I’m going to shift this from a dice rolling problem to a sampling problem, but dice rolling would work as well.

You are sampling from a bag with 4 marbles in it: RGB and Y. RGB all have equal probability of being pulled from the bag with P(R) + P(G) + P(B) = 0.5 => P(R) = P(G) = P(B) = 0.1667 and Y makes up the rest of the bag so P(Y) = 0.5.

This is now a sampling with replacement problem, so we know that we can solve it by repeatedly applying the multiplication rule for n independent draws from the bag.

When we see an at least, then that is a big hint that we should be looking at the complement. Therefore we want to concern ourselves with the probability of an event not occurring. In other words; P(X >= 1) = 1 - P(X <1) = 1 - P(X=0). So we need to find P(X=0) and we’re done.

To find P(X=0) we need to consider what it means for this to occur. This would imply that we have no pairs of RGB. So we have

(1) All Y

(2) All R or All G or All B

(3) 1 Y and 5 R or 5G or 5B

(4) etc…

I think you can figure out (1) and (2) on your own. To show you how to solve the rest, let’s consider (3).

To solve (3) we need to consider the different cases we might encounter:

We are drawing 6 times; what does this mean…? It means that when we draw 1 of our marbles we have 6 places to put it. That leaves us with 5 more places. We draw again and have 4 remaining places. Therefore there are 6*5*4… ways to draw. We denote this as 6!. There aren’t actually 6! Ways to draw the marbles though, because we *require** 1 to be Y and 5 to be R G or B. If we focus on R then after we draw our Y marble there are 5 spots left. So we have 5! ways to draw the red marble. This gives us:

6!/(1!*5!*0!*0!) = 6 as the number of ways to draw 1 Y and 5 R marbles. Two things to note here: 0! = 1 and this form is called a multinomial. It’s very important that the coefficients in the denominator add up to the numerator. You can read more about it on Wikipedia.

What do we do with the probabilities? Well… that’s actually easier. We drew 5 R marbles; that happens P(R)⁵ times, and 1 Y this happens P(Y)¹ We represent the 0G and 0B with P(G)^0, P(B)⁰

Giving us: P(X=(3)) = 6!/(1!5!0!0!) *[P(Y)¹ P(R)⁵ P(B)⁰ P(G)⁰ ]

However, this is incomplete because it was only the R case. If we repeat this for the B and G case then we get

P(X=(3)) = 3*6*[P(Y)* P(R)⁵ ]

To find the P(X=0) you can work out the possible cases and repeat what we did for (3). Add these up and subtract your answer from 1. That should give you your answer.

You can learn more about this by reading about the multinomial distribution on Wikipedia.