# Expected value of number of carps pool

1. Jan 17, 2007

### twoflower

1. The problem statement, all variables and given/known data

There is k carps in the pool, m of them are marked. I randomly fish out n carps and see that x of them are marked.

What is the expected value of number of carps in the pool? (ie. expected value of number of the carps in the pool in the beginning)

How will the expected value change as we fish the carps out one by one? (ie. not together, but one carp at the time).

To sum it up, I know m, n and x and I don't know k.

2. The attempt at a solution

I'm not sure how to solve it. For the first case, I think that ratio of marked carps in the group of the carps drawn out should be (averagely) the same as the ratio or marked carps in the whole pool, ie.

$$\frac{x}{n} = \frac{m}{k}$$

so

$$k = \frac{m.n}{x}$$

But will it really determine the expected value of number of all carps?

For the second case, I'm even more out-of-idea.

From definition, I'd say expected value of number of the carps is

$$\sum_{i=0}^{k} X_i p_i$$

Where $X_i = i$ and $p_i$ is probability that there are $i$ carps in total.

But this doesn't seem to be an effective approach.

Could someone give me some hint, please?

Thank you.

2. Jan 17, 2007

### Dick

Your answer for k is correct. I think the gist of the second question is just to realize that the number of carp pulled out increases your estimates of k will keep changing. E.g. after one is pulled out what are the possibilities for k? What they are after is what do you think will happen to the estimates for k after a LARGE number are pulled out?

3. Jan 17, 2007

### twoflower

Now I draw it, but I still don't know...In the beginning, I have m carps marked in the pool. I draw one. Probability that it will be the marked one is $\frac{m}{m+u}$ where u is unknown and it denotes the number of unmarked carps in the pool.

But I still can't figure out how can I conclude or guess the total count of carps in the pool considering the result of this one carp which I just fished out.

4. Jan 17, 2007

### Dick

I know there are 100 marked carp in the pool. I pull out 20 and 10 are marked. What's your guess for the total? Admittedly, a sample of one gives you pretty poor estimates. But then I think that is the point of the 'how does the expected value change' part of the question.

5. Jan 17, 2007

### twoflower

I still get the estimation for k:

$$k = \frac{m.n}{x}$$

where n is number of carps I have fished out so far and x is count of marked ones among them, so this expression gives me estimation of total carps count in n-th step. But I guess this is not what I was asked for in the original problem...

6. Jan 17, 2007

### Dick

Hmmm. I think the point is that the estimate of how many fish, while poor if you've only pulled a few fish out, gets better and better as you pull more out. In fact, it is perfect when you've pulled them all out. Can you agree with that? I think that is the sort of thing they want you to observe.

7. Jan 17, 2007

### twoflower

Maybe, but it seems rather obscure to me, I think they expect some kind of $EX = ...$, ie. expected value based only on current number of carps drawn out and the probability of drawing the marked carp in one individual step. When we were discussing this problem on our school forum, one of the ideas was to consider expected value of hypergeometric distribution.

8. Jan 17, 2007

### Dick

It looked to me like the question was asking for a qualitative answer, since it wasn't asking for any exact error distributions on your estimated value. If you want to push it to another level then you can ask questions like 'how many carp do I have to pull out before I'm 95% confident that my answer is within 5% of the correct total'. Then it's just the same problem as defining error margins in sampled opinion polls. But I don't think the question as you stated it asks for that level of detail.