It occurred to me that I had improperly counted the amount of possible starting combinations of the so-called “Chess 60”. I had overlooked that there are two knights, and so many of the positions will simply be repeats of another, but with Knight A and Knight B flipping positions. So is it true that we can simply halve the amount of positions, and start calling the game Chess 30? Perhaps not.

Here we have a chess board, set up with the classical chess starting position.

The classic chess starting position

Imagine that we introduced a new piece (purely for illustrating this point), called the Archbishop and put that new piece somewhere on the second rank, with the pawns. There are 8 possible locations for this new piece, so we have 8 different starting positions. Now if we add a second new piece, called the Jester, and added that to the pawn rank. Now we have 6 pawns, and 8*7=56 possible starting combinations. If we add another piece, called the Seer, then we have 8*7*6 = 336 starting combinations. So far so good, but what if the third piece we add is just another Jester, how many combinations do we have?

To make this easier, lets pretend we’re back to two new pieces, both Archbishops. If we have two Archbishops, then our 56 positions are going to have some duplicates. How many? Let’s make this even easier. Pretend that the chess board is only four squares wide. That means that the second rank contains two pawns, and two Archbishops. It’s easy to visualize that we have 4*3=12 positions, with 6 duplicates. I’ll even write out all of them.

I’ve highlighted one of the Archbishops to differentiate between him and the other Archbishop:

A=Archbishop, P=Pawn

1: *A*,A,P,P

2: *A*, P, A, P

3: *A*,P,P,A

4: A,*A*,P,P (duplicate)

5: P,*A*,A,P

6: P,*A*,P,A

7: A,P,*A*,P (duplicate)

8: P,A,*A*,P (duplicate)

9: P,P,*A*,A

10: A,P,P,*A*(duplicate)

11: P,A,P,*A* (duplicate)

12: P,P,A,*A* (duplicate)

If these were two different pieces, we would have 12 different positions. Since the two Archbishops are functionally equivalent, we only have half as many positions, or 6.

To bring this back, yes we need to divide by two to get the true amount of positions in “Chess 60”. Since “Chess 60” does not actually have sixty positions the name is pretty horrible. “Chess 30” might be more appropriate, but I have a different idea…

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