How many times can you win a game with limited chances but increasing odds after a failure?

Question

I am stumped as to how to figure this out, so hoping someone way better at statistics and probability can help!

Say you have a game where you pay 1 coin for each attempt, and your odds of winning a prize on the first attempt are 2%. If you fail, you can play again and the odds of winning go up to 8%. If you continue to fail, your odds of winning continue to go up at the increments of 12%, 25%, 40%, 50%, 75%, and finally on the 8th try they would reach 100% and you would be guaranteed to win.

My question is, how many prizes could you expect to win if you had 100 coins to spend?

Edit: Yes, the odds go back to 2% each time you win.

Answer 1

Suppose that you have a number of coins $c$ in your pocket. You can then play $c$ times, and the expected number of victory is $c$ times the chance of winning.

I am supposing that when you win the price, your odds go back to 2% on the next attempt

Denote $p = (0.02, 0.08, 0.12, 0.25, 0.40, 0.50, 0.75, 1)$

Furthermore, the number of tries before winning the price will be :

• $1$ with probability $q_1 = p_1$
• $i$ with probaility $q_i = (1-q_{i-1})*p_i$ for all $i \in {2,...,8}$

Indeed the maximum number of try before winning is 8 since on the 8th try you always win.

Denoting N this number of tries before winning (i.e $\mathbb{P}(N = i) = q_i \forall i$), the chance of winning is $\frac{1}{N}.

So the mean number of prize for each try is :

$$\sum\limits_{i} q_i \frac{1}{i}$$