Two players rolling a $20$-sided die; player B can re-roll; how to decide when to reroll

Question

This is somewhat related to my earlier question What is the probability that player A rolls a larger number if player B is allowed to re-roll (20-sided die)? and somewhat related to 30 sided die and 20 sided die.

I made up this question as a follow-up to the previous one. Let's consider a game where player $A$ and $B$ tosses a 20-sided die. Player $B$ is allowed to re-toss and plays optimally. Player $B$ can observe his first toss before deciding whether to re-toss again, but he is not allowed to observe player $A$'s toss before deciding whether to re-toss. Player $A$'s toss is compared to player $B$'s LAST toss (we're NOT taking the maximum of 2 tosses by player $B$).

How does $B$ decide the threshhold at which he should re-toss for each of the following:

(1) If player $A$ tosses a strictly larger number, then player $B$ pays \$1 to player $A$. Otherwise, player $A$ pays player $B$ \$1.

(2) If player $A$ tosses a strictly larger number, then player $B$ pays \$X to player $A$, where $X$ is the value player $B$ tossed. Otherwise, player $A$ pays player $B$ \$Y, where $Y$ is the value player $B$ tossed.

For both parts, assume player $B$ plays optimally and that he wants to maximize his profit, i.e., minimize player A's profit.

For part (1), the problem of deciding whether to re-toss or not is equivalent to minimizing the probability of $A$ winning, and we find that player $B$ will re-toss if the first toss is $\leq 10$ (this was proven in my previous post, and I confirmed it via monte carlo).

At first, I thought the threshhold at which player $B$ decides to re-toss is the same for both. But it seems to not be, but it is not intuitive to me why this is not the case. I did not solve part (2) analytically yet, but Monte Carlo is telling me that $\leq 11$ is the threshhold at which player $B$ should re-toss.

I may solve this analytically tomorrow (I think perhaps the approach would be to use conditional expectation to solve this rather than only using probabilities as in part (1)), but before I do that, I was wondering if someone can give me an intuitive explanation of why (2) isn't equivalent to minimizing the probability of $A$ winning or maximizing the probability of $B$ winning?

Answer 1

The easiest way to do this is to work out B's expected profit if B sticks on a value $r$, call this $p(r)$. Since A's roll is independent of what B does, we can do this. Then if B rerolls he gets a random number with distribution $U$ which is uniform on $1,\ldots,20$, and his expected profit is therefore $E(p(U))$. We should reroll if and only if $p(r)<E(p(U))$.

There seem to be some typos, but I assume that the loser pays the winner the amount the loser rolled.

Now $p(r)=-r\times \frac{20-r}{20}+\sum_{s\leq r}\frac{s}{20}$. Calculating these (I hope correctly) in python gives:

1   -0.9
2   -1.65
3   -2.25
4   -2.7
5   -3.0
6   -3.15
7   -3.15
8   -3.0
9   -2.7
10  -2.25
11  -1.65
12  -0.9
13  0
14  1.05
15  2.25
16  3.6
17  5.1
18  6.75
19  8.55
20  10.5

The average of these values is $0.525$, so you should reroll if and only if the actual value is higher, i.e if and only if you get less than $14$.

If the payment is always what B rolled, then the function is simpler: $p(r)=r\times\frac{r}{20}-r\times\frac{20-r}{20}=\frac{10r-r^2}{10}$. The average works out to be $3.85$, which is between the values for $r=12$ and $r=13$, so reroll if you get less than $13$.