# 20 and 30 Sided Dice Game

Two people, A and B, have a 30-sided and 20-sided die, respectively. Each rolls their die, and the person with the highest roll wins. (B also wins in the event of a tie.) The loser pays the winner the value on the winner's die.

1)What is expected value for player A?

2)How does the expected value of the game for player A change when player B can re-roll (AFTER seeing player A’s roll)?

My struggle with this question comes in part 2. As player B will only reroll in the case when A's number is higher, here is my calculation for A's expected value.

For $$\frac{10}{30}$$ of the cases, A's roll will be between 21 and 30 and hence automatically higher. For the remaining $$\frac{20}{30}$$ cases, $$\frac{210}{400}$$ times B will not reroll as B would have either higher or tied score to A. Finally, in the remaining $$\frac{190}{400}$$ cases, B will have a lower score, and will hence reroll. Here, the probability of B's rerolled score being higher than A's score should be $$\frac{1}{3}$$, as it represents the ordering B2>A1>B1. Putting this all together, I get:

E(A) = $$\frac{10}{30}(\frac{21+30}{2}) + \frac{20}{30}((\frac{210}{400})(\frac{-21}{2}) + \frac{190}{400}(\frac{2}{3}(\frac{21}{2})-\frac{1}{3}(\frac{21}{2})) = 5.933$$

However, the answer appears to be 5.4725. What is wrong in my approach, or am I misinterpreting the problem?

From later duplicate question

For $$\frac{10}{30}$$ of the cases, A's roll will be between 21 and 30 and hence automatically higher. For the remaining $$\frac{20}{30}$$ cases, $$\frac{210}{400}$$ times B will not reroll as B would have either higher or tied score to A. Finally, in the remaining $$\frac{190}{400}$$ cases, B will have a lower score, and will hence reroll. Here, the probability of B's rerolled score being higher or equal to A's score should be $$\frac{210}{400}$$. Putting this all together, I get:

E(A) = $$\frac{10}{30}(\frac{21+30}{2}) + \frac{20}{30}((\frac{210}{400})(\frac{-21}{2}) + \frac{190}{400}(\frac{190}{400}(\frac{21}{2})-\frac{210}{400}(\frac{21}{2})) = 4.66$$

However, the answer appears to be 5.4725. What is wrong in my approach, or am I misinterpreting the problem?

Both of the proposed answers are wrong for the problem as stated. To reiterate the problem: both players A and B roll their die. B observes both rolls, and is then able to choose whether or not she wants to reroll. If she decides to reroll, she does so and must keep the value of the second roll. The player in the end with the higher roll obtains that roll's value from the other. In the case of a tie, player B reaps the roll value.

The major flaw in your solution is this:

As player B will only reroll in the case when A's number is higher

This isn't true, at least not if B is playing to maximize her expected winnings. Think about an extreme case where A and B both roll $$1$$. A's number is not higher, but surely B would be remiss to pass up rolling again!

To find the correct strategy, we need to more carefully consider when rerolling increases B's expected winnings. Suppose A rolls $$a \in \{1, ..., 30\}$$ and B rolls, initially, $$b_1 \in \{1, ..., 20\}$$.

We start by computing the expectation of a reroll. Note that this value only depends on the value of $$a$$, since the value of B's first roll is essentially forgotten (once B commits to rerolling, they must accept the new value). We have for A's value

$$E_r (a) = \sum_{b_2 = 1}^{\min(a - 1, 20)} \left( \frac{1}{20} \right) (a) + \sum_{b_2 = a}^{20} \left( \frac{1}{20} \right) (-b_2)$$

Note that for $$a \ge 21$$ (which happens with probability $$\frac{1}{3}$$), $$E_r (a) = a$$. Therefore, it does not really matter whether or not B decides to reroll. This makes sense because it's impossible to top a score between $$21$$ and $$30$$ (incl.) with a $$20$$-sided die. The value for A contributed by this case is straightforward:

$$\left( \frac{1}{3} \right) \left( \frac{1}{2} (21 + 30) \right) = \color{blue}{\frac{17}{2}}$$

For $$a \le 20$$, we have

$$E_r (a) = \frac{a (a - 1)}{20} - \frac{\frac{1}{2} (a + 20) (21 - a)}{20} = \frac{1}{40} (3 a^2 - 3 a - 420)$$

Consider the case where $$a \le 20$$ and $$a > b_1$$. If B does not reroll, she will get $$-a$$. On the other hand, if B does reroll, she will get $$-E_r (a)$$. It is easy to see that $$a > E_r (a) \implies -a < -E_r (a)$$ for all relevant $$a$$. Therefore, if $$a > b_1$$, then B should choose to reroll (this is the converse of your original statement). The value for A contributed by this case is

$$\sum_{a = 1}^{20} \sum_{b_1 = 1}^{a - 1} \frac{1}{600} E_r (a) = \frac{1}{600} \sum_{a = 1}^{20} (a - 1) E_r (a) = \color{blue}{\frac{1197}{800}}$$

The final case is the most complicated and interesting; it is where $$a \le 20$$ and $$a \le b_1$$. If B does not reroll, she will get $$b_1$$ (having won the first round). If B rerolls, she will (just as in the previous case) get $$-E_r (a)$$. B should therefore reroll if $$b_1 \le -E_r (a)$$. What you will find if you plot these values is that for when $$8 \le a (\le b_1)$$, B is not inventivized to reroll. For $$a \le 7$$, there is a maximum value $$f (a)$$ such that if $$(a \le) b_1 \le f (a)$$, then B will reroll, and if $$b_1 > f (a)$$, then B will not reroll. You can check that $$f$$ is described by $$\{ (1, 10), (2, 10), (3, 10), (4, 9), (5, 9), (6, 8), (7, 7) \}$$. The value contributed to A by this case is

$$\sum_{a = 1}^7 \sum_{b_1 = a}^{f (a)} \frac{1}{600} E_r (a) + \left( \sum_{a = 1}^7 \sum_{b_1 = f (a) + 1}^{20} + \sum_{a = 8}^{20} \sum_{b_1 = a}^{20} \right) \frac{1}{600} (-b_1)$$

$$= \frac{1}{600} \left( \sum_{a = 1}^7 (f (a) - a + 1) E_r (a) - \sum_{a = 1}^7 \frac{1}{2} (f (a) + 21) (20 - f (a)) - \sum_{a = 8}^{20} \frac{1}{2} (a + 20) (21 - a) \right)$$

$$= \color{blue}{-\frac{4027}{800}}$$

The total expected value for A is therefore

$$\color{blue}{\frac{17}{2}} + \color{blue}{\frac{1197}{800}} \color{blue}{- \frac{4027}{800}} = \boxed{\frac{397}{80}} = 4.9625$$

which is slightly under $$5$$.

Alternate Related Problem

Besides the point, but things change a bit (i.e., the problem becomes considerably easier) if B is only allowed to observe her own roll before making a decision to reroll. The problem becomes simpler because B's decision only depends on $$b_1$$, the value of her first roll. B will obviously reroll if she expects to obtain a higher value on the reroll. Since the expected value of a roll of the $$20$$-sided die is $$\frac{1}{2} (1 + 20) = \frac{21}{2}$$, B will reroll when $$b_1 \le 10$$.

When $$a \ge 21$$, A's expected winnings remains $$\frac{17}{2}$$, since nothing that B does can change the fact that A wins.

When $$a \le 20$$ and $$b_1 \le 10$$, then B will reroll. This contributes

$$\sum_{a = 1}^{20} \sum_{b_1 = 1}^{10} \frac{1}{600} E_r (a) = \frac{1}{60} \sum_{a = 1}^{20} E_r (a) = -\frac{7}{40}$$

to A's expected value, where $$E_r (a)$$ is identical to the one used in the original problem. Finally, when $$a \le 20$$ and $$b_1 \ge 11$$, B will not reroll. If $$a \le 11$$, then A's expected value is

$$\sum_{b_1 = 11}^{20} \frac{11}{600} (-b_1) = -\left( \frac{11}{600} \right) \left( \frac{1}{2} (31) (10) \right) = -\frac{341}{120}$$

If $$a \ge 12$$, then A's expected value is

$$\sum_{a = 12}^{20} \sum_{b_1 = 11}^{a - 1} \frac{1}{600} (a) + \sum_{a = 12}^{20} \sum_{b_1 = a}^{20} \frac{1}{600} (-b_1)$$

$$= \frac{1}{600} \left( \sum_{a = 12}^{20} a (a - 11) - \sum_{a = 12}^{20} \frac{1}{2} (a + 20) (21 - a) \right) = 0$$

Bringing everything together, we obtain an overall expected value for A of

$$\frac{17}{2} - \frac{7}{40} - \frac{341}{120} = \boxed{\frac{329}{60}} \approx 5.4833$$

which is interestingly close to what you claimed is the correct answer:

However, the answer appears to be $$5.4725$$

Can you provide a reference for that result? I do not see a way to justify it.

Anyways, it's informative to compare the results of these distinct but related problems. Both have B acting in a way to maximize her profit ("rational"), but the difference is in the amount of information B has. With more information (which applies to the original problem), B has more advantage, and A's expected value for the game is lower, by more than $$0.5$$.

• Thank you for the explanation! The result 4.725 was actually from the assumption that B only sees her roll before making the choice, as done here: math.stackexchange.com/questions/1236805/…. What is the the cause for the small discrepancy between your answer and the one provided in this thread?
– Anon
Aug 16, 2023 at 12:02
• I did not see where 4.725 was mentioned in that thread, could you check again? Aug 16, 2023 at 16:02
• Yes, my bad. It is mentioned in the original post here: math.stackexchange.com/questions/2487975/… and I believe the proposed solution evaluates to 5.4725. I mistyped in my previous comment. I was talking about the discrepancy between your 5.4833 and 5.4725.
– Anon
Aug 16, 2023 at 17:49
• Ok, please understand that that number is mentioned there just as in your problem, with no calculation justifying it. I cannot read minds; I cannot tell you how 5.4725 was obtained. My solution is identical to the one proposed in the first answer (even though there are issues with that answer, for instance on the strategy in the case that A's roll is known to B before her decision). I urge you to try the problem yourself, see if you can justify "5.4725", and then I can better help you. Aug 16, 2023 at 19:17

I see two problems. First, your assessment that the chance B2 is highest ($$B2 \gt A1 \gt B1)$$ as $$\frac 13$$ assumes that none of the rolls are equal. The fact that they can be changes the chance as you want the chance of $$B2 \ge A1 \gt B1$$. Second, you are using the $$\frac 13$$ only in the case $$A1 \gt B1$$. This biases $$A1$$ high, so the $$\frac 13$$ is not correct.

• Could you please suggest the modified calculation?
– Anon
Aug 15, 2023 at 11:28