Question

1. A standard six-sided die has a different number from 1 through 6 on each side, with thoe average roll being a 3.5. Grime dice, on the other hand, have a different set of numbers of each side, with the same average roll as shown in the table. In this problem (or generally unless otherwise stated), we treat the dice as fair, such that any side is likely to be the top face when rolled Die Normal Red Grime Side 1Side 2 Side 3|Side 4 |Side 5 Side 6Mean Roll 2 3 2 3.5 3.5 3.5 3.5 6 6 3 Green Grime2 Blue Grime (a) Suppose you and a friend each roll one of the three Grime dice, and the higher roller wins. As you are clever, you ask your friend to pick a die first, then you pick one of the remaining dice, to guarantee on average you will win. Calculate the probabilities that red beats blue, blue beats green, and green beats red. The odd result is why these are an example of "non-transitive" dice.] (b) Calculate these probabilities if you roll two identically colored dice. [Note you should see that the order of which die beats which die is reversed.] (c) (OPTIONAL) What happens with these probabilities if you roll three dice?

Answer 1

Let us first represent each case in the form of a probability distribution to make it easy to understand. Let N, R, G, and B denote the number appearing on top while rolling the normal dice, red grime, green grime, and blue grime respectively. The respective probability distributions can then be written as:

Normal dice

N	P(N)
1	1/6
2	1/6
3	1/6
4	1/6
5	1/6
6	1/6

Red Grime

R	P(R)
3	5/6
6	1/6

Green Grime

G	P(G)
2	1/2
5	1/2

Blue Grime

B	P(B)
1	1/6
4	5/6

a) Case I: Red beats Blue

Red will beat blue in one of the following situations:

red shows 3 & blue shows 1 or

red shows 6 & blue shows 1 or

red shows 6 & blue shows 4.

Since the events N, R, G, and B are independent, the required probability can thus be calculated as:

$P(Red \hspace{.1cm} beats \hspace{.1cm} Blue) = \frac{5}{6}*\frac{1}{6}+\frac{1}{6}*\frac{1}{6}+\frac{1}{6}*\frac{5}{6}=\frac{11}{36}$

Case II: Blue beats Green

Blue can beat green if and only if blue shows 4 and green shows 2. Therefore the required probability is:

$P(Blue \hspace{.1cm} beats \hspace{.1cm} Green) = \frac{5}{6}*\frac{1}{2}=\frac{5}{12}$

Case III: Green beats Red

Green can beat red if and only if green shows 5 and red shows 3. The required probability is thus:

$P(Green \hspace{.1cm} beats \hspace{.1cm} Red) = \frac{1}{2}*\frac{5}{6}=\frac{5}{12}$

b) This can be solved in a similar manner as above by first creating the probability distribution when two identically colored dice are rolled (in each case).