Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a fascination with tabletop sports games, and through that, I've developed an interest in probability. That said, it's not my strong suit, so I wanted to pose this question because I think involves a few different probability principles to solve it.

Here's the premise...the game uses a roll of 3d6 to determine the result of a play. The dice are read from lowest to highest, as in:

a roll of 3, 6, 2 would be 2-3-6 a roll of 1, 4, 1 would be 1-1-4 etc.

That roll is then looked up on a chart to determine the result.

I was curious about the probability of different results coming up so that if I wanted to make a house rule, I would know which result would be the best place to modify.

So, what I do know is that for probability, you multiply chances together, correct? So without the low to high rule I discussed above, any number would have a 1/6 * 1/6 * 1/6 chance of occurring, correct?

How would I apply this principle to any result? My guess is something like:

1-1-4 = 1/6 * 1/6 * 5/6

only one chance for the first 2 die, and the last can be anything but 1, so 5 remaining numbers

2-3-6 = 1/6 * 4/6 * 4/6

first number can be 2 only = 1/6

second number can be any number but 2 or 6 = 4/6

second number can be any number but 2 or 3 = 4/6

Am I understanding this correctly?

share|cite|improve this question
This is the kind of thing that's really easy to get the computer to do. Complete results are here. – MJD Oct 28 '13 at 18:32
Wouldn't $3,6,2$ be $2-3-6?$ – Ross Millikan Oct 28 '13 at 18:37
Ross, you're correct, I swear I had that typed correctly when I first posted it incorrectly on the main stack overflow page. I'll fix it... – tjans Oct 29 '13 at 12:53
up vote 3 down vote accepted

Your answer's not correct. For 1-1-4, you have multiplied by $\frac56$, but it's not clear to me why. The correct calculation goes like this: there is a $\frac16$ probability of getting a 1 on the first die, a $\frac16$ probability of getting a 1 on the second die, and a $\frac16$ (not $\frac56$) probability of getting a 4 on the third die. This multiplies out to $\frac1{216}$. But there are three different orders in which the rolls could occur to give a result of 1-1-4, since 1-4-1 or 4-1-1 give the same result. So you must multiply the $\frac1{216}$ by 3, for a final result of $\frac1{72}$.

Similarly, the result for 2-3-6 is again $\frac1{216}$, but this time multiplied by 6 because there are 6 different orders in which the 2, 6, and 3 can appear; the final result is $6\cdot\frac1{216}=\frac1{36}$.

In general, the answer is:

  • If the pattern you're looking up is XXX, with all three dice the same, the probability is $\frac1{216}$.
  • If the pattern is XXY or XYY, the probability is $\frac1{72}$.
  • If the pattern is XYZ, the probability is $\frac1{36}$.
share|cite|improve this answer
The 5/6 was me overthinking things. I'll save me the embarrassment by not explaining why I thought it'd be 5/6 :) – tjans Oct 29 '13 at 12:56
for 1-1-4, how are there 3 possibilities? Wouldn't there be 6 different ways for that as well, as each die is independent of the other? 1a-1b-4, 1a-4-1b, 1b-1a-4, 1b-4-1a, 4-1a-1b, 4-1b-1a – tjans Oct 29 '13 at 13:03
Suppose the three dice were red, blue, and green. The 4 can be on the red die, the blue die, or the green die; that's 3 ways. Saying that a green 4, a red 1 and a blue 1 is somehow different from a green 4, a blue 1 and a red 1 makes no sense at all. – MJD Oct 29 '13 at 14:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.