(Sorry if the question is badly titled, I had a hard time to find a concise yet correct way of expressing the situation I'm in)
Hello,
Let's say I have a game in which I must roll a value between 1 and 100 (inclusive). If I roll 75 or less, I have a gain of 25; and I roll again the value with the same threshold to win the gain. I can go up to 3 rolls for a total maximum gain of 75. If I roll more than 75, I gain nothing more (but I keep what I earned with the precedent rolls) and can't roll anymore. Each roll is equally probable
I would like to compute the expected value of such a game, and my problem is that I have the method of doing this manually, but not the formula
In this game, there are 4 outcomes possibles
| 1st roll | 2nd roll | 3rd roll |
|---|---|---|
| S | S | S |
| S | S | F |
| S | F | F |
| F | F | F |
For each roll, probabilities are as the following
| 1st roll | 2nd roll | 3rd roll | P(outcome) |
|---|---|---|---|
| 75% | 75% | 75% | 42% |
| 75% | 75% | 25% | 14% |
| 75% | 25% | 100% | 19% |
| 25% | 100% | 100% | 25% |
I assumed that, since you can't roll again if you fail, we could say that your subsequent probability of failing for the next roll is 100%
Since I have the probability and the gain of every outcome, I can thus compute the expected value, which should be (I if made no mistake), about 43.36 ($\frac{P(outcome_1)\times75 + P(outcome_2)\times50 + P(outcome_3)\times25 + P(outcome_4)\times0}{\sum_{i=1}^nP(outcome_n)}$)
With all that said, what I would like is to find the formula allowing me to compute the expected value from the initial data, rather than having to pass through all theses intermediate steps
Thanks, Meta
Let $p = 3/4, q = 1 - p.$
Chance of return of $0 = q.$
Chance of return of $25 = pq.$
Chance of return of $50 = p^2q.$
Chance of return of $75 = p^3.$
Overall computation is
$$q(0 + 25p + 50p^2) + 75p^3.$$