Fix $0\leq\delta\leq1.$ Bob rolls a die repeatedly in the hopes of rolling a six.

HINT : Let $K$ denote the random variable taking integer values, which denotes at which turn Bob will stop rolling if he has not got a six yet. For us, this is a geometric random variable with parameter $\delta$. (For $\delta = 0,1$, we can work the problem out obviously, so assume $0<\delta<1$)

With this $K=k$ fixed, we have fixed the turn at which Bob will quit if he does not roll a six by then. Conditioned on this, find the probability that Bob does not roll a six. Of course, this is just equal to the probability of him not rolling a six in $k$ turns.

Now, return to the distribution of $K$, which depends on $\delta$, to get the desired probability.

In symbols, if $A$ denotes the event that Bob quits before getting a six, then $P(A) = \sum_{k} P(K=k) P(A | K=k)$.