# Dice Probability ?

• July 19th 2010, 06:02 PM
Delbert
Dice Probability ?
When rolling 5 dice, what is the probability of rolling at least one 6, one 5, and one 4.
Is there a Formula Such as P=(1/6)x(1/6)x(1/6)+(1/6)x(1/6). I know the probability
of rolling one of each with 3 dice is 1/216. I don't know how to compute the
additional probability of the other two dice. I am looking for the most basic way to do the math.

Thanks
Del
• July 19th 2010, 08:14 PM
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Quote:

Originally Posted by Delbert
When rolling 5 dice, what is the probability of rolling at least one 6, one 5, and one 4.
Is there a Formula Such as P=(1/6)x(1/6)x(1/6)+(1/6)x(1/6). I know the probability
of rolling one of each with 3 dice is 1/216. I don't know how to compute the
additional probability of the other two dice. I am looking for the most basic way to do the math.

Thanks
Del

The simplest thing I can think of is using inclusion-exclusion.

Let A = event of rolling at least one 6, one 5, and one 4.

Let C = event that a 6 is not rolled.
Let D = event that a 5 is not rolled.
Let E = event that a 4 is not rolled.

Then by inclusion-exclusion,

$\displaystyle P(\overline{A}) = P(C) + P(D) + P(E) - P(C\cap D) - P(C\cap E) - P(D\cap E) + P(C \cap D \cap E)$

Then you take $\displaystyle P(A) = 1 - P(\overline{A})$

P(C) = P(D) = P(E) = ? (think about it)
P(C ∩ D) = P(C ∩ E) = P(D ∩ E) = ? (think about it)
P(C ∩ D ∩ E) = ?

See how far you can get.