How would I calculate a combination of the Binomial and Geometric Distributions?

Question

To be specific with my problem, I'm calculating a formula for a game. There's 6 independent trials, each with an independent probability of success = 0.34. I know that from the Geometric distribution, it'll take about 3 trials before I get a success, so should expect about 2 successes for every 6 trials?

However, my issue is that anytime I get a success, I get another trial, so if I get 2 successes in 6 trials, then I can get 2 more trials, and I'm having trouble determining the amount of successes I can expect on average, although I have a feeling it's around 3.

Answer 1

The way the game is structured, a success, in effect, doesn't count against the six original trials, because you get a new trial with each success. Therefore you are really playing the game until you get six failures.

The appropriate distribution for the number of successes before some fixed number of failures from i.i.d. Bernoulli trials is the negative binomial distribution. The shape parameter $r$ would in this case be equal to six and the probability parameter $p = 0.34$ (in the parameterization on the Wikipedia page). The expected value of the number of successes is $rp/(1-p)$, which equals 3.09, pretty close to your intuition!