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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: $$ 1 - \sum_{i=1}^{6}\sum_{j=1}^{6} \left( P(d_i \mid d_i\cap d_j) + P(d_j\mid d_i\cap d_j) \right) $$ I am pretty sure this is not correct--any suggestions?

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How about $P(X \neq Y, X \neq Z, Y \neq Z)$. Of course this isn't an answer, it's just a statement of the problem. –  goblin Feb 10 '13 at 6:32
@user18921 That looks correct as a statement of what I am looking to express. It has been some time since I have studied statistics, so I am trying to put together an expression for this that actually makes sense :) –  quannabe Feb 10 '13 at 6:49
I think I am erring in using the \cap notation. Would an addition/multiplication of the probabilities be a more correct method? –  quannabe Feb 10 '13 at 7:14
I dont know - i currently know very little probability. Why not rephrase the question like this: Let X,Y and Z denote independent random variables representing rolls of three different six-sided die. How would i go about finding P(...) (what i wrote earlier) –  goblin Feb 10 '13 at 7:45
And youre right to emphasize that the dice may be biased differently. –  goblin Feb 10 '13 at 7:47
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1 Answer 1

For a fair die with $n$ sides, each side has an equal probability of $1/n$.

For a biased die, each side has a probability $p_i$, with $\sum_{i=1}^np_i=1$.

For two biased dice $A$ and $B$, the probability that they will roll the same number is:


For 3 dice, the chance that they are the same is the sum of the chance of each pair are the same less twice the sum that all 3 are the same. So the chance they are different is:

$$1 - \sum_{i=1}^np_{Ai}p_{Bi} - \sum_{i=1}^np_{Ai}p_{Ci} - \sum_{i=1}^np_{Bi}p_{Ci} + 2\sum_{i=1}^np_{Ai}p_{Bi}p_{Ci}$$

Note that this expression also works with a fair die (and gets much simpler).

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The alternating sum at the end is an example of the inclusion-exclusion principle. –  Did Mar 26 '13 at 6:31
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