# Fish in a Pond

There are 5 fish in a pond. What is the probability that you can split the pond into 2 halves using a diameter, so that all fish end up in one half?

**SOLUTION**

Let us generalize the problem to N fish in a pond. We can assume that all fish are on the boundary of the pond, which is a circle, and we need to find the probability that all of them are contained within a semi-circle.

For every fish **Fᵢ**, consider the semi-circle **Cᵢ** whose left end-point is at **Fᵢ**. The probability that all fish belong to **Cᵢ** is equal to **1/2ᴺ⁻¹**. Since it is impossible to have 2 fish **Fᵢ** and **Fⱼ**, such that the semi-sircles **Cᵢ** and **Cⱼ** contain all fish, we see that the probability that all fish belong to **Cᵢ** for some **i** is equal to **N/2ᴺ⁻¹**.

When N = 5, we get that the answer is 5/16.

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This does not consider the size of fish neither relative to each other, nor relative to the diameter of the pond. This problem seems to be assuming that all fish are dots. If so, it needs to be stated.

Also you can think of the question of asking, what are the odds that if you randomly place 5 fish in circle, that they all end up on the desired side.

Then you could use the binomial probability formula P(X=x) = C(n,x)*p^x*q^(n-x), where:

X = # of occurrences of desired outcome: a fish being on the desired side

n = total number of fish (5)

x = number of fish we want to be on the correct side (5)

p = probability of X occurring (0.5 because it has only 2 options)

q = probability of X not occurring (0.5 for same reason as above)

Hi Alexander, thank you for your comment. That’s right, you can use this formula, but only once you fix the semi-circle. Then, you will get C(4,4)(0.5)^4=1/16, which is what is implicitly done in the provided solution:) The main trick is to realize that: