Probability in a Dice Game (Zombie Dice)

Question

In the game of Zombie Dice (Rules) there exist 13 dice:

• 6 Green - 3 Brains, 2 Footprints, 1 Shotgun
• 4 Yellow - 2 Brains, 2 Footprints, 2 Shotguns
• 3 Red - 1 Brain , 2 Footprints, 3 Shotguns

A footprint forces the user to roll the dice again. A shotgun counts as a 'X', and 3 result in the loss of a turn.

I question what the probability of rolling 13 brains, in a single turn, assuming 13 dice total.

When calculating this myself, I assumed a probability of: ((1/6)^3) * ((1/3)^4) * ((1/2)^6) However, a friend of mine brought up that the footprints would have an influence on the odds, which I disputed.

TLDR; What is the impact of the footprints, and what is the total probability of rolling 13 brains?

Answer 1

Let's look at a green die.

Let p=the odds of getting brains on a green die. What are the cases when tossing one?

1. Brains: $\dfrac12$
2. Footprint: $\dfrac13\cdot p$

So, we get that $p=\dfrac12+\dfrac13p$. Thus $p=\dfrac34$.

Similarly, you can get the probabilities for the other two colors of dice. Then multiply them out like you did with the other probabilities.

Answer 2

The fact that some sides are "roll again" allows us to consider only the sides that would cause us to stop rolling, that is, the Brains and Shotgun.

So for the Green dice, we have a $\frac{3}{4}$ chance of ending on a Brains, on the Yellow it is $\frac{1}{2}$, and on the Red it is $\frac{1}{4}$. The final answer is $\left(\frac{3}{4}\right)^6\left(\frac{1}{2}\right)^4\left(\frac{1}{4}\right)^3$.