There are $2$ fair dice:

  • An $11$-sided die, valued from $-5$ to $5$,
  • A $41$-sided die, valued from $-20$ to $20$.

You pick one, and I'll take the other one. We roll the dice, and whoever rolls the larger number wins (if tied, we roll again.)

Which die will you choose to maximize your winning probability?


8 Answers 8


The answer is

The chances are equal


Suppose we consider all the possible outcomes. Then I propose, there is a one-to-one mapping from winning outcomes for you to winning outcomes for me.

To see this, consider a scenario where you roll $i$ and I roll $j$ and you win ($i > j$). Then, in the case, that you roll $-i$ and I roll $-j$, I win (since ($-i < -j$). The mapping $(i,j) \rightarrow (-i, -j)$ is clearly 1-1 for the given problem, it inverts the winner, and the set of draws maps onto itself.

If we were to enumerate overall all possible outcomes, I would win the same number of times as you so the probability of winning must be the same.

NB: This additionally show that the chances would be equal if we replaced $5$ and $20$ by any integer values.

Alternative proof

The probability of winning with the $41$-sided dice in a single go can be summed as the probability of winning given that the $11$-sided die rolls a $5,4,3,\ldots$ multiplied by the probability of rolling a value in an $11$-sided die that is $$ P(41\text{ wins}) = \frac{1}{11}\displaystyle \sum_{j=15}^{25} \frac{j}{41} = \frac{220}{451}$$ Similarly, the probability of the $41$-sided die losing is given by calculating a similar sum $$ P(41 \text{ loses}) = \frac{1}{11}\displaystyle \sum_{j=25}^{15} \frac{j}{41} = \frac{220}{451}$$ where the sum is descending over the integers in this case. The probability of a single outcome being a draw is just $\frac{11}{451}$.



Doesn't matter.


The D11 wins with probability $\frac{220}{451}$, where $220=16+17+\dots+24+25$. It's a tie with probability $\frac{11}{451}=\frac1{41}$, and the D41 also wins with a probability of $\frac{220}{451}$.


Mapping out the 2-dimensional probability space you can see that the area in which each die wins is symmetrical, therefore it does not matter which die you choose, your odds of winning are the same.

 A -20 ... -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6 ... 20
-5   B      B  .  A  A  A  A  A  A  A  A  A  A  A      A
-4   B      B  B  .  A  A  A  A  A  A  A  A  A  A      A
-3   B      B  B  B  .  A  A  A  A  A  A  A  A  A      A
-2   B      B  B  B  B  .  A  A  A  A  A  A  A  A      A
-1   B      B  B  B  B  B  .  A  A  A  A  A  A  A      A
 0   B      B  B  B  B  B  B  .  A  A  A  A  A  A      A
 1   B      B  B  B  B  B  B  B  .  A  A  A  A  A      A
 2   B      B  B  B  B  B  B  B  B  .  A  A  A  A      A
 3   B      B  B  B  B  B  B  B  B  B  .  A  A  A      A
 4   B      B  B  B  B  B  B  B  B  B  B  .  A  A      A
 5   B      B  B  B  B  B  B  B  B  B  B  B  .  A      A

I'd take the

11-sided die


it has less sides, therefore it has less tendency to roll and thus it is easier to "aim" for a number with a controlled roll, at least after some practice. Even more so if the numbers are grouped with a cluster of positive numbers somewhere. Just like it is easier to throw a 4-sided die on a predetermined side than doing the same with a D20 which will inevitably roll away and land on a random number.

  • 1
    $\begingroup$ An interesting take. The question talks about "fair dice" though. I'd think this implies no such tricks as well. $\endgroup$
    – Gassa
    Dec 18, 2018 at 10:13
  • 1
    $\begingroup$ @Gassa AFAIK "fair dice" means only that each of the faces have equal chance, i.e. it is not skewed, leaded, drilled, sanded or otherwise modified. A "fair roll" on the other hand would indeed imply no such tricks. As a corollary, my approach is of course not fair, but as the proper analytical answer has already been given and accepted, I ventured to offer a more unusual solution. $\endgroup$
    – zovits
    Dec 18, 2018 at 10:24

The average value of each die rolled an infinite number of times will be zero for either die. So it makes no difference as neither has an advantage over the other for gaining a larger number.

  • 10
    $\begingroup$ Considering the long term average won't work in general. For example, consider the case where the first die has 40 on one side and -1 on every other side and the second die has -10 on one side and 1 on every other side. The average value is 0 for both dice but one of them is more likely to win. $\endgroup$
    – hexomino
    Dec 17, 2018 at 17:12
  • 1
    $\begingroup$ That's not an even distribution of frequency though, OPs example is. $\endgroup$
    – Shardj
    Dec 17, 2018 at 17:17
  • 12
    $\begingroup$ Sure, but do you see that the first sentence of your answer is true for my choice of dice as well as the OP's? $\endgroup$
    – hexomino
    Dec 17, 2018 at 17:21

The numbers of the smaller die are included in the bigger die, the possibilities are hence the same so we have to care only about the bigger die.
All higher integers will win, all smaller ones will lose.
There is an equal number of higher and smaller integers so the possibility is the same.


It doesn't which dice you pick.

Easy proof:

For each combination of a=the number on the first dice and b=the number on the second dice (where a and b are not equal), there is an equally likely combination -a and -b. If a beats b then -b beats -a, and conversely. Example 31 on the first dice beats 7 on the second dice, but it's equally likely to get -31 on the first dice and -7 on the second dice, where the second dice beats the first.


I would take the 11 sided dice

because P(getting a higher number in 20 sided dice ) = 1/11*25/41 + 1/11*24/41 + .. 1/11*15/41 = 220/451=10/41

P(getting a higher number in 11 sided dice ) = P(getting -20 in dice2)*P(getting a higher number in dice2) + P(getting -19 in dice2)*P(getting a higher number in dice2) + .. =1/41*1 + 1/41*1 + .. etc = (1/41 )*15 + etc .. > 10/41

Clearly 11 sided dice wins..

  • $\begingroup$ Okay but you have the exact opposite on the negative side. The numbers range from -20 to 20 not 0 to 40 $\endgroup$
    – Shardj
    Dec 21, 2018 at 16:12

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