A Mathematical Analysis of Exploding Dice March 22, 2009

Posted by eric22222 in General

(Session 66 has been posted)

I’m guessing the title caught your eye? I wrote this essay a while back, but decided to hold off on posting it until I had my power cord back. So… here we go:

With a normal die roll, the result is either 1, 2, 3, 4, 5, or 6. This will later be generalized to N-sided dice, but for now, the standard six-sided variety will suffice. Some games (dice-based tactics games, usually) use a system called “exploding dice.” With this system, rolling the maximum value on a die allows the roller to roll the die again and add the new result to the previous one. There is usually no limit on the number of rerolls.

For example, a player rolls a six. He rerolls the die, which lands on six once again. Once more, he rolls, but this time gets a two. His final result for this roll is effectively fourteen, even though the die has only six faces .

Put simply, expected value is what the average result for an experiment would be if were tried an infinite number of times. For example, if a coin flip was worth one point if heads lands and two points for tails, the expected value would be 1.5 since heads and tails are equally likely.

To compute expected value, the sum of each possibility times its respective chance is taken. For the coin flip example, this would be . That is, a 50% chance of 1 point plus a 50% chance of 2 points.

So the problem to be tackled here today is this: how does the expected value of an exploding die differ from its standard expected value?

The first step will be to find a pattern. Here is a quick comparison of values and their respective probabilities:

x, Pr(x)

1, 1/6

2, 1/6

3, 1/6

4, 1/6

5, 1/6

7, 1/36

8, 1/36

9, 1/36

10, 1/36

11, 1/36

13, 1/216

14, 1/216

…

So how can these be put into an elegant formula? First of all, a simple sum of products will be used:

The factors preceeding each sum are powers of six, so they can be consolidated into a nested sum:

Each sum, such as , is equal to the average value times the number of elements. The number of elements remains a constant 5, but the average value is increased by 6 for each subsequent sum, beginning with a value of 3. The formula can now be rewritten like so:

Simplifying some of the products, the formula can be rewritten once again:

Now an obvious pattern begins to develop. By expanding these nested products, the pattern becomes even more clear:

Six is factored out of all terms:

This new formula can be rewritten in sumation notation:

For each term added, our error will be reduced by a factor of 6, so just a few terms will return a very accurate result:

N, sum to N

0, 2.5

1, 3.75

2, 4.0972222

3, 4.178240741

4, 4.194058642

5, 4.197595165

The sum is swiftly approaching exactly 4.2, but what is the relation of that value to a six-sided die’s normally expected value?

Adding the exploding modifier to a six-sided die changes its expected value by a factor of . That is, the ratio of the number of sides on the die to the number of sides that do not “explode.”

I have performed the same calculations on four and eight-sided dice to compare results, and the forumla holds.

For any N-sided die numbered 1 to N with all sides equally likely, the exploding modifier will increase the die’s expected value by a factor of .

Interestingly enough, this formala holds for a hypothetical one-sided die. Heuristically, since the highest number will always be rolled, the total will constantly increase, returning an infinite result. Mathematically, the expected value of 1 will increase by a factor of , which approaches positive infinity from positive values.

This essay was written by Eric Dobbs, an undergraduate student of computer science at the University of Tennessee, Knoxville. He can be reached at erictdobbs -at- gmail -dot- com.