I would like to sum the probability that given 3 (potentially biased) die rolls, all 3 rolls will be different-- what is the correct way to do this?

So far, I have:

alt text

I am pretty sure this isn't correct--any suggestions?

asked 10 Feb '13, 02:04

quannabe's gravatar image

quannabe
112

edited 10 Feb '13, 02:12

Any ideas??
(10 Feb '13, 03:36) quannabe quannabe's gravatar image

I have trouble matching up your post title with the associated text. I don't have time to fully work this out right now, but here is the basic idea:
P(all 3 rolls different) = 1 - P(2 or more of the rolls are the same)
Call the three dice d1, d2, and d3 then we have:
P(2 or more of the rolls are the same) = P(d1=d2) + P(d1=d3) + P(d2=d3) - 2*P(d1=d2=d3)
(the last term corrects for multiple counting of d1=d2=d3 in the first three cases)
Then each of those terms can be expressed as the number of possible ways the case can occur times the probability of each. For example, the last term is (assuming 6 sided die):
P(d1=d2=d3) = 6 * (1/6)^3

Hopefully this is enough to get you started. If not, you might try working out a simpler case (say 3 x d3 or 2 x d6) and try to cast it in the form I described.

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answered 10 Feb '13, 18:43

rseiter's gravatar image

rseiter ♦
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