determine the probability of each value multiple rolls of dice. stop once same number shown N times

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

• This seems like an application of the multinomial distribution see [multinomial distribution] and [multinomial theorem] for more info. I can't see a simple closed form for this since it involves the sum of multinomial coefficients where only the term representing our desired sum has an exponent greater than or equal to $N$ and all others are less than $N$. : en.wikipedia.org/wiki/Multinomial_distribution : en.wikipedia.org/wiki/Multinomial_theorem – law-of-fives May 2 '17 at 16:12
• @RossMillikan: That seemed an appealing thought, but if you have a (usually unfair) coin that flips H with probability $p$ and T with probability $q=1-p$, you get two H first with probability $p^2(p+3q)$, and two T first with probability $q^2(q+3p)$, and those are not in the ratio $p^2$ to $q^2$. (For instance, with $p = 2/3$, the probabilities are $20/27$ and $7/27$.) This seems a rather thorny problem for $N = 8$. – Brian Tung May 3 '17 at 0:29
• @MatthewConroy: Confirmed by direct numerical analysis of the $N = 2$ case. The two probabilities are $2464012795318281/36^{11} \doteq 0.0187$, and $58197398895009288/36^{11} \doteq 0.442$, with a ratio of about $23.6$. (I'd have to write my code better to extend it to $N = 8$.) – Brian Tung May 3 '17 at 1:04
• Unfortunately, the winning numbers are not conditionally independent; that is, you cannot find the ratio of $7$ winning to $2$ winning by pretending those are the only two possibilities, and then normalizing the probabilities. Knowing that (for instance) $4$ didn't win changes the ratio of $7$ winning to $2$ winning. – Brian Tung May 3 '17 at 1:08
• @MatthewConroy: You'll go blind looking at this code; it's awful (and it's in C). If I get it into a state where it's reasonable to extend to $N = 8$, I may post it. – Brian Tung May 3 '17 at 2:06