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.

Could you please provide a hint, no solutions please, just a hint on how to start thinking about this kind of problem.

Thank you for time, I appreciate any feedback.

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    $\begingroup$ $\delta$ 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 $1$. $\endgroup$ – 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 $p$.


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