What is the expected sum of the numbers that appear on two dice, each biased so that a 3 comes u

What is the expected sum of the numbers that appear on two...

Key Concepts

Biased Probability Distribution

A biased probability distribution is one in which the outcomes do not all have the same likelihood of occurring. Instead, some outcomes are more probable than others. This type of distribution requires assigning different weights to outcomes when computing measures like the expected value.

Normalization in Probability

Normalization is the process of adjusting the weights in a probability distribution so that the total probability sums to one. When outcomes are weighted unequally, it is important to divide each weight by the sum of all weights to convert them into proper probabilities.

Expected Value

This is a measure of the central tendency of a random variable, calculated as the weighted average of all its possible outcomes, with weights being the probabilities of those outcomes. It represents the long-run average value you would expect after many trials.

Linearity of Expectation

This principle states that the expected value of the sum of random variables is equal to the sum of their expected values, regardless of whether the variables are independent. This is useful for calculating the expected sum of multiple random events, such as dice rolls.

Transcript

00:01 Okay, so for this problem, we are determining the expected sum when numbers appear on two dice, which are biased.

00:09 So three is twice as likely.

00:11 And so if we let x and y each be the random variable representing the numbers on each die, then it's important to note that the expected value of the sum of these numbers is going to be equal to the sum of their expected values.

00:31 Okay, and figuring out the expected value of just 1 to i is certainly going to be a little bit less complicated.

00:39 So to do that, we'll consider that the probability of rolling a 3, so let's work with variable x, is going to be equal to two times the probability of rolling anything else, okay, whether that was a 1, a 2, a 4, 5, or 6.

01:00 And so if we let, if we just let the probability that x equal to 3, if you're represented by q, right, then we can divide by two on both sides here, right? so you can divide by two on this side to get rid of it and then divide by two on this side.

01:18 We'll get that the probability of rolling anything else, right, is then equal to q over two.

01:31 Okay, i'm going to use the fact that the probability of these events.

01:33 Must sum to 1.

01:36 So the probability of getting a 1 plus the probability of getting a 2, plus all the probabilities up till 6 together must equal 1.

01:46 So what that means is that q, which is our probability of getting a 3, plus q over 2, but plus q over 2 when we get a 1, and we get a 2, a 4, 5, or 6, right? so that's happening 5 times must equal 1.

02:05 Well, we can just take this and solve for q.

02:09 And when we do that, we'll get that q is actually equal to two sevenths.

02:17 Right? and then since the probability of, and q is our probability of getting a three, right? but the probability of getting anything else is half of that...

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