 + 1156/1296 + 68/1296) = 1292/1296 = 323/324
But this problem allows me to keep rolling the dice until I get the same value come up N times. Still assuming N = 2, that would the maximum would be rolling all the numbers once, then rolling the 1*11 + 1 = 12. I would need to roll the dice between two and 12 times to guarantee to get two matching values. I assume I need to add up the various probabilities for rolling 2 to 12 times.
I start having trouble because I don't have/know a formula which calculates the probabilities of the different permutations for the different number of rolls.
e.g The value 3 would win if you rolled 12 times getting 2,3,4,5,6,7,8,9,12,11,10,3 probability of ((2/36)^2 * (34/36)^12), but how many permutations?
The value 3 also wins in these cases rolling only four times. 1,2,3,3 probability of ((2/36)^2 * (34/36)^2) is the same as 3,1,2,3 and so on
But value 3 doesn't win in this case 3,1,1,3, so I can't just apply ((2/36)^2 * (34/36)^2) * NumberOfPermutations.
Ultimately want to solve for N=8