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## Statistics and probability

### Course: Statistics and probability>Unit 9

Lesson 8: More on expected value

# Getting data from expected value

Expected value refers to the average outcome you would expect from repeating an experiment over and over. It is calculated by multiplying each possible outcome by its probability of occurring, and summing those products together. If we know the expected value, we can go backwards and solve for frequency. Created by Sal Khan.

## Want to join the conversation?

• I got my feeling of not understanding a bit and frustrated
when, at he divides 67.4 by 20. I can´t get the intuition behind the steps he makes. I could memorize it by the sake of memorizing but i really want to grasp the understanding in order to be able to solve problems on my own •   Suppose that the average number of children per person is 1.5. This is the expected value. In a group of 100 people you would expect to find 1.5 * 100 = 150 chilldren. This is just the expected sum of children among 100 people. If you want to go back to the expected value, you need to divide the expected sum (150) by the number of observations (# of people) you are considering: 150/100 = 1.5.

Sal is doing the same thing: dividing the expected sum (67.4) by the number of observations (20) to get the expected value (3.37).
• why do you multiply by the die value? • So for the dice problem you can think of it like getting an average of the results.
So if you rolled say 2, 5, 5 then your average value would be (2 + 5 + 5)/3 = 4
So you need to include the face value of the dice roll.

As for your comment where for the di face "there was a color" that's not really going to work the same way. In the dice problem rolling a 5 and 1 is the same as rolling 3 and 3. But if you're dice were colored and you roll yellow then blue then how would you add or average those values? In order to make it work for colors you'd probably need to assign the colors number values or do something else to make the problem make sense.
• Why couldn't he just add all the known frequency numbers (110, 95, et cetera), subtract it from the total (500), and divide it by 2? That is what I did when he asked us to pause the video at the beginning, and I found the same answer that he did in the end. • Because it was a coincidence that both values were 75. The first part of your reasoning is correct, if you sum the know frequencies and subtract from 500 you get a number that must be divided between the 2 missing frequencies, but you didn't know how the 150 were distributed before the rest of the procedure.

For example, with exactly the same data, but changing the expected value of the sum of 20 rolls to be 52.4, you would find that the first number is 150 and the second number is 0.
• I don't really get why do we multiply by the die value? It doesn't make any sense to me... Can anyone explain that as simple as possible? I'm really confused! :( • Expected value of a dice roll assumes that the face of the dice is the value.
So notice you can't really do this with a coin. Unless you assign a number value to Heads and Tails a coin has no number value.

So for example,
If you roll 2 dice and get 2 and 4 then the average value is 3.
If you roll 3 and 4 then the average is 3.5.

So you could experimentally roll a hundred dice or a thousand or whatever and calculate the average and that would be an estimate for the Expected Value.

That's exactly the same as the problem in this video. By summing up all of the probabilities * values he finds the average or the estimate for the Expected Value.

Hope that helps.
• At you mention you are taking a weighted sum of the values, isn't there an equal chance of any number between 1-6 coming up when you role a dice? If so why should any number be weighted more than another? • The numbers 1-6 have an equal theoretical probability. That is, we know that the six sides are equally likely, and so we expect 1/6 of the rolls to come up as each number. For example, in 600 rolls we'd expect 100 of each number.

In practice, the observed results will vary slightly from the theoretical probability. The weighted sum Sal is calculating is the observed expected value (that is, the sample mean), he's using the sample data. So in 600 rolls there might be only 95 1's, and 105 2's, and those determine the weights for the sample mean.

It's somewhat odd to use "expected value" for observed results instead of theoretical results, but that's what Sal is doing.
• I dont understand the difference between The expected Value (E(x)) and Mean Value (μ). Thank you.
(1 vote) • The mean of a set of values is simply the average of those values. For instance, if you had the set of numbers {1, 5, 7}, then the mean would be (1+4+7)/3 = 4.

Expected value can be very similar to (and sometimes even equal to) the mean. However, it requires an extra step: you must take into account that different outcomes may have different probabilities. Here's an example:

Let's say you had a machine that gave you a random amount of money each day, and that on any day it had a 20% chance of giving you \$50, a 50% chance of giving you \$100, and a 30% chance of giving you \$300. How much money could you expect to make after many many days? (Say, 1,000 days for example).

Well, based on the numbers above, we could say that we would probably make \$50 on about 200 of our days (because 20% of 1000 is 200), \$100 on about 500 of our days, and \$300 on about 300 of our days. Therefore, our total profits would be \$50 x 200 + \$100 x 500 + \$300 x 300 = \$150,000. So, based on this, how much money did we make each day on average? Well, \$150,000 / 1000days = \$150/day. This is the expected value.

The formula above can be simplified if we use 1 day instead of 1000 days. Essentially, this is done the same way, except that you'd substitute in the number 1 wherever you see the number 1000, and you'd get 20% x 1 day x \$50 + 50% x 1 day x \$100 + 30% x 1 day x \$300 = \$150/day.

In some cases, expected value can actually equal the mean. This happens when every possible outcome has the same probability of occurring. For example, rolling a die.

If you roll a die, each outcome has a 1/6 chance of occurring (assuming it's a fair die). Therefore, the expected value would be 1/6 x 1 + 1/6 x 2 + 1/6 x 3 + 1/6 x 4 + 1/6 x 5 + 1/6 x 6. This can be simplified to 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6, which can then be simplified to (1+2+3+4+5+6)/6. This is exactly the same as the mean.

Anyways, great question! And I hope my comment helped clear things up a bit.
• Why does an expected value of the sum of 20 rolls being 67.4 mean that the expected value of one roll is 67.4/20? • Why isn't expected value of 1 roll 3.5? Was it from experiment? I thought you could calculate expected value before experiment, just from probability of each event happening.
(1 vote) • In this case, Sal is calculating the expected value according to the data that Jamie collected.

If the die is in fact a 'fair' die then you would be exactly right that the expected value of a given roll is 3.5. In this case however, Sal is calculating the expected value according to the data collected because that will provide information to figure out what the two washed away values were.  