# Exercise 3.1: Probabilities when Rolling Dice

Sum  $S= R + B$  of two dice

We consider the random experiment  »rolling one or two dice«.  Both dice are fair (the six possible outcomes are equally probable) and distinguishable by their colors:

• The random quantity  $R = \{1, \ 2,\ 3,\ 4,\ 5,\ 6 \}$  denotes the number of eyes of the red cube.
• The random quantity  $B = \{1,\ 2,\ 3,\ 4,\ 5,\ 6 \}$  denotes the number of eyes of the blue cube.
• The random quantity  $S =R + B$  stands for the sum of both dice.

In this exercise different probabilities are to be computed with reference to the random variables  $R$,  $B$  and  $S$,
whereby the scheme indicated above can be helpful.  This includes the sum  $S$  as a function of  $R$  and  $B$.

Hints:

### Questions

1

Give the following probabilities:

 $\text{Pr}(R = 6)\ = \$ $\text{Pr}(B ≤ 2)\ = \$ $\text{Pr}(R = B)\ = \$

2

What are the following probabilities?

 $\text{Pr}(S = 3)\ = \$ $\text{Pr}(S = 7)\ = \$ $\text{Pr(odd sum)}\ = \$

3

State the following probabilities:

 $\text{Pr}\big [(R = 6)\ \cup \ (B =6)\big]\ = \$ $\text{Pr}\big[(R = 6)\ \cap \ (B =6)\big]\ = \$

4

What is the probability that the first  "6"  will be on the  $L$–th double roll?

 $L = 1\text{:}\hspace{0.5cm}\text{Pr(first „6”)} \ = \$ $L = 2\text{:}\hspace{0.5cm}\text{Pr(first „6”)} \ = \$ $L = 3\text{:}\hspace{0.5cm}\text{Pr(first „6”)} \ = \$

5

What is the probability of the event  »You need an even number of double rolls, to get the first "6"«  ?
Using the nomenclature given in subtask  (4):

 $\text{Pr(}L\text{ is even | first „6”)}\ = \$

### Solution

#### Solution

(1)  Assuming fair dice, the probability that a  "6"  is rolled with the red die:

$$\underline{{\rm Pr}(R=6) = 1/6} = 0.1667 \hspace{0.05cm},$$
• that a  "1"  or a  "2"  is rolled with the blue die:
$$\underline{{\rm Pr}(B\le 2) = 1/3} = 0.3333 \hspace{0.05cm},$$
• that both dice show the same number of points:
$$\underline{{\rm Pr}(R=B) = 6/36} = 0.1667 \hspace{0.05cm}.$$

The latter is based on the 2D representation on the specification sheet and on the "Classical Definition of Probability" corresponding to  $K/M$:

• $K = 6$  of the total  $M = 36$  equally probable elementary events  $R \cap B$  can be assigned to the event  $R=B$  derived from this.
• These lie on the diagonal.  Dice players speak in this case of a  "double".

(2)  The solution is again based on the Classical Definition of Probability:

• In  $K = 2$  of the  $M = 36$  elementary fields there is a  "3"   ⇒   ${\rm Pr}(S = 3) = 2/36\hspace{0.15cm}\underline{ = 0.0556}.$
• In  $K = 6$  of the  $M = 36$  elementary fields there is a  "7"  ⇒   ${\rm Pr}(S = 7) = 6/36\hspace{0.15cm}\underline{ = 0.1667}.$
• In  $K = 18$  of the  $M = 36$  fields there is an odd number   ⇒   ${\rm Pr}(S\text{ is odd}) = 18/36\hspace{0.15cm}\underline{ = 0.5}.$

• This last result could be obtained in another way:
$${\rm Pr}(S\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}) = {\rm Pr}\big [(R\hspace{0.12cm}{\rm is \hspace{0.15cm} odd}) \cap (B\hspace{0.12cm}{\rm is\hspace{0.12cm} even}) \big ] + {\rm Pr}\big [(R\hspace{0.12cm}{\rm is\hspace{0.12cm} even}) \cap (B\hspace{0.12cm}{\rm is\hspace{0.12cm} odd})\big ]\hspace{0.05cm}.$$
• With  ${\rm Pr}(R\hspace{0.12cm}{\rm is\hspace{0.12cm} even}) = {\rm Pr} (R\hspace{0.12cm}{\rm is\hspace{0.12cm} odd}) = {\rm Pr}(B\hspace{0.12cm}{\rm is\hspace{0.12cm} even})= {\rm Pr}(B\hspace{0.12cm}{\rm is\hspace{0.12cm} odd}) = 1/2$  it also follows:
$${\rm Pr}(S\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}) = 1/2 \cdot 1/2 + 1/2 \cdot 1/2 = 1/2 \hspace{0.05cm}.$$

(3)  The probability for the event that at least one of the two dice shows a  "6" is:

$${\rm Pr}\big [(R= 6) \cup (B= 6) \big ] = K/M = 11/36 \hspace{0.15cm} \underline{= 0.3056} \hspace{0.05cm}.$$
• The second probability stands alone for the  "double sixes":
$${\rm Pr}\big [(R= 6) \cap (B= 6) \big ] = K/M = 1/36 \hspace{0.15cm} \underline{= 0.0278} \hspace{0.05cm}.$$

(4)  The result for  $L = 1$  has already been determined in subtask  (3) :

$$p_1 = {\rm Pr}\big [(R= 6) \cup (B= 6) \big ] = {11}/{36} \hspace{0.15cm} \underline{= 0.3056} \hspace{0.05cm}.$$
• The probability  $p_2$  can be expressed with  $p_1$  as follows:
$$p_2 = (1 - p_1) \cdot p_1 = \frac{25}{36} \cdot \frac{11}{36} \hspace{0.15cm} \underline{= 0.2122} \hspace{0.05cm}.$$
In words, the probability that a  "6"  is rolled for the first time in the second roll is equal to the probability that no  "6"  was rolled in the first roll   ⇒   probability  $1-p_1$, but there is at least one  "6"  in the second roll   ⇒   probability  $p_1$.
• Correspondingly, for the probability "first 6 in the third throw":
$$p_3 = (1 - p_1)^2 \cdot p_1 = \frac{25}{36} \cdot \frac{25}{36} \cdot\frac{11}{36} \hspace{0.15cm} \underline{= 0.1474} \hspace{0.05cm}.$$

(5)  By extending the sample solution to subtask  (4) , we obtain:

$$\text{Pr(}L\text{ is even | first „6”)}\ = p_2 \hspace{-0.05cm}+ \hspace{-0.05cm}p_4 \hspace{-0.05cm}+ \hspace{-0.05cm} p_6 \hspace{-0.05cm}+ \hspace{-0.05cm} \text{...} = (1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1) \cdot p_1 \hspace{-0.05cm}+ \hspace{-0.05cm} (1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1)^3 \cdot p_1 \hspace{-0.05cm}+ \hspace{-0.05cm}(1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1)^5 \cdot p_1 \hspace{-0.05cm}+ \hspace{-0.05cm} \text{...} = (1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1) \cdot p_1 \cdot \left [ 1 \hspace{-0.05cm}+ \hspace{-0.05cm} (1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1)^2 \hspace{-0.05cm}+ \hspace{-0.05cm} (1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1)^4 +\text{...}\hspace{0.05cm} \right ] \hspace{0.05cm}.$$
• Accordingly, we obtain for the probability of the complementary event:
$${\rm Pr}(L\text{ is odd | first „6”)} = p_1 + p_3 + p_5 + \text{...} = p_1 \cdot \left [ 1 + (1 - p_1)^2 + (1 - p_1)^4 + \text{...} \hspace{0.15cm} \right ] \hspace{0.05cm}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{{\rm Pr}(L\text{ is odd | first „6”)}) } {{\rm Pr}(L\text{ is even | first „6”)})} = \frac{1}{1 - p_1} \hspace{0.05cm}.$$
• Further, must hold::
$${\rm Pr}(L \text{ is even | first „6”)} + {\rm Pr}(L \text{ is odd | first „6”)} = 1$$
$$\Rightarrow \hspace{0.3cm} {\rm Pr}(L \text{ is even | first „6”)} \cdot \left [ 1 + \frac{1}{1 - p_1} \right ] = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Pr}(L\hspace{0.15cm}{\rm is\hspace{0.15cm} even}) = \frac{1 - p_1}{2 - p_1} = \frac{25/36}{61/36} = \frac{25}{61} \hspace{0.15cm} \underline{= 0.4098} \hspace{0.05cm}.$$