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### Course: Statistics and probability > Unit 9

Lesson 8: More on expected value- Term life insurance and death probability
- Getting data from expected value
- Expected profit from lottery ticket
- Expected value while fishing
- Comparing insurance with expected value
- Expected value with empirical probabilities
- Expected value with calculated probabilities
- Making decisions with expected values
- Law of large numbers

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# Expected value while fishing

Sal walks through an example where he multiplies probability by values and sums to find expected value. Created by Sal Khan.

## Want to join the conversation?

- The problem is quite simple when we only need to list all 8 possibilities in Bet 2 (at least 2 sunfish of the next 3 fish). What if we have to deal with a bigger bet, for example: at most 4 sunfish of the next 15 fish?(19 votes)
- You can still calculate the number of unique options without listing them all out. In your example, there would be 2^15 total options. The number of ways that you could pick 4 things from 15 (15C4) could be multiplied by 1/(2^15) to find the probability of getting exactly 4 out of 15. You could repeat that for 0 out of 15, 1 out of 15, 2 out of 15 and 3 out of 15. The sum of those would provide the probability of getting at most 4 out of 15.(3 votes)

- is there a way to solve the P(bet 2) without listing all of the outcomes? I know you love to draw out everything but i find blatantly ignoring the equations is counter productive(13 votes)
- There is indeed. There are different ways to catch at least 2 sunfish so we can calculate the probability of one of those ways and multiply it by the number of ways. So what are the odds of catching a sunfish then a trout then a sunfish? (10/20)*(10/20)*(10/20)=(1/2)^3=1/8. And how many ways are there to catch at least 2 sunfish? There are 8 possible combinations of fish and half of those have at least 2 sunfish. 8/2=4

4*1/8=1/2(0 votes)

- at3:11sal said there will be chance of 9/20 chance for sunfish. I think in that case there will be chance of 9/19. Am I correct?(8 votes)
- Yes, he made a mistake or misspoke, but does not affect his solution of the problem posed (since he was discussing a how the problem would differ if they were not replacing the fish after catching them).(4 votes)

- How do you find the probability of catching at least two sunfish of the next three you catch without listing the outcomes?(4 votes)
- Every fish has two equally likely possibilities – either it's a sunfish or it's not a sunfish.

So, with three fish there are 2³ = 8 possible and equally likely outcomes.

Of those, there are 3C2 = 3 outcomes with exactly two sunfish, and 3C3 = 1 outcome with three sunfish.

So, the probability that at least two of the three fish are sunfish is

(3 + 1)∕8 = 1∕2(11 votes)

- For Bet 2, I was thinking why we couldn't calculate the possibility of catching at least 2 sunfish as (3C2)*(1/2)*(1/2). The explanation for this calculation is to randomly choose two out of the three catches to get two sunfish. For the other catch, we do not consider what it is since it would not affect the result. I understand the right calculation is (3C2)*(1/8)＋(3C3)*(1/8), but why the other way doesn't work out?(3 votes)
- > "why we couldn't calculate the possibility of catching at least 2 sunfish as (3C2)*(1/2)*(1/2). The explanation for this calculation is to randomly choose two out of the three catches to get two sunfish. For the other catch, we do not consider what it is since it would not affect the result."

But we do care about the result - kind of. The`(3C2)`

part of this calculation is saying that we catch three fish, and exactly two of them are sunfish. That means that one of the fish is a trout, which means we need to account for the probability of catching that trout.

If the "unknown" fish was a sunfish, then the`(3C2)`

would be incorrect. The case of catching 3 sunfish is handled in the part of the solution with the`(3C3)`

calculation.(4 votes)

- At1:12, Sal defines the random variable X as "what your profit is from bet 1." He then goes on to define a function E(X) that gives the expected profit from taking the first bet. Shouldn't X be defined as the outcome of Jeremy's next three fish?

Again at6:55, Sal similarly defines Y as the expected profit from the second bet, which is what E(Y) is. Shouldn't Y be defined as the outcome of my next three fish?(3 votes)- He makes a small mistake at6:55when he says Y is the "expected" profit.

Y should be defined the same as X as just the profit of the bet.

E(Y) is the expected profit.(2 votes)

- Would there really every be a scenario where you wouldn't take a bet all of the time (assuming both are positive profits)? Technically no?(2 votes)
- Only if you don't have enough money to play through (hopefully temporary) negative results, or if the payoff is not enough to justify the time it takes you. There is always a small probability of unlikely results.(1 vote)

- At3:40ish wouldn't the probability be a dependent event since there is a set amount of fish in the pond/lake?(1 vote)
- Fish after catching are being released back into the pond that is why the probability is independent of what was caught.(1 vote)

- If I have to calculate the expected number of points earned in a test of 4 questions (each one with 2 options), how can I know the number of relevant outcomes without listing all the posibilities? (The test is randomly answered and I know the number of points as a function of the number of correct answers).(1 vote)
- But how is each possibility equally likely? In Bet 2 , there are three ways to catch 2 fish. If we look at case where 2nd and 3rd catch is a fish then probability becomes 10/20 * 10/19 * 10/ 19. Similarly probability of each possible outcome differs slightly. So how can we just count number of required outcomes(catching 2 or more fish) and say its probability is 4/8?(1 vote)
- You are correct that not all outcomes in Bet 2 are equally likely. The statement that each of the eight possibilities is equally likely is incorrect. In fact, as you pointed out, some outcomes have different probabilities depending on the order in which the fish are caught. For example, catching two sunfish in the first two attempts and then a trout in the third attempt has a different probability than catching a sunfish, a trout, and then another sunfish. To accurately calculate the probability of catching at least two sunfish, you need to consider all possible combinations of catching sunfish and trout and calculate their respective probabilities.(1 vote)

## Video transcript

You and your friend Jeremy are fishing in a pond that contains ten trout and ten sunfish. Each time one of you catches a fish you release it back into the water. Jeremy offers you the choice of two different bets. Bet number one. We don't encourage betting but I guess Jeremy wants to bet. If the next three fish he catches are all sunfish you will pay him 100 dollars, otherwise he will pay you 20 dollars. Bet two, if you catch at least two sunfish of the next three fish that you catch he will pay you 50 dollars, otherwise you will pay him 25 dollars. What is the expected value from bet one? Round your answer to the nearest cent. I encourage you to pause this video and try to think about it on your own. Let's see. The expected value of bet one. The expected value of bet one where we'll say bet one is -- Let's just define a random variable here just to be a little bit better about this. Let's say x is equal what you pay, or I guess you could say , because you might get something, what your profit is from bet one. It's a random variable. The expected value of x is going to be equal to, let's see. What's the probability, it's going to be negative 100 dollars times the probability that he catches three fish. The probably that Jeremy catches three sunfish, the next three fish he catches are going to be sunfish, times 100 dollars. Or I should, well you're going to pay that. Since you're paying it we'll put it as negative 100 because we're saying that this is your expected profit, so you're going to lose money there. That's going to be one minus this probability, the probability that Jeremy catches three sunfish. In that situation he'll pay you 20 dollars. You get 20 dollars there. The important thing is to figure out the probability that Jeremy catches three sunfish. Well the sunfish are 10 out of the 20 fish so any given time he's trying to catch fish there's a 10 in 20 chance, or you could say one half probability that it's going to be a sunfish. The probability that you get three sunfish in a row is going to be one half, times one half, times one half. They put the fish back in, that's why it stays 10 out of the 20 fish. If he wasn't putting the fish back in then the second sunfish you would have a nine out of 20 chance of the second one being a sunfish. In this case they keep replacing the fish every time they catch it. There is a one eighth chance that Jeremy catches three sunfish, so this right over here is one eighth. And one minus one eighth, this is seven eighths. You have a one eighth chance of paying 100 dollars and a seven eights chance of getting twenty dollars so this gets us to ... Your expected profit here, there's a one eighths chance, one eighth probability, that you lose 100 dollars here, so times negative 100. But then there is a seven eighths chance that you get -- I'll just put parentheses here to make it clear. I think the order of operations of the calculator would have taken care of it but I'll just do it so that it looks the same. Seven eighths, there's a seven eighths chance that you get 20 dollars. Your expected payoff here is positive five dollars. Your expected payoff here is equal to five dollars. This is your expected value from bet one. Now let's think about bet two. If you catch at least two sunfish of the next three fish you catch he will pay you 50, otherwise you will pay him 25. Let's think about the probability of catching at least two sunfish of the next three fish that you catch. There's a bunch of ways to think about this but since there's only three times that you're trying to catch the fish and there's only one of two outcomes you could actually write all the possible outcomes that are possible here. You could get sunfish, sunfish, sunfish. You could get, what's the other type of fish that you have? Oh, trout. You could have sunfish, sunfish, trout. You could have sunfish, trout, sunfish. You could have sunfish, trout, trout. You could have trout, sunfish, sunfish. You could have trout, sunfish, trout. You could have trout, trout, sunfish. Or you could have all trout. You see here that each of these, each time you go there's two possibilities, each time you try to catch a fish there's two possibilities, so if you're doing it three times there's two times two times two possibilities. One, two, three, four, five, six, seven, eight possibilities here. Now out of these eight equally likely possibilities how many of them involve you catching at least two sunfish? You catch at least two sunfish in this one, in this one, in that one, in this one and I think that is it. Yep, this is only one sunfish, one sunfish, one sunfish and no sunfish. In four out of the eight equally likely outcomes you catch at least two sunfish. Your probability of catching at least two sunfish is equal to four eighths or one half. Let's see, what's the expected value? Let's say Y is the expected profit from bet. Let's let Y equals, another random variable, is equal to expected profit from bet two. The expected value of our random variable Y, you have a one half chance that you win. You have a one half chance of getting 50 dollars and then you have the one half chance, the rest of the probability. If there's a one half chance you win there's going to be a one minus one half or essentially a one half chance that you lose. So you have a one half chance of having to pay 25 dollars. Let's see what this is. This is one half times 50 plus one half times negative 25. This is going to be 25 minus 12.50, which is equal to 12.50. Your expected value from bet two is 12.50. Your friend says he's willing to take both bets a combined total of 50 times. If you want to maximize your expected value what should you do? Well bet number two -- Actually both of them are good bets, I guess your friend isn't that sophisticated, but bet number two has a higher expected payoff, so I would take bet two all of the time. I would take bet two all of the time.