I've written below my answer for the following expectation question:
Two tetrahedral dice (red & blue) are rolled in each turn. If blue shows more than red, you are paid the difference, otherwise paid nothing. If the blue and red dice show the same value, the game is immediately over, otherwise, you take another turn. What is the fair value of the game?
Let $X$ be the expected value of the game as described above, $t$ to be short for "tie", $b>r$ to be short for "blue roll is greater than red roll" and similarly $r>b$. We know that $X = P(t)V_{t} + P(b>r)V_{(b>r)} + P(r>b)V_{(r>b)}$
There are 4 situations in which red and blue roll the same number and therefore $P(t) = \frac{1}{4}$. Since the game is over when this happens $V_t = 0$.
We then enumerate the situations in which blue is greater than red. There are specifically 6 situations out 16 in which this happens:
| Red rolled: | Blue rolled: |
|---|---|
| 1 | 2, 3, 4 |
| 2 | 3, 4 |
| 3 | 4 |
| 4 | - |
So $P(b>r) = \frac{6}{16} = \frac{3}{8}$. To find $V_{b>r}$, we first find the expected winning in one round (given $b>r$) by calculating the difference between dices $\times$ the probability they happen. Given that $b>r$, there is a $\frac{1}{6}$ chance for each possibility in the above table. We calculate $\frac{1}{6}((4-1)+(3-1)+(2-1)+(4-2)+(3-2)+(4-3)) = \frac{1}{6}(3+2+1+2+1+1) = \frac{10}{6} = \frac{5}{3}$. In addition to the $\frac{5}{3}$, $V_{b>r}$ also embodies possible winning in the next round, so $V_{b>r} = \frac{5}{3} + X$.
By subtracting to find the probability complement, we get $P(r>b) = (1 - P(b>r) - P(t)) = \frac{3}{8} = P(b>r)$. Since when $r>b$ happens we just roll a gain, $V_{r>b}$ is just $X$.
Putting everything together, we have $X = P(t)V_{t} + P(b>r)V_{(b>r)} + P(r>b)V_{(r>b)} = \frac{1}{4}(0) + \frac{3}{8}(\frac{5}{3}+X) + \frac{3}{8}X$. Combining like terms to solve for $X$ we have $X = \frac{5}{8} + \frac{6}{8}X \Rightarrow \frac{1}{4}X = \frac{5}{8} \Rightarrow X = \frac{5}{2} $.
So the fair value of the game is 2.5. Did I miss something? I solved this question completely on my own and have no idea if this line of solution would be correct. Any tips would be appreciated. Thanks!
On each roll:
$\begin{array}{} Winnings & Probability\\ 3 & \frac {1}{16}\\ 2 & \frac {2}{16}\\ 1 & \frac {3}{16}\\ 0 & \frac {10}{16} \end {array}$
With each roll your expectation is $\frac {3*1+2*2+1*3}{16} = \frac {10}{16}$
The game has a probability of ending of $\frac {4}{16} = \frac {1}{4}$ on any roll.
We expect the game to last for 4 rolls.
$4*\frac {10}{16} = \frac {5}{2}$