Consider two dice. One $10$-sided, one $6$-sided, both fair. Guess a number between $2$ and $16$, if the sum of the dice shows that number, you get that number of dollars. What is the best number to guess to maximize winnings?
The answer is $11$, but I'm not sure why. I thought that since $\mathbb{E}[D_{10}] = 5.5$ and $\mathbb{E}[D_{6}] = 3.5$, the best choice would just be $5.5 + 3.5 = 9$, but I am wrong.
Can someone please explain the solution to this problem for me?
On
\begin{align} Pr(D_6 + D_{10}=x) &= \sum_{y=1}^6Pr(D_6 + D_{10}=x|D_6=y)Pr(D_6=y) \\ &= \frac16 \sum_{y=1}^6 Pr(D_{10}=x-y)\\ &= \frac16 \sum_{y=\max(1,x-10)}^{\min(x-1,6)} Pr(D_{10}=x-y) \\ &= \frac1{60}[\min(x-1,6)-\max(1,x-10)+1] \end{align}
Hence, we want to maximize
$$\frac{x}{60}[\min(x-1,6)-\max(1,x-10)+1]$$
On
The expected number of winnings when guessing a number $n$ is going to be the dollar value $n$ multiplied by the probability that the sum of the two dice is $n.$ So even though $9$ is the expected value for the sum of the dice, the expected winnings by guessing $9$ is going to be $9 \cdot \frac{1}{10} = \$0.90,$ because we can compute that the probability of the sum being $9$ is $\frac{6}{60} = \frac{1}{10}.$ Using this methodology, we can find that the highest expected winnings occur at $n=11,$ with $11 \cdot \frac{1}{10} = \$ 1.10.$
The number $9$ is the expected value $\mathbb E[D_6 + D_{10}]$, but in general that's not necessarily the likeliest value.
After all, if you were playing the same game with just a single $D_6$, you wouldn't guess $3.5$. If you did, it would take a very long time before that number was rolled.
Here, it turns out that both $9$ and $11$ are equally likely. The distribution of sums is
\begin{array}{rcccccc} 2: & (1,1) \\ 3: & (1,2) & (2,1) \\ 4: & (1,3) & (2,2) & (3,1) \\ 5: & (1,4) & (2,3) & (3,2) & (4,1) \\ 6: & (1,5) & (2,4) & (3,3) & (4,2) & (5,1) \\ 7: & (1,6) & (2,5) & (3,4) & (4,3) & (5,2) & (6,1) \\ 8: & (1,7) & (2,6) & (3,5) & (4,4) & (5,3) & (6,2) \\ 9: & (1,8) & (2,7) & (3,6) & (4,5) & (5,4) & (6,3) \\ 10: & (1,9) & (2,8) & (3,7) & (4,6) & (5,5) & (6,4) \\ 11: & (1,10) & (2,9) & (3,8) & (4,7) & (5,6) & (6,5) \\ 12: & (2,10) & (3,9) & (4,8) & (5,7) & (6,6) \\ 13: & (3,10) & (4,9) & (5,8) & (6,7) \\ 14: & (4,10) & (5,9) & (6,8) \\ 15: & (5,10) & (6,9) \\ 16: & (6,10) \end{array}
The likeliest value is not necessarily the one you should be picking.
A high value might be unlikely to win, but give you a higher reward if you win, since the number of dollars you get is the number you guessed. We should maximize the expected winnings, not the probability.
From the table above, we see that there's no point in guessing the numbers $2$ through $10$. Rather than a small number like $2$, you should guess a big number with the same probability (like $16$): it has the same chance of winning, but gives you more money if you do. Similarly, since all of $7$ through $11$ have the same chance of winning, the only one of them worth guessing is $11$.
Now we can compute the expected payoffs, multiplying the reward by the probability of getting it.
You can see that these expected payoffs keep decreasing as the guess becomes higher, and so $11$ is the best value to guess.
(Once you spot that, for values in the range $11\dots16$, the expected value is $n \cdot \frac{17-n}{60}$, you can predict this behavior: to maximize the product of two values with a fixed sum, like $n$ and $17-n$, you want to make the values as close to each other as possible.)