# Expected cost of eight successes where we can 'purchase' more attempts

So we want to achieve $$8$$ successes where each trial has a probability of succeeding of $$0.3$$,

$$P(S) = 0.3.$$

We have $$8$$ "free" trials, and assuming we don't get lucky and get $$8$$ successes from our free trials, we can pay $$1.00$$ dollar for a $$10\%$$ chance to regain one trial. So say we get $$3$$ successes from the free trials, then we would need to pay at least $$5$$ dollars [but expected value of $$50$$ dollars because $$50\times0.1 = 5$$ successful $$10\%$$ chances] to regain the $$5$$ unsuccessful attempts. What is our expected cost to obtain $$8$$ successes?

In my attempt I start with the expected number of successes from the $$8$$ free trials ($$n \times p = 8 \times 0.3 = 2.4$$) and then start a cycle; we have $$5,6$$ trials to be regained that we missed on the first trials, which would take an expected $$\56.00$$ (or fifty-six $$10\%$$ chances) to regain.

Then from that we get an expected ($$5.6 \times 0.3 =$$) $$1.68$$ successes and have $$4.08$$ successes total. Wash and repeat until we hit $$8$$ successes. I'm getting around $$\185.00$$ as an answer but I'm not sure of my answer and also curious if there's a better or less tedious way of doing this problem.

Any thoughts are appreciated. Thanks!

Yes there is a simpler way. $$\frac{8}{0.3} = 26.667$$ which rounds up to $$27$$ trials to get an expected $$8$$ successes. So you have to pay for another $$19$$ trials for which the expected number of $$\1$$ payments would $$\frac{19}{0.1} = 190$$ hence the expected cost would be $$\190$$.