Question

You have two six-sided dice. Die A has 2 twos, 1 three, 1 five, 1 ten, and 1 fourteen on its faces. Die B has a one, a three, a five, a seven, a nine, and an eleven on its faces.A. What are the average expected values for die A and die B?B. On any given roll of both dice, what is the probability that the number showing on die A will be greater than the number on die B? What is the probability that the number showing on die B will be greater?C. Considering your answers for parts A and B, which die would you choose to roll, and why, if your goal was (i) to achieve a higher score than an opponent rolling the other die, and (ii) to achieve the highest possible score on a single roll of the die?

Answer 1

Answer:

a) $E(A) = 2 \frac{2}{6} + 3 \frac{1}{6} + 5\frac{1}{6} + 10 \frac{1}{6} + 14\frac{1}{6}=6$

$E(B) = 1 \frac{1}{6} + 3 \frac{1}{6} + 5\frac{1}{6} + 7 \frac{1}{6} + 9\frac{1}{6} + 11 \frac{1}{6}=6$

b) $P(A>B) =\frac{16}{36}= \frac{4}{9}$

$P(B>A) =\frac{18}{36}= \frac{1}{2}$

c) (i)

If the goal is to obtain a higher score than an opponent rolling the other die, It's better to select the die B because the probability of obtain higher score than an opponent rolling the other die is more than for the die A. Since P(B>A) > P(A>B)

(ii)

We see that the expected values of both the dies A and B were equal, so then a roll of any of the two dies would get us the maximum value required.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Solution to the problem

Part a

For this case we can use the following formula in order to find the expected value for each dice.

$E(X) =\sum_{i=1}^n X_i P(X_i)$

Die A has 2 twos, 1 three, 1 five, 1 ten, and 1 fourteen on its faces. The total possibilites for die A ar 2+1+1+1+1= 6

And the respective probabilites are:

$P(2) = \frac{2}{6}, P(3)=\frac{1}{6}, P(5) =\frac{1}{6}, P(10)=\frac{1}{6}, P(14) = \frac{1}{6}$

And if we find the expected value for the Die A we got this:

$E(A) = 2 \frac{2}{6} + 3 \frac{1}{6} + 5\frac{1}{6} + 10 \frac{1}{6} + 14\frac{1}{6}=6$

Die B has a one, a three, a five, a seven, a nine, and an eleven on its faces

And the respective probabilites are:

$P(1) = \frac{1}{6}, P(3)=\frac{1}{6}, P(5) =\frac{1}{6}, P(7)=\frac{1}{6}, P(9) = \frac{1}{6}, P(11)=\frac{1}{6}$

And if we find the expected value for the Die A we got this:

$E(B) = 1 \frac{1}{6} + 3 \frac{1}{6} + 5\frac{1}{6} + 7 \frac{1}{6} + 9\frac{1}{6} + 11 \frac{1}{6}=6$

Part b

Let A be the event of a number showing on die A and B be the event of a number showing on die B.

For this case we need to find P(B>Y) and P(B>A).

First P(A>B):

$P(A>B) = P(A -B>0)$

$P(\frac{(A-B)-E(A-B)}{\sqrt{Var(A-B)}} > \frac{0-E(A-B)}{\sqrt{Var(A-B)}})$

We can solve this using the sampling space for the experiment on this case we have 6*6 = 36 possible options for the possible outcomes and are given by:

S= {(2,1),(2,3),(2,5),(2,7),(2,9),(2,11), (2,1),(2,3),(2,5),(2,7),(2,9),(2,11), (3,1),(3,3),(3,5),(3,7),(3,9),(3,11), (5,1)(5,3),(5,5),(5,7),(5,9),(5,11), (10,1),(10,3),(10,5),(10,7),(10,9),(10,11), (14,1),(14,3),(14,5),(14,7),(14,9), (14,11)}

We need to see how in how many pairs the result for die A is higher than B, and we have:  (2,1), (2,1), (3,1), (5,1), (5,3), (10,1), (10,3), (10,5), (10,7), (10,9), (14,1), (14,3),(14,5),(14,7),(14,9), (14,11). so we have 16 possible pairs out of the 36 who satisfy the condition and then we have this:

$P(A>B) =\frac{16}{36}= \frac{4}{9}$

And for the other case when B is higher than A we have this: (2,3), (2,5), (2,7), (2,9), (2,11), (2,3), (2,5), (2,7), (2,9), (2,11), (3,5), (3,7), (3,9), (3,11), (5,7), (5,9), (5,11), (10,11). We have 18 possible pairs out of the 36 who satisfy the condition and then we have this:

$P(B>A) =\frac{18}{36}= \frac{1}{2}$

Part c

(i)

If the goal is to obtain a higher score than an opponent rolling the other die, It's better to select the die B because the probability of obtain higher score than an opponent rolling the other die is more than for the die A. Since P(B>A) > P(A>B)

(ii)

We see that the expected values of both the dies A and B were equal, so then a roll of any of the two dies would get us the maximum value required.