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### Course: Statistics and probability>Unit 9

Lesson 8: More on expected value

# 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?
• 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.
• 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
• 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
• at sal 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?
• 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).
• How do you find the probability of catching at least two sunfish of the next three you catch without listing the outcomes?
• 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
• 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?
• > "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.
• At , 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 at , 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?
• He makes a small mistake at when 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.
• 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?
• 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)
• At ish 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)