What Are The Odds of Rolling a Six?

I've been playing the new Third Reich with its new combat system, which uses Bucket of Dice (BoD) instead of odds / Combat Results Tables (CRTs) to resolve battles. In CRT system, you add up the attacking factors and defending factors and divide attackers by defenders to get a ratio. You then roll once on a table. This is fast and useful, but it delivers extreme results too often.

In the BoD system, you roll one die per combat factor, and each roll of "6" is generally a hit. So knowing the odds of rolling sixes is important.

The odds are not simple.

Many of us simply multiply the odds of rolling six on one die (1/6) by the number of dice to get the odds of actually rolling a six. This system seems intuitively to be right, but careful thought will show that it is not: if you roll six dice, you are not guaranteed to roll a six.

Simplify the problem: what are the odds of rolling a 6 on two dice? Intuitively, you might guess the odds are 1:3. You'd be wrong.

Simplify it further: what are the odds of getting "heads" at least once on two coin flips? It's certainly not 1/2 + 1/2 = 1. The matrix of possibilities of two coin flips are:

HH HT TH TT

There are four possible outcomes of two coin flips and only one of them does not have "heads" at least once.

The odds of getting "heads" at least once on two coin flips is 3:4 or 75 percent. No matter how many coins you flip, it is possible, though very unlikely, that you'll never get "heads" at least once.

The matrix of possibilities for two dice

While there are only 4 possible results from flipping a coin twice (2 x 2 =4), there are 6 x 6 =36 possible results of rolling two dice.

If you roll a six on the first die, there are six possible outcomes for the second. However, if you roll a six on the second die, there are only five unique possibilities for the first die, because we have already covered the possibility of rolling a six on both dice. The odds of rolling at least one six on two dice are 11/36, slightly less than 1/3 (and there's a 1/36 chance of rolling two sixes).

The matrix of other results contains all the numbers between 1 and 5. It is, in fact, 5 x 5 = 25 possible results (1 and 1-5, 2 and 1-5, 3 and 1-5, 4 and 1-5, and 5 and 1-5 = 5 x 5). 25 + 11 = 36, so our math checks out. The odds of _not_ rolling at least one six on two dice is 25/36, slightly over 2/3.

What are the odds of rolling one six on three dice? Now we're dealing with 6 x 6 x 6 = 216 possible outcomes. It's easier to calculate the odds of not rolling at least one six. Those odds are 5 x 5 x 5 / 6 x 6 x 6 = 125 / 216. That means that the odds of rolling at least one six on three dice are 89 / 216 or about 41 percent.

If you visualize the matrix of possibilities as a 6 x 6 square, with columns 1-6 across the top and rows 1-6 across the sides, you'll see that the outcomes containing a six are in the 6 row and 6 column. There are six possibilities in each, but rolling a 12 (two sixes) is a duplicate, so there are 11 possibilities. To confirm the truth of this visualization, imagine the 5 x 5 grid made up of the 1-5 rows and 1-5 columns that contain no sixes. That's 5 x 5 = 25 out of 36, as described above.

What are the odds of most people's favorite attacks, trying for one six on six dice or two sixes on twelve dice?

The odds of not rolling one six on six dice are 5 ^ 6 / 6 ^ 6 = 15,625 / 46,656 = very close to one-third. This is counter-intuitive if you're calculating odds through addition. If you are doing it that way, you'll think you need six combat factors to be guaranteed to get one hit. As I show here, it's not true. There's about a 1/3 chance of failure.

Nine factors give you a better guarantee of at least one hit: 5 ^ 9 / 6 ^ 9 = 1,913,125 / 10,077,698. There's still a 1/5 chance of failure (actually 18.98 percent, which is very slightly less than 1/5).

But I need two or three hits

Calculating the odds of getting two or three hits is much tougher. We're no longer rolling two dice, so the matrix would be a six-dimensional or nine-dimensional. So let's start with the odds or rolling two hits on three dice.

The matrix of possibilities is a cube with 216 entries (6 x 6 x 6). There are three sets of six outcomes containing two sixes, with a duplicate at the remote possibility (1/216) of rolling three sixes. The odds of rolling two hits on three dice are 5 + 5 + 6 = 16/216 or almost 7.5 percent. It is exactly 2 / 27. It is very unlikely.

Math check: the odds of not getting two hits on three dice is 6 x 5 x 5 = 150 possibilities if the first die is a six, 25 possibilities if the second die is a six, and 25 possibilities if the third die is a six.

It is 6 x 5 x 5 + 1 x 5 x 5 + 1 x 5 x 5 = 200 / 216. Math check complete.

So what are the odds of getting two hits on twelve dice?

The odds of not getting two hits on twelve dice are, I think, (6 x 5 ^ 11 + 11 x 5 ^11) / 6 ^ 12 =

(292,968,750 + 48,828,215 * 11) / 6 ^ 12 = (292,968,750 + 537,109,375) / 2,176,782,336

= 830,078,125 / 2,176,782,336 = 38 percent, which is very close to one third, just like the odds of not getting one hit on six dice. I am not certain that my math is exact here, but I am certain that it's close to accurate.

6 x 5 ^ 11 represents the possibilities in which the first die is any number and the subsequent ones have no sixes. The remaining 11 x 5 ^ 11 represent the possibilities that involve a six rolled on any one of the other elevent dice.

What are the odds of rolling three sixes with eighteen dice?

The odds of not getting three hits on eighteen dice, I believe, are made up of

6 x 6 x 5 ^ 16, which are the possibilities of rolling a six on the first two dice

x 16, representing the possibilties that involve rolling sixes on the first die and any of the remaining sixteen dice.

But in fact, we want to use a different mathematical function. We want to calculate the exact number of ways we can roll two sixes on eighteen dice. The number of possibilities are 17 given a six on the first die, 16 given a six on the second die (with a six on the first die already covered) ... the number of possibilties is 17 + 16 + 15 + 14 + 13 + 12 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 153.

I think that the odds of not rolling three sixes on eighteen dice is 153 x (6 x 6 x 5 ^ 16) / 6 ^ 18 =

153 x (36 x 25 x 25 x 25 x 25 x 25 x 25 x 25 x 25) / 216 x 216 x 216 x 216 x 216 x 216) =

153 x (5.49316406 × 1012) / 1.01559957 × 1014 = which doesn't help at all, so I'll just plug the numbers above into Google and I get:

a number that does not work, so I take out the "36" and I get

153 x (25 x 25 x 25 x 25 x 25 x 25 x 25 x 25) / 216 x 216 x 216 x 216 x 216 x 216) = 23 percent.

Again, I'm not certain that this is correct, but it will, I think, explain your actual experience with bucket of dice games. You need to know the odds.