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

a)
`E(A) = 2 (2)/(6) + 3 (1)/(6) + 5(1)/(6) + 10 (1)/(6) + 14(1)/(6)=6`

`E(B) = 1 (1)/(6) + 3 (1)/(6) + 5(1)/(6) + 7 (1)/(6) + 9(1)/(6) + 11 (1)/(6)=6`

b)
`P(A>B) =(16)/(36)= (4)/(9)`

`P(B>A) =(18)/(36)= (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.

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) = (2)/(6), P(3)=(1)/(6), P(5) =(1)/(6), P(10)=(1)/(6), P(14) = (1)/(6)`

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

`E(A) = 2 (2)/(6) + 3 (1)/(6) + 5(1)/(6) + 10 (1)/(6) + 14(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) = (1)/(6), P(3)=(1)/(6), P(5) =(1)/(6), P(7)=(1)/(6), P(9) = (1)/(6), P(11)=(1)/(6)`

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

`E(B) = 1 (1)/(6) + 3 (1)/(6) + 5(1)/(6) + 7 (1)/(6) + 9(1)/(6) + 11 (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(((A-B)-E(A-B))/(√(Var(A-B))) > (0-E(A-B))/(√(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) =(16)/(36)= (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) =(18)/(36)= (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.