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Some days ago I read this problem, and I haven't arrived yet at a solution: Imagine you have 3 kinds of regular number die:

d1 has 12 faces (from 1 to 12), d2 has 6 faces (from 1 to 6), d3 has 4 faces (from 1 to 4).

Now, you can choose between using one d1, two d2's, or three d3's. Which configuration of dice would you choose if the one who gets the higher number wins?

I would choose the first one, because the probability of winning (getting the highest number) is 1/12. Instead, throwing two d2's the probability is (1/6)^2, and for three d3's it's (1/4)^3. But I'm not sure.

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  • $\begingroup$ What do you think? Also, what numbers are in the dice? $\endgroup$
    – Arthur
    Oct 7, 2017 at 16:49
  • $\begingroup$ There are some subtleties. With one D1, you will have nonzero chance to roll a 1. But you will always roll something higher than 1 if you roll two D2 or three D3. This kind of consideration counters your observation about what it takes to roll a 12. And ultimately, these considerations outweigh your observation, as @Austin's answer shows $\endgroup$
    – 2'5 9'2
    Oct 7, 2017 at 16:56

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Simply calculate the expected value of the sum of these dice:

$$\Bbb E[d_1]=\sum^{12}_{n=1}\frac1{12}n=\frac{13}2=6.5$$ $$\Bbb E[2d_2]=2\left(\sum^6_{n=1}\frac16n\right) = 2\cdot\frac72=7$$ $$\Bbb E[3d_3]=3\left(\sum^4_{n=1}\frac14n\right) = 3\cdot\frac52=7.5$$

Therefore, $3d_3$ is to be the best choice.

EDIT:

I might have misinterpreted this question. My answer is correct if the number referred to in "highest number" is the sum. Alternatively, with two players, this could mean the highest number rolled on any die. I will answer this now:

Expected value of the highest die rolled is relatively simple:

$$\Bbb E[d_1]=\sum^{12}_{n=1}\frac1{12}n=\frac{13}2=6.5$$ $$\Bbb E[2d_2]=\sum^6_{n=1}\frac{n^2-(n-1)^2}{6^2}n = 4.47\bar2$$ $$\Bbb E[3d_3]=\sum^4_{n=1}\frac{n^3-(n-1)^3}{4^3}n = 3.4375 $$

So the answer in the case of the maximum die winning, you should choose the $12$-sided die.

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    $\begingroup$ The expected values can also be calculated thusly with no summation. D1's distribution is symmetric, ranging from 1 to 12. So $\frac{1}{2}(1+12)$ is its expected value. And D2+D2's distribution is symmetric, ranging from 2 to 12. So $\frac{1}{2}(2+12)$ is its expected value. And similarly for three D3s. $\endgroup$
    – 2'5 9'2
    Oct 7, 2017 at 17:01
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    $\begingroup$ The question cannot be answered simply by expectation values. As far as I understand, the OP is considering a game in which two players will each roll (with their chosen set of dice) and the higher score wins (quote: "the one who gets the higher number wins"). One should really consider the probability of getting a higher score than an opponent, not simply as high as possible. This might depend on the opponent's choice. I admit that here the expected values are sufficiently different to make this unlikely, but I can certainly devise a set-up in which the highest E is not the optimal choice. $\endgroup$
    – Nick Pavlov
    Oct 7, 2017 at 18:54
  • $\begingroup$ @Nick check out the edit. I have answered the question both ways. $\endgroup$
    – Austin Weaver
    Oct 7, 2017 at 19:17
  • $\begingroup$ @AustinWeaver Actually, that is not what I meant. I meant that the aim might not be to maximize an expected value, but to maximize probability of getting more than the result of another random event (the opponent's roll). As in: player A rolls one of the options (12-sided, or two 6-sided, or three 4-sided) and gets a sum of $x$ (a random variable with a certain distribution which depends on what A has chosen to roll). What choice of dice for player B gives him a better ratio of getting more than $x$ (winning) vs. less than $x$ (losing)? $\endgroup$
    – Nick Pavlov
    Oct 7, 2017 at 19:57
  • $\begingroup$ @AustinWeaver In this situation (and I suspect whenever the distributions are symmetric), the distribution with the higher expected value has an advantage. But it is conceivable that it is not always the case (in fact I have a counter-example with non-symmetric distributions) $\endgroup$
    – Nick Pavlov
    Oct 7, 2017 at 20:00

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