Suppose we have the a game with a 5-sided unfair die (just to make the probabilities easier to sum to 1), each side having a different payout.
For each side $ x \in \{1,2,3,4\}$ we have the corresponding payouts $ \$1,…,\$4$. However, if the die lands in $5$ there's no payout and the game is over.
Each side, respectively, has probability $P(X=x) = \{0.15, 0.2, 0.25, 0.3, 0.1\}$
What's the expected payout for the game?
If we directly apply the expected value, we would get something like
$$\sum_x x_ip(x_i)= \$2.5 $$
However, the meaning behind calculating the expected value this way seems to only capture the expected value for the game for 1 trial.
How would one approach such a problem?
Best Answer
I would have liked to comment but I still can't... so I'll give a complete answer hoping that I'm not spoiling any homework.
I'd start saying that the "winning" sides of this five-sided unfair die are a distraction. We can re-arrange the calculations and obtain the expected value for a single roll:
$$ E_s = 0.9 \cdot \sum_{i=1}^{4}{x_i \cdot \frac{p(x_i)}{\sum_{i=1}^{4}{p(x_i)}}} + 0.1 \cdot 0 = 0.9 \cdot E_w$$
where $E_w$ is the expected value for the win under the assumption that we win (which happens with probability $0.9 = \sum_{i=1}^{4}{p(x_i)} = 1 - 0.1$).
It's like having a loaded coin, where you win $E_w$ with probability $0.9$ and get nothing otherwise.
In the "extended game" case (i.e. winning allows us to continue), if we win the first roll we are getting an expected value of $E_w$ (for the first successful roll) plus our expected value for... an undefined number of rolls, i.e. what we are after. In other terms, the expected value $E_m$ for multiple rolls will be:
$$ E_m = 0.9\cdot(E_w + E_m)$$ $$ E_m = \frac{0.9 \cdot E_w}{0.1} = \frac{E_s}{0.1} = \frac{E_s}{p(x_5)}$$
From another angle, we might observe that the number of trials to get one "success" (in this case, losing!) when repeatedly tossing our loaded coin can be modeled by a geometric distribution. Hence we might just multiply the expected value for a single toss $E_s$ by the average number of tosses needed to get this "successful failure", which is $\frac{1}{p(x_5)}$, and obtain the same result.