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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?

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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

1 Answer 1

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 as follows: 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 $\frac1{72}$. And if the pattern is XYZ, the probability is $\frac1{36}$.

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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

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