Expected rolls to get n result k times non-consecutively

I've searched through several questions trying to get the answer to this but been unable to find the answer I need. Most questions deal with consecutive results, but I am looking for non-consecutive results.

Basically, what is the formula needed to calculate the expected number of rolls needed to roll the number 6 n times. For instance, if a game requires me to roll the number 6 five times, how many times can I expect to roll the die before I get the result of 6 five times? They do not need to be consecutive.

To add difficulty to the problem, suppose I have a six sided die and a 10 sided die. What would be the formula to find the expected number of rolls to achieve a the number 6 five times on each die. You would roll both dice at once and record the results. If one die reaches the result five times before the other, it doesn't matter and you continue to roll both dice until each has achieved the result five times.

I'm still somewhat new to working out probabilities and having difficulty with this one.

• Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. – gung - Reinstate Monica Oct 4 '16 at 23:39
• @Carl, please don't add the [self-study] tag for the OP. Ask them to add it themselves & to read its wiki. That way there is a better chance they will be familiar w/ our policies. If they don't add the tag, we can close the question. – gung - Reinstate Monica Oct 4 '16 at 23:41

1 Answer

To answer your first question, let $X_i$ be the number of rolls it takes you to get your first 6. Then $X_i$ is geometric with parameter (1/6) and has an EV of 6. You're asking for the sum of running this experiment 5 times, i.e. convince yourself that drawing $X_i$ 5 times and adding would also give you precisely one realization of $Y_i$ the number of times it takes you to hit 6 five times. So your answer is 6*5=30. In particular, your random variable is Negative Binomial, but shifted to the right by 5 since the Wikipedia definition involves the 'failures' before you stop, not counting that your 5 successful rolls of 6 count.

To answer your second question, you are looking for the distribution of the maximum of two negative binomials (plus 5); the 6-sided dice being $X \sim NB(r=5,p=5/6)$ and the 10-sided dice second being $Y \sim NB(r=5,p=9/10)$. This is a lot trickier to do by pencil and the way I would proceed is to use the formula $$E[max(X,Y)]=\sum_{k=1}^\infty P(X\leq k, Y \leq k)=\\ \sum_{k=1}^\infty P(X\leq k) P(Y \leq k)$$ due to independence of X and Y, and taking the cdf of the negative binomial for $P(X \leq k)$ and for Y.