Previously, on Minus Two Fish:

P.A.M. Dirac was presented with a puzzle.

Three fishermen come back from the sea. Each collapses in their respective tent. Fisherman #1 wakes up and decides to get his share of the catch. He counts the fish, realizes it is not a number divisible by three, throws away one fish to the sea correcting the situation, and takes a third of the remaining fish into his tent. Fisherman #2 wakes up later and also decides he is going to get his share of the catch. Fisherman #2 wants a third of the fish he sees. It is not a multiple of three, but he throws away one fish and takes a third of the fish and goes to sleep. Fisherman #3 wakes up after, and does the same: he throws away one fish, takes a third of the fish, and goes into his tent.

What is the smallest number of fish for which this would happen?

The real solution was not revealed, until now.

[Dirac’s Fish Puzzle Solution Spoiler Warning]

The want to find what is the smallest number of fish $$f$$ at the beginning.

$$f-1$$ is the number of fish after Fisherman #1 throws away one. $$left(f-1right)frac{1}{3}$$ is the number of fish that Fisherman #1 takes for himself, leaving only $$left(f-1right)frac{2}{3}$$. $$(f-1)frac{2}{3}-1$$ is the number of fish after Fisherman #2 throws away one. $$left( left(f-1right)frac{2}{3}-1right)frac{2}{3}$$ is the number of fish that Fisherman #2 leaves after he takes his share. $$left( left( f-1 right) frac{2}{3} -1 right)frac{2}{3}-1 $$ is the number of fish after Fisherman #3 throws away one. $$left(left(left(f-1right)frac{2}{3}-1right)frac{2}{3}-1right)frac{2}{3}$$ is the number of fish that Fisherman #3 leaves after he takes his share.

The last number must be an integer $$i$$, as it is implicit that only a whole number fish are left.

$$i=(((f-1)frac{2}{3}-1)frac{2}{3}-1)frac{2}{3}=ffrac{8}{27}-frac{38}{27}$$. Solving for $$f$$,

$$f=ifrac{27}{8}+frac{38}{8}$$.

Now, $$i$$ must be a small integer. The easiest way to explain the solution is by trial and error. So, try $$i=1$$, and you will see $$f$$ is not an integer. Try $$i=2$$, $$i=3$$ … still $$f$$ is not an integer. Until you reach $$i=6$$, where $$f=25$$.

So, the smallest number of fish $$f$$ is **25**.

Also, try in the equation to plug in $$f=-2$$, Dirac’s solution and you will see $$i=-2$$. That is why it is so weird, Dirac assumes that there are **negative fish** to start, and negative fish at the end.

This solution permits for infinite number of fishermen to come and throw away $$1$$ fish and take $$frac{1}{3}$$ of the remaining fish as the number of fish at the end is the same as the beginning.