Author Topic: Maths question: How many possible setups...?  (Read 27 times)

ebinola

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Maths question: How many possible setups...?
« on: September 22, 2018, 06:35:09 pm »
Returning to chess variants, I've got an idea in the works that so far has posed a number of questions. One particular question requires some maths work that I can't wrap my head around, and I'm wondering if someone could help me.

The starting position for this game is identical to orthodox chess (pppppppp/rnbqkbnr...), but the player has some power over how their army is set up. The game has 5 piece classes (Pawn, Knight, Bishop, Rook, Queen), and 18 piece sub-classes (with their own unique properties). This makes for 90 unique pieces (excluding the King): 18 different Pawns, 18 different Knights, and so on. Multiples of pieces are allowed, i.e. you can have multiple Pawns of one sub-class. The King in this game has no special properties like the other pieces, it's just the same as any old King.

This setup phase is done blindly before the game begins, so neither player knows their opponent's setup until they see the starting position. Which leaves me with the question: how many possible setups exist with the above parameters?
« Last Edit: September 22, 2018, 06:37:05 pm by ebinola »
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Asher Hurowitz

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Re: Maths question: How many possible setups...?
« Reply #1 on: September 23, 2018, 10:56:57 pm »
This is an awesome question - working on it currently!  ;D
Indeed it is certain that Chess Variants make me happy.

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joejoyce

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Re: Maths question: How many possible setups...?
« Reply #2 on: September 29, 2018, 09:58:55 pm »
Let's start with rooks. How many rook combinations are there? There are 2 rooks, each of which can be 18 similar but not identical pieces. SO there are 18x18=324 possible combinations for rooks. This number will be the same for the knight and bishop pairs, so that will give you 324x324x324 possible combos for the B, N, R pieces. Since the king is always just a standard king, it's multiplicative value is 1, and the queen's is 18. So the back rank is 324x324x324x1x18.

Okay, now the pawns; each pair of pawns gives us 324 possibilities, and there are 4 pairs, giving us 324x324x324x324, which will multiply with the back rank to give the answer. Since so far I'm doing this in my head, I'm keeping the numbers simple... 324x324=90000+12000+2400 (300x300+2x20x300+2x4x300) = 104.400, using European notation, or 104,400 for US notation.

Now I'm gonna cheat! call 104400 ~= 100000, so the pawns have ~100000x100000=10000000000 possibilities, roughly (it's a little low). Now, the back rank is 324x324x324x18, or 100000x324x18 ~= 100000x300x20=600000000, giving a final estimate of  10000000000x600000000=6000000000000000000, or 6x10^18, and this is a little low.

Grin, somebody want to check my numbers? It's been a long time since I did probabilities!