Statistics Expected Variable Question

[thomas49th]

Homework Statement

Y is the larger score showing when two dice are thrown
Calculate E(Y)

Homework Equations

Presume the dice are fair

The Attempt at a Solution

Hmm not sure. I've listed combinations of throw events on paper but I don't think that will help. Y can never be 1. Y can never be the same as the other die?
Do I want to draw a probability distrubution table?

I've been messing around with it for a while with not much progress

Thanks :)

 

[Mark44]

If both dice come up with 1s, if you don't count the 1, you won't have anything to count for that roll.

With two dice, there are 36 possible outcomes: {(1, 1), (1, 2), ... , (1, 6), (2, 1), ... (2, 6), (3, 1), ..., (3, 6), (4, 1), ..., (4, 6), (5, 1), ..., (5, 6), (6, 1), ..., (6, 6)}.

I think that for the expected value of Y, E(Y), you add together the larger number in all 36 pairs, and divide by the number of pairs, which is 36. For the 6 pairs that have the same value, pick either number to be the larger; it's certainly not the smaller of the two.

 

[Architeuthis Dux]

I would approach this problem by using the basic principles of probability and statistics. Since the dice are fair, we can assume that each number on the dice has an equal chance of being rolled. Therefore, the possible outcomes for Y are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12. To calculate E(Y), we will use the formula E(Y) = Σ(y * P(y)), where y represents the possible outcomes and P(y) represents the probability of each outcome.

First, we need to determine the probability of each outcome. This can be done by creating a probability distribution table, as you mentioned. The table would look like this:

Outcome (Y) | Probability (P(Y))
2 | 1/36
3 | 2/36
4 | 3/36
5 | 4/36
6 | 5/36
7 | 6/36
8 | 5/36
9 | 4/36
10 | 3/36
11 | 2/36
12 | 1/36

To calculate the expected value, we will multiply each outcome by its probability and then sum up the results. This can be represented as Σ(y * P(y)) = (2 * 1/36) + (3 * 2/36) + (4 * 3/36) + (5 * 4/36) + (6 * 5/36) + (7 * 6/36) + (8 * 5/36) + (9 * 4/36) + (10 * 3/36) + (11 * 2/36) + (12 * 1/36) = 7.

Therefore, the expected value of Y is 7. This means that if we were to roll the two dice many times, the average value of Y would be 7. I hope this helps you in your calculations.