There's this question: You roll a dice three times, what's the probability of getting 2 "6" in a row?
I have a solution for the problem that says: either getting 2 six in a row (first pair) or getting 2 six in a row (last pair), or getting 3 six in a row.
Since the first two cases get: 1/6 * 1/6 * 5/6 * 2 Then the last case: 1/6 * 1/6 * 1/6 This gives a probability of 11/216.
My first try on this exercice I wanted to use combinatorics. My intuition led me to calculate all the possible outcomes which is 6^3 = 216 (6 numbers possible, 3 times) This gives the denominator. Yet I've been stuck to find the number of favorable outcomes (nominator) using combinatorics. I understand that it is: (6,6, 1to5) which gives 5 combinations or (1to5, 6,6) which gives again 5 combinations or (6,6,6) which is 1 combination
this gives a probability of 11/216.
But I can't find a mathematical formula that fits, and in a case where the combinations are too big, I wouldn't have found it. Can someone tell me the way to solve this using combinatorics formula for any n and p