I'm rusty on probability and, even so, I don't think I know how to deal with this situation. Here's the problem.

Let's say I have a cup of $n$ dice. Each dice face has either a shotgun, a brain, or a runner on it (images on the dice.) There are three types of dice: red, green, and yellow. The red dice have 3 shotguns. The greens dice have 1 shotgun. The yellow dice have 2 shotguns.

There are 13 dice total, and dice that on the table in front of me have been drawn from the cup. The player rolls 3 dice at a time. There are 3 red dice, 6 green dice, and 4 yellow dice.

Let's say I have one of the dice in front of me, and the rest are in the cup. I know the colour of the dice in front of me. I have to draw 2 more dice from the cup at random. I won't know what colour they are.

I then roll all three dice.

What are the chances that I roll 2 or more shotguns?

There are other situations in the game where I have 0-2 dice in front of me, and I would want to know the chance of rolling 1-3 shotguns. If someone can explain the calculation to me, that would be great.

I can calculate for when I'm rolling 3 dice of known colour, but when I'm drawing from the cup, I can't figure it out.


If you calculate the probability that you get a certain color combination of chosen dice (which you can do), and then calculate the probability that you get 2 or more shotguns given that choice of colored dice, and then multiply and take the sum over all possible dice color outcomes, you will get your answer. It may be a bit tedious to calculate however without a computer.

  • $\begingroup$ Okay, I'll try that. Thanks. $\endgroup$ – Korgan Rivera Mar 16 '15 at 21:50
  • $\begingroup$ I wrote the code to calculate it and it works. Thanks! $\endgroup$ – Korgan Rivera Mar 27 '15 at 19:58

The expected number of shotguns per dice are:

$$\text{Red:}\frac{1}{2}$$ $$\text{Green:}\frac{1}{6}$$ $$\text{Yellow:}\frac{1}{3}$$

Due to linearity of expectation, the expected number of shotguns on a dice picked at random is:


For 3 dice this is $\approx0.885$.

If you already have 1 red dice the equation becomes:


You can work the others out yourself.

You should also check out www.anydice.com


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