Suppose you roll a fair dice with 6 sides and you start with 0 money. Every time you roll 1, 2 or 3 you earn 1 dollar, but if you roll 4 or 5 the game ends and you get paid what you have earned and finally, if you roll a 6 game ends and you get paid 0.
The expected payoff is $\frac{2}{3}$ – calculated using the definition of expectation $\mathbb{E}[X]=\sum_{n=1}^{n=\infty}n\mathbb{P}(X=n)$. However, I am unsure whether the following solution works and would very much appreciate if someone can check it.
$\mathbb{P}(\text{game ends on 4,5})=2/3$ and $\mathbb{P}(\text{game ends on 6})=1/3$ by symmetry.
$$\mathbb{E}[X]=\mathbb{E}[X|\text{game ends on 4,5}]\mathbb{P}(\text{game ends on 4,5}) + \mathbb{E}[X|\text{game ends on 6}]\mathbb{P}(\text{game ends on 6})$$
$\mathbb{E}[X|\text{game ends on 6}]=0$ because we don't earn anything if game ends on 6 so
$$\mathbb{E}[X]=\mathbb{E}[X|\text{game ends on 4,5}]\cdot \mathbb{P}(\text{game ends on 4,5})$$
Now, let A and C be any events. Then $\mathbb{P}(C|A)=\frac{\mathbb{P}(C)}{\mathbb{P}(A)}\mathbb{P}(A|C)$ by the Bayes' formula.
$\mathbb{P}(\text{game ends on 4,5}|\text{we roll 1, 2 or 3 in next roll})=\mathbb{P}(\text{game ends on 4,5})$ since conditioning on getting 1, 2 or 3 we don't really gain any information about what the game ends on.
Therefore, $$\mathbb{P}(\text{we roll 1, 2 or 3 in next roll}|\text{game ends on 4,5})=\mathbb{P}(\text{we roll 1, 2 or 3 in next roll})=1/2$$
Therefore, let $\mathbb{E}[X|\text{game ends on 4,5}]=\mu$.
$$\mu=\mathbb{P}(\text{we roll 1, 2 or 3 in next roll}|\text{game ends on 4,5})(1+\mu)+0\cdot \mathbb{P}(\text{we roll 4,5}|\text{game ends on 4,5}).$$
Thus, we get that $\mu=1/2 \cdot (1+\mu)$ so $\mu=1$.
Finally, $$\mathbb{E}[X]=2/3 \cdot \mu=2/3.$$
Best Answer
Yes. You are correct. Here is another way to compute the expected pay-off. Let $X$ denote the total earnings before the game ends. This is not same as expected pay-off in the sense, even if game ends with a $6$ or $4, 5$ this earning $X$ remains the same.
$$\ E(X) = E(G) - 1 = 1 \text{ where } G \text{ denotes the rounds played} $$
$$\ E(\text{pay-off}) = \frac{2}{3}E(X) + \frac{1}{3}\times0 = \frac{2}{3}E(X) $$
This is because the game may end with a $4, 5 \implies P= \frac{2}{3}$ or with $6 \implies P = \frac{1}{3}$