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.
Best Answer
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 $P(A) = \sum_{k} P(K=k) P(A | K=k)$.