Gfrktyl

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?

The answer is

The chances are equal

Proof

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 $$ P(41text{ wins}) = frac{1}{11}displaystyle sum_{j=15}^{25} frac{j}{41} = frac{220}{451}$$ 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 $frac{11}{451}$.

It:

Doesn't matter.

Why?

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 $frac{220}{451}$.

I'd take the

11-sided die

because

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.

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

 B

-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

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.

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..