Yes - you got it right.

You can think this through by considering that on any throw the probability of a six is 1/6, and the probability of anything but six is 1 - 1/6 = 5/6. Hence, the probability of getting a 6 on the first throw and no more sixes for the next 5 throws is: (1/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6). But you can also win this game by rolling the 6 on the second throw instead of te hfirst, or the the third, forth, fifth, of sixth. The probability of any one of these occurring is the same as above. So give that anyone of these 6 outcomes is a winner, the total probability is: 6*(1/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6).

In math terms, for independent events like throwing a dice, if the probability of something occuring isp, then the probability of that thing occurringktimes out ofnthrows is equal to:

Where C(n,k) means the combination ofnthings takenkat a time, which is equal to:

So for your problem you have: