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Probability of rolling at least 2 double 6s in 3 rolls

Posted By: Michael Petch
Date: Saturday, 10 May 2014, at 12:43 p.m.

In Response To: Probability of rolling at least 2 double 6s in 3 rolls (leobueno)

Casper, Daniel, Bob et all are correct as well as your method (with the typo corrected).

With such a small number of trials and successes the answer is pretty trivial. I only wanted to point out that your problem resolves down to simple binomial probabilities. There is a standard formula for the binomial probability:

P(k,n)=(n choose k)*(p^k)*(q^(n-k))
P(k,n)=probability of k successes in n trials.
n = number of trials
k = number of successes

n – k = number of failures
p = probability of success in one trial
q = 1 – p = probability of failure in one trial

So in your trivial case you could use this formula twice. Adding P(2,3)+P(3,3)
So you'd end up with calculations that pretty much look like Casper had:

P(2,3)+P(3,3) = (3 choose 2)*(1/36)^2*((35/36)^(3-2)) + (3 choose 3)(1/36)^3*((35/36)^(3-3))
P(2,3)+P(3,3) = 3*(1/36)^2*(35/36) + (1/36)^3

I'm not giving anything different that what others did, but just provided the well known equation that can be used to answer questions similar to this.

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