For the example you just gave the distribution would negative binomial.

X~negative binomial(k=3, p=\(\displaystyle \frac{1}{6}\))

\(\displaystyle pr(X=8){x-1\choose k-1}(p)^{k}(p-1)^{(x-k)}\)

For this one, you have to consider all eight rolls because no "sixes" have happened yet. If we take this same problem that you just posted and say that George has already rolled 1 "six" and you want to find the probability that he will win on his 8th roll then we'd be looking for:

X~negative binomial(k=**2**, p=\(\displaystyle \frac{1}{6}\))

\(\displaystyle pr(X=7){x-1\choose k-1}(p)^{k}(p-1)^{(x-k)}\)

Because 1 six has already occurred.

If we take this same problem that you just posted and say that George has already rolled 1 time and you want to find the probability that he will win on his 8th roll then we'd be looking for:

X~negative binomial(k=**3**, p=\(\displaystyle \frac{1}{6}\))

\(\displaystyle pr(X=7){x-1\choose k-1}(p)^{k}(p-1)^{(x-k)}\)

Because 1 roll has already occurred.

I guess I'm just having a hard time explaining this for you. Sorry