Main content

### Course: Precalculus > Unit 8

Lesson 9: 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
- Probability and combinatorics: FAQ

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Expected payoff example: lottery ticket

We can find the expected payoff (or the expected net gain) of a certain lottery ticket by taking the weighted average the outcomes. Created by Sal Khan.

## Want to join the conversation?

- if you paid 10,000 tickets would there be a 9/10000 probability of winning (like 1111, 2222, etc)? Why is it only 1/10000(9 votes)
- It is 1/10,000 since it is saying that they win if they match all 4 digits in the correct order. It isn't saying that all 4 digits are the exact same number like 3333, it is saying that the numbers could be any possibility (like 1234), but there is only 1/10,000 probability for that exact number in the same order. Hope you understand it. The wording also made me confused for a while. If you have any questions do let me know.(10 votes)

- In the second way that Sal calculates the net value. He says that the win is NOT guaranteed, but if you buy a ticket for every possible outcome would that not guarantee a win? The timestamp is3:22

In this example, you would spend the $10,000 and "guess" 0000, 0001, 0002 ... 9997, 9998, and lastly 9999. I am saying that this would be the same a 9999/9999 = 1 the total number of possible outcomes there for a guaranteed win, once.(4 votes)- If you were to bet randomly, then you wouldn't be guaranteed to pick all the different outcomes exactly once. That's probably what Sal meant, even though it obviously makes more sense to mindfully pick each possibility once.(2 votes)

- How can the probability of win be 1/10000?(2 votes)
- As the instructor explains, there are 10000 possible combinations. 0000 (combination 1), 0001 (combination 2), all the way too 9998 (combination 9999) and 9999 (combination 10000). If you match all four numbers (which there are 10000 combinations, as we have established) then you will win.(2 votes)

- I'm a bit confused, isn't the probability of winning 10/1000?

Here are the 10 possibilities: 0000, 1111, ... 9999.

So those are 10 of 10 000.(2 votes) - why do you keep making this harder(1 vote)
- What does x3 y3 z3 k mean?(1 vote)
- I know its given by the questions that possible selection is 10,000. But what if its not given, like its ask this way:
**A "pick 4" lottery game... labeled from 0~9. Play can choose**... (*the rest are the same as question*)

Shouldnt the possible outcome be 10 chose 4, which equals to 5040?

0~10 (n=10)

r=4

nCr= 10C4=5,040

How do we get 10,000?

update:`nvm, i figured it out. I misunderstood that the # 0~9 are not without replacement. Since this doesnt say so, combinations is just 10*10*10*10=10,000`

(1 vote)

## Video transcript

- [Instructor] We're told
a Pick 4 lottery game involves drawing four numbered
balls from separate bins, each containing balls labeled from 0 to 9. So there are 10,000 possible
selections in total. For example, you could get a 0, a 0, a 0 and a 0, a 0, a 0, a 0 and a 1, all the way up to 9,999, four nines. Players can choose to play a
straight bet, where the player wins if they match all four
digits in the correct order. The lottery pays $4,500 on a
successful $1 straight bet. Let X represent a player's
net gain on a $1 straight bet. Calculate the expected net gain. And they say, hint, the expected
net gain can be negative. So why don't you pause this video and see if you can calculate
the expected net gain? All right. So there's a couple of ways
that we can approach this. One way is to just think about
the two different outcomes. There's a scenario where you
win with your straight bet. There's a scenario where you
lose with your straight bet. Now let's think about the net gain in either
one of those scenarios. The scenario where you win, you pay $1, we know it's a $1 straight
bet, and you get $4,500. So what's the net gain? So it's going to be $4,500 minus one. So your net gain is going to be $4,499. Now what about the net gain in
the situation that you lose? Well, in the situation that you lose, you just lose a dollar. So this is just going to be
negative $1 right over here. Now let's think about the probabilities of
each of these situations. So the probability, so
the probability of a win we know is 1 in 10,000, 1 in 10,000. And what's the probability of a loss? Well, that's going to be 9,999 out of 10,000. And then our expected net gain is just going to be the weighted
average of these two. So I could write our expected net gain is going to be 4,499 times the probability of that, 1 in 10,000 plus negative 1 times this, so that I could just write that as minus 9,999 over 10,000. And so this is going to
be equal to, let's see, it's going to be 4,499 minus 9,999, all of that over 10,000. And let's see, this is
going to be equal to negative 5,500 over 10,000, negative 5,500 over 10,000, which is the same thing
as negative 55 over 100, or I could write it this way. This is equal to negative .55. I could write it this way, 0.55. So that's one way to calculate
the expected net gain. Another way to approach
it is to say, all right, what if we were to get 10,000 tickets? What is our expected net
gain on the 10,000 tickets? Well, we would pay $10,000 and we would expect to win once. It's not a guarantee, but
we would expect to win once. So expect 4,500 in payout. And so you would then, let's
see, you would have a net gain of, it would be negative $5,500, negative $5,500. Now this is the net gain
when you do 10,000 tickets. Now, if you wanted to find the
expected net gain per ticket you would then just divide by 10,000. And if you did that, you would get exactly what we
just calculated the other way. So any way you try to approach this this is not a great bet.