# expected value of a fishing strategy

Suppose there is a pond with infinite number of fish. Weights of the fish are iid uniform $$(0,1)$$. We catch fish from this pond with the following rules:

1. Each day we catch at most one fish from the pond.
2. We need in total 3 fish and we are given 10 days.
3. Each day (before catching), if we have already 3 fish, we can choose to release one of the fish and catch a new one.
4. Our goal is to maximise the expected sum of the weight of the 3 fish we have on day 10.

What's the optimal strategy and what's the expected weight under this strategy?

I think the difficulty is that we only catch one per day. If we are allowed to release and catch any number of fish, the problem becomes that famous dice problem (see here).

Here's my attempt under the current setup. Let $$Y_k$$ be the expected total weight of the 3 fish by the end of day $$k$$. Then we have $$Y_3=1.5$$, and \begin{align} Y_4=&P\left(X_1+X_2+X_3\ge Y_3\right) E\left(X_1+X_2+X_3|X_1+X_2+X_3\ge Y_3\right)\\ &+P\left(X_1+X_2+X_3< Y_3\right)E\left(X_1+X_2+X_3+X_4-\min(X_1,X_2,X_3)\right). \end{align} Unfortunately this formula cannot be generalised to $$Y_5$$ and above.

• The solution depends on what you gain at the end. As an example, suppose you need to feed a group of starving people on day 10 and nothing less than 2 units of fish will do: then you will want to maximize the chance of making the total exceed 2. On the other hand, if the value of the fish is in proportion to their total weight, perhaps you want to maximize the expected total. But if you are risk-averse, you will want to maximize the expected utility of that total. Which of these many possible variations of the problem do you want answered? – whuber Aug 11 '20 at 19:53
• bad wording in my part. I want to maximize the expected total weight under the given rules. – dynamic89 Aug 11 '20 at 19:56
• I suspect that a key observation, then, is to notice that if you have any fish weighing less than 0.5, then the expectation of the total increases when you release that fish; whereas if all your fish weigh more than 0.5, then the expectation can only decrease. This makes an arbitrage argument attractive: given the weights of three fish in hand, what would you be willing to pay for the opportunity to release one (the smallest, of course) and catch another? Once you can answer that for day 9 (and any combination of fish), then you can recursively answer for all earlier days in the same manner. – whuber Aug 11 '20 at 20:00
• @whuber So for any day after day 3, let $x$ be the current minimum weight out of the three. The expected gain from releasing it and catch a new one is $P(X>x)((1+x)/2-x)=(1-x)^2/2$, and the expected loss from releasing it and catch a new one is $P(X<x)x/2=x^2/2$. Expected gain is greater than loss when $x<1/2$. This proves your observation. Similarly we can show that if I am willing to pay $a$ for the opportunity to release and re-catch, I should do it when $x<1/2-a$. How do I frame a strategy from here? – dynamic89 Aug 11 '20 at 20:25
• I believe you need, unfortunately, to keep track of all three weights, not just the smallest. Consider starting with a simpler version of this problem, such as two fish and four days (where you have two decisions to make), so you can work through the details. – whuber Aug 11 '20 at 20:29

## 1 Answer

Following whuber's suggestion, let's look at a simple example. Say we need to catch 2 fish and we are given 4 days. To begin we denote the two fish we have on day two $$x_1$$ and $$x_2$$, and the minimum denoted by $$x$$. On the third day, I release the smallest fish and re-catch if the following holds \begin{align} (1-x^2)\frac{1+x}{2}+x^2\frac{x}{2}-x>0, \end{align} where $$1-x^2$$ is the probability that in the remaining days I can catch a bigger fish than my current minimum, and $$x^2$$ is the probability that in the remaining days I will not catch a bigger fish than my current minimum. $$(1+x)/2$$ and $$x/2$$ are the conditional expectations in the two cases. The $$-x$$ is the fish I have to give up in order to re-catch. Solve this inequality on day three, we will release and re-catch if the minimum of the two fish we have on the second day is less than $$0.6183$$.

Now let's turn to the fourth day. Similarly we have \begin{align} (1-x)\frac{1+x}{2}+x\frac{x}{2}-x>0, \end{align} solving this we will release and re-catch if the minimum of the two fish we have on the third day is less than $$0.5$$. This makes sense since we this is our last chance to catch a fish.

Under this strategy we notice

1. if the release and re-catch condition is not met on day $$k$$, it will not be met on day $$k+1$$.
2. this formulation does not depend on the total number fish required, but only depend on the number of days left, and $$x$$, the current minimum of course. In general we should release and re-catch if \begin{align} (1-x^k)\frac{1+x}{2}+x^k\frac{x}{2}-x>0, \end{align} where $$k$$ is the number of days left. So if there are 10 days, we should release and re-catch on the third days if the minimum from the first two days is less than $$0.81$$.

Following this strategy, what's the expected sum of our fish on the last day? I haven't quite figure it out. Let's come back to the simple example (2 fish, 4 days). On the second day, the expected minimum is $$1/3$$, and we will need $$1.5$$ catches on average to catch a fish bigger than $$1/3$$. So If I round it up to $$2$$ catches, then on the final day, the expected sum would be $$4/3$$... These are very rough ideas.