# Stacks of coins

A very simple endgame model for Go.

Stacks of coins stand on a table. The two players alternately take one stack, until the table is empty.

This is a useful starter model. One stack on its own is entirely trivial, so anything you can get out of this shows the use of disjunctive sums of games, a piece of general theory.

It illustrates the idea that you look to play big points when everything is gote. The best play is to take the largest stack, and if you know enough to ask 'how do you prove that?', you know enough to prove it.

It also illustrates miai, namely two stacks of the same value. They might as well not be there, since they will be divided one to Black and one to White if play is correct.

If the sizes of the stacks are s, t, u, v ... in decreasing order, the value of starting is

s - t + u - v + ...

which is at most s.

If we aren't given full information about the stacks, we can guess this value as s/2. That's a more meaty use of the model, and requires some discussion.

The value depends upon how many stacks of coins there are. If the number is even then s/2 is the best guess. If the number is odd it will vary between s for small numbers and s/2 at the limit for large numbers. -- Saesneg

Charles Matthews: The discussion depends on what one accepts as a decent probability model. Look at the hypothesis that the number of stacks of a given value v is equally likely to be even or odd. If this is OK up to v = s then by removing miai (pairs of stacks of equal size) one can look at cases with at most one stack of each value, and do a calculation. Which gives the answer s/2 as expected value.

I'm reasonably happy with that hypothesis as a model of small endgame plays. I don't see that it is so good for the larger endgame plays - given there is one of (*deiri*) value 15, there is quite a good chance there are no others (tedomari). So I'd have thought this calculation has a certain, limited application, because the 19x19 board has only restricted scope.

This argument then affects what one thinks about reverse sente.

victim: I disagree with Saesneg. The value will be somewhere between 0 and s, but it won't be 0 if it's odd and won't be s if it's even. s/2 is the expected value, and you'll get extreme values if the stacks are miai-heavy.

E.g. if there are five stacks, s=t and u=v and w=1, the value is 1.

For five stacks where t=u and v=w, the value is s.

For four stacks, s=t and u=v, the value is 0.

For four stacks where t=u and v=1, the value is s-1.