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

Fix a parameter $$0\leq\delta\leq1.$$ Bob rolls a die repeatedly in the hopes of rolling a six. However, after each failure to roll a six he gives up with probability $$1-\delta$$ and decides to try again with probability $$\delta$$. What is the probability that Bob will never roll a six?

Let $$A$$ denote the event that Bob does not roll a six, and let $$B$$ be the event that he gives up after a failure. Then $$P(A\cap B)=1-\delta$$ and $$P(A\cap B^{C})=\delta.$$ Now after the first roll, the probability that Bob did not roll a six is $$5/6$$.

I am having difficulty with understanding how the parameter $$\delta$$ comes into the calculation of the probability that Bob does not get a six given that he failed and tried again.

Thank you for time, I appreciate any feedback.

• $\delta$ comes up because Bob may quit before he gets the $6$. For instance, if $\delta =0$ then the probability he never throws a $6$ is $\frac 56$ since he is sure to give up after one failure. If $\delta =1$ then he never gives up so he'll eventually get his $6$ with probability $1$. – lulu Feb 15 at 13:51

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 $$P(A) = \sum_{k} P(K=k) P(A | K=k)$$.
Hint: Work recursively. Let $$p$$ denote the probability that Bob eventually throws a $$6$$. Then consider the first toss. Either he gets the $$6$$, in which case he stops, or he doesn't and quits, or he doesn't but he tries again (in which case the probability of success is $$p$$ again). Write that out algebraically and solve for $$p$$.