Rolling at least $k$ on $n$ d6s – with a twist!

Question

I am putting together a table of dice probabilities for a project I am working on and have found myself intimidated by a little "special case" I'm trying to work with.

For determining the probability of, for example, rolling at least three (k) 6's with five (n) dice I am using the following:

P(at least three 6's) = P(exactly three 6's) + P(exactly four 6's) + P(exactly five 6's)

using this: P(k) = nCk * p^k * q^(n-k)

This has worked perfectly and I can use this generalized method to solve for any number of successful rolls across any number of individual rolls.

Here is my twist and issue:

How do I find the probability of rolling at least k 6's with n dice if one of the dice has two 6's on it?

At first I thought I could just unsimplify the (p^k * q^(n-k)) part of the formula to account for the new case of a 2/6 chance for (and 4/6 chance against) a 6 on one of the dice, but how will this new condition affect the nCk portion of the formula?

...this is where I lose the mental grasp

If possible, as this could greatly help me and others understand, could an answer be given that also includes a solution to the specific situation seen below with all work shown?

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You have five six-sided dice. Four of the dice are regular, but the fifth die has two 6's on it. What is the probability of rolling at least two 6's if you roll all five dice?

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Answer 1

Assuming all dice are standard six-sided dice with the numbers 1-6 on each face:

To determine the probability of rolling at least k 6's with n dice, where p = the probability of rolling a 6, and q = the probability of not rolling a 6:

$P(\ge k \ 6's) = P(k) + P(k+1) + P(k+2) + ...$

using this: $P \left(k\right) = nCk * p^k * q^ \left (n-k\right)$

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To incorporate a dice with two 6's on it:

Let's use a specific example here: We will be rolling 6 dice in total. Five of these dice are regular, they are six-sided and have a 6 on one face. One of these dice is irregular, it is six-sided but has a 6 on two of its faces. What is the probability of rolling all six dice and getting at least three 6's?

n = the rolling of the five regular dice

m = the rolling of the one irregular die

k = a "successful" roll (rolling a 6)

$P_{n+m} \left( k \ge 3 \right) = P_{n} \left( k \ge 3 \right) + \left( P_{n} \left( k = 2 \right) * P_{m} \left( k = 1 \right) \right ) $

Answer 2

It seems you could look at the probability of rolling at least (k-1) 6's with (n-1) dice. i.e. P(exactly 2) + P(exactly 3) ... Then you just need to use independence of dice and multiply P(exactly 2)*1/3, to represent the last die.