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## High school statistics

### Course: High school statistics > Unit 7

Lesson 4: Expected value- Mean (expected value) of a discrete random variable
- Mean (expected value) of a discrete random variable
- Interpreting expected value
- Interpret expected value
- Expected payoff example: lottery ticket
- Expected payoff example: protection plan
- Find expected payoffs

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# Interpreting expected value

AP.STATS:

VAR‑5 (EU)

, VAR‑5.D (LO)

, VAR‑5.D.1 (EK)

CCSS.Math: We can interpret expected value as a long term average outcome. This example looks at expected value in the context of a lottery ticket. Created by Sal Khan.

## Want to join the conversation?

- At0:49, how is he getting the probability the probability of wining to be1:51instead of1:50?(4 votes)
- The question states that the odds of winning are 1∶50,

which means that if 𝑝 is the probability of winning, then the probability of losing is 50𝑝.

Since winning or losing are the only possible outcomes when playing the game,

we know that 𝑝 + 50𝑝 = 1

⇒ 51𝑝 =1

⇒ 𝑝 = 1∕51(7 votes)

- If anyone is curious on how much you would win.

Suppose you win $x. Then it follows that 1/50 * x - 2 = 0.95. Solve to get x = 147.50.(1 vote)

## Video transcript

- [Instructor] We're told a
certain lottery ticket costs $2 and the back of the ticket says, "The overall odds of winning
a prize with this ticket are one to 50, and the expected return
for this ticket is $0.95." Which interpretations of the
expected value are correct? Choose all answers that apply. Pause this video, have a go at that. All right, now let's go
through each of these choices. So choice A says the probability that one of these tickets wins
a prize is 0.95 on average. Well, I see where they're
getting that 0.95. They're getting it from right over here, but that's not the probability
that you're winning, that's the expected return. The probability that
you win is much lower. If the odds are one to 50, that means that the probability
of winning is one to 51. So it's a much lower probability
than this right over here. So definitely rule that out. Someone who buys this ticket
is most likely to win $0.95. That is not necessarily the case either. We don't know what the different
outcomes are for the prize. It's very likely that there's no outcome for that prize where
you win exactly $0.95. Instead, there's likely to be
outcomes that are much larger than that with very low probabilities, and then when you take
the weighted average of all of the outcomes, then you get an expected return of $0.95. So it's actually maybe even
impossible to win exactly $0.95. So I would rule that out. If we looked at many of these tickets, the average return would
be about $0.95 per ticket. That one feels pretty interesting, 'cause we're looking at
many of these tickets. And so across many of them, you
would expect to, on average, get the expected return as your return. And so this is what we are seeing here. The average return would be about that. It would be approximately that. So I like that choice. That is a good interpretation
of expected value. And then choice D, if 1,000 people each bought
one of these tickets, they'd expect a net gain
of about $950 in total. This one is tempting. Instead of net gain,
if it just said return, this would make a lot of sense. In fact, it would be completely
consistent with choice C. If you have 1,000 people,
that would be many tickets, and if on average, if their average return is about $0.95 per ticket, then their total return
would be about $950, but they didn't write return
here, they wrote net gain. Net gain would be how much you
get minus how much you paid. And 1,000 people would have to pay, if they each got a
ticket, would pay $2,000. So they would pay 2,000. They would expect a return of $950. Their net gain would
actually be negative $1,050. So we would rule that one out as well.