# Throwing Dice

1. Nov 26, 2007

### Beowulf2007

1. The problem statement, all variables and given/known data

Given $$S = \{(r,w)| r,w = 1,2,\ldots, 6\}$$

Deduce the following three probability functions.

Probability that the number of eyes are red

(1)$$P_{R}(t) = \frac{1}{6}$$ for $$t \in \{1,2,\ldots 6 \}$$

Probability that the number of eyes are either red or white

(2)$$P_{Y}(t) = \frac{13-2t}{36}$$ for $$t \in \{1,2,\ldots 6 \}$$

Probability that the number of eyes are either red and white

(3)$$P_{Z}(t) = \frac{2t-1}{36}$$ for $$t \in \{1,2,\ldots 6 \}$$

3. The attempt at a solution

My Proof (1):

Since there is 6 sides on each dice the combined space $$S = 6 \cdot 6 = 36$$ and since there is 6 sides on each sides of red dice, then

$$\frac{6}{36} = \frac{1}{6} = P_{R}(t)$$

My Proof(2):

The Events of throwing the two dice are describe in the schema:

$$\begin{array}{|c| c| c| c| c| c| } \hline (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6)\\ \hline (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6)\\ \hline (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6)\\ \hline (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6)\\ \hline (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6)\\ \hline (6,1) & (6,2) & (5,3) & (5,4) & (6,5) & (6,6)\\ \hline \end{array}$$
Thus by in the schema:

$$\begin{array}{ccc} P(x = 1) = \frac{11}{36} & P(x = 2) = \frac{9}{36} & P(x = 3) = \frac{7}{36}\\P(x = 4) = \frac{5}{36} & P(x = 5) = \frac{3}{36} & P(x = 6) = \frac{1}{36} \end{array}$$

which can be describe by the function:

$$P_{Y}(t) = \frac{13-2t}{36}$$ for $$t \in \{1,2,\ldots 6 \}$$

Proof(3)

Thus by in the schema:

$$\begin{array}{ccc} P(x = 1) = \frac{1}{36} & P(x = 2) = \frac{3}{36} & P(x = 3) = \frac{5}{36}\\P(x = 4) = \frac{7}{36} & P(x = 5) = \frac{9}{36} & P(x = 6) = \frac{11}{36} \end{array}$$

which can be describe by the function:

$$P_{Y}(t) = \frac{2t-1}{36}$$ for $$t \in \{1,2,\ldots 6 \}$$

What You Guys say I have deduced the probability functions correctly?? Am I on the right track??

SIncerely Yours
Beowulf

Last edited: Nov 26, 2007
2. Nov 27, 2007

### HallsofIvy

Staff Emeritus
I am totally baffled by this! What in the world does the probability of a die coming up a particular number, or anything you have done, have to do with "eyes" being red or white?
Your definition of S says nothing about "eyes" and there is no mention of "eyes" in anything except the questions! What "eyes" are you talking about?

3. Nov 27, 2007

### Beowulf2007

Hello Hallsoft:

Thank You for Your reply. I am sorry that the context of the problem was not clear.

Here is the problem in its full context:

Suppose we throw a red and white die simultaneously. The possible outcome of an experiment such as this can be recorded as follows:

$$S = \{r,w\}|r,w = 1,2,\ldots, 6\}$$

where r is the number of eyes the die shows and w is the number of eyes that the white die shows.

R, Y, Z are random variables on the Space S and are defined as follows:

R(r,w) = r (number of eyes that the red dice shows)

Y(r,w) = $$r \wee w$$ (lowest number of eyes)

Z(r,w) = $$r \land w$$ (largest number of eyes).

What I am tasked with showing is that these random varibles can be expressed by probability functions below:

$$p_{R} (t) = p_{W}(t) = \frac{1}{6}$$ where $$t \in \{1,2,\ldots,6\}$$

$$p_{Y} (t) = \frac{13-2t}{36}$$ where $$t \in \{1,2,\ldots,6\}$$

$$p_{Y} (t) = \frac{2t-1}{36}$$ where $$t \in \{1,2,\ldots,6\}$$

This was what I am was trying to show in my orginal post. Does the assigment justify my atempted solution?