What's the expected value of this problem (random variable)?

[EdmureTully]

Homework Statement

What's the expected value of this problem (random variable)?

X: represent the result of dice number 1 - result of dice number 2

example dice 1 first roll = 2; second roll = 3
dice 2 first roll = 1; second roll = 2

X = 2+3 -(1+2) = 2

what's the expected value of X

Homework Equations

The Attempt at a Solution

I think it's 0, because the Expected value of dice 1 is the same as that of dice 2

 

[Seydlitz]

I assume you know how to create a table for discrete random variable, showing the outcome of $X$ and it's probability $P(X)$. Do that first before you try to calculate the $E(x)$

 

[EdmureTully]

I don't even know that. Our teacher didn't show us any of that. He just skipped over a lot of material.

It's not a homework problem. I just tried to think of an example that may eventually help me in doing my assignment.

Also, if you happen to have a document with a lot of examples that shows you how to compute E(X) that is easy to understand, don't hesitate. Thank you.

 

[Seydlitz]

I don't even know that. Our teacher didn't show us any of that. He just skipped over a lot of material.

It's not a homework problem. I just tried to think of an example that may eventually help me in doing my assignment.

Also, if you happen to have a document with a lot of examples that shows you how to compute E(X) that is easy to understand, don't hesitate. Thank you.

I'm referring to this table:

Ok why don't you look at this page here: (You can no doubt search more in google also)
http://gwydir.demon.co.uk/jo/probability/calcdice.htm

Or this video from Khan's Academy
https://www.khanacademy.org/math/probability/random-variables-topic/random_variables_prob_dist/v/expected-value--e-x

 

[haruspex]

I'm referring to this table:

I don't understand the relationship between that table and the OP. That table appears to be more to do with tossing a coin 5 times than with rolling a die twice.
More usefully, if you have two random variables, X and Y, then E(X+Y) = E(X)+E(Y). You can use this both to get the expected value of the sum of two rolls of a die, and to get the expected difference from the other sum.

 

[Seydlitz]

I don't understand the relationship between that table and the OP. That table appears to be more to do with tossing a coin 5 times than with rolling a die twice.
More usefully, if you have two random variables, X and Y, then E(X+Y) = E(X)+E(Y). You can use this both to get the expected value of the sum of two rolls of a die, and to get the expected difference from the other sum.

I'm just giving an example of discrete distribution table from the internet. Maybe he can recall if it was previously taught.

 

[HallsofIvy]

The smallest possible value of a single roll of a single die is 1. The largest possible value is 6. The smallest possible value of "the result of die #1- the result of die #2" is 1- 6= -5 and that is possible only with "1" and "5". What is the probability of that?

You can get -4 in two different ways: 1- 5 and 2- 6. What are the probabilities of "1 and 5" and "2 and 6"? So what is the probability of "1 and 5 or 2 and 6"?

Continue that until you get to +5, given by 6- 1. You should quickly see a pattern that will simplify the calculations.

 

[EdmureTully]

Didn't see that. I don't even know what that means.

 

[EdmureTully]

I don't understand the relationship between that table and the OP. That table appears to be more to do with tossing a coin 5 times than with rolling a die twice.
More usefully, if you have two random variables, X and Y, then E(X+Y) = E(X)+E(Y). You can use this both to get the expected value of the sum of two rolls of a die, and to get the expected difference from the other sum.

What did you do? Did you use the Discrete Binomial Distribution and why do you have 10/32 in the middle?

 

[EdmureTully]

The smallest possible value of a single roll of a single die is 1. The largest possible value is 6. The smallest possible value of "the result of die #1- the result of die #2" is 1- 6= -5 and that is possible only with "1" and "5". What is the probability of that?

You can get -4 in two different ways: 1- 5 and 2- 6. What are the probabilities of "1 and 5" and "2 and 6"? So what is the probability of "1 and 5 or 2 and 6"?

Continue that until you get to +5, given by 6- 1. You should quickly see a pattern that will simplify the calculations.

The pattern I see leads me to believe it's 0.

 

[HallsofIvy]

I was hoping for a pattern for the probabilities of "-5", "-4", ..., "4", "5": probability of -5 is 1/36, -4 is 2/36= 1/18, -3 is 3/36= 1/12, etc. But if you mean that the expected value is 0, yes, that is correct because of the symmetry: a- b and b- a are equally likely.