960 Calculated

For anyone who has wondered about where the number 960 came from as the number of possible opening positions in Chess960, I will demonstrate the calculation here. I’m a math guy, so it was actually fun for me to work this out.

Let’s start with the Bishops, who need to occupy different colors. On the first rank, there are 4 light squares which the LSB could occupy, and there are 4 dark squares which the DSB could occupy. Thus, there are 4 x 4 = 16 ways in which the bishop pair could be placed.

Next, let’s focus on the Rook-King-Rook trio where the King must be in between the Rooks. For any given placement of the bishops, there are 6 remaining squares to place this trio of 3 pieces. This amounts to 6 choose 3, or 6! / 3! / 3! = 720 / 6 / 6 = 20 possible placements of the Rook-King-Rook trio once the Bishop pair is placed.

Next: The Queen placement. At this point, 5 pieces have been placed into a Chess960 opening position (the Bishops, Rooks, and King). There are 3 squares left for the Queen to occupy.

Lastly, the Knights. The placement of this pair is trivial. The other 6 squares have been occupied, so there is only 1 way to place the pair of Knights at this point.

So, the number of possible starting positions can be computed by:

(Bishop pair possibilities) x (Rook-King-Rook possibilities) x (Queen Possibilities) x (Knight pair possibilities) = 16 x 20 x 3 x 1 = 960.

There you have it in case anybody was curious. Let me know if anyone wants clarification.

I have explained on more than one occasion in the past the correct method to calculate the number.

Your approach does not follow the procedure for setting up the position on the board in practice, you do K + R + R after the Bishops which is wrong.

 

First the basics, you have to derive a formula that works for everything.

You use only 3 squares say A1 + B1 + C1 and you start with K + Q + R.

The total combinations are King (3) x Queen (2) x Rook (1) = 3 * 2 * 1 = 6.

 

Now you take away the Queen and replace it with the second Rook, is it still 6 positions?

For mathematics it is but for chess it is not because the 2 Rooks are considered to be exactly the same, switch them around and nothing really changes.

 

So you modify the formula as: number of squares / number of pieces

So for K + Q + R = (3/1) * (2/1) * (1/1) = 3 * 2 * 1 = 6

So for K + R + R = (3/1) * (2/2) * (1/1) = 3 * 1 * 1 = 3

 

Note that a standard formula has been derived.

Now you apply it in practice, exactly as you set it up on the board.

First you do the bishops and because of the special rule the 2 bishops are not considered to be exactly the same, so the total for the bishops is:

Bishops = (4/1) * (4/1) = 4 * 4 = 16.

 

Then you do the Queen which is easy.

Queen = (6/1) = 6

 

Then you do the Knights which are considered to be exactly the same.

Knights = (5/2) * (4/1) = 2.5 * 4 = 10.

 

Then you do the K + R + R which is very easy, the King is placed between the Rooks.

King + Rook + Rook = 1.

 

The calculation is: Bishops (16) * Queen (6) * Knights (10) = 16 * 6 * 10 = 960.

 

Warlord

Either way works. Six steps one way, half a dozen the other, so to speak. The key is starting the calculation with the Bishops.

When I was first curious about it, I used tree diagrams that started with the King at various positions (b1 to g1), then branched off to the Rooks, and only then to the Bishops. I eventually did get to 960, but it was a rather non-elegant method, particularly because I had to take note of colors occupied by the King and Rooks, which then affected the number of possible squares for each Bishop.

On a side note, the main method that either of us described can be used to randomly select a 960 starting position without the aid of a computer. All you need is the aid of a six-sided die. I will walk through the process using your order of piece consideration. Note that any squares referred to will indicate 1st-rank squares.

1. Number the light squares 1-4. Roll the die until a number 1-4 shows up, and that assigns the placement of the LSB.

2. Repeat step one with the dark squares to place the DSB.

3. Number the remaining squares 1-6. Roll the die until you get three unique results. The first result places the Queen, and the second and third unique results place the Knights.

4. Place the Rooks and King on the remaining Squares with the King in between the Rooks.

I will use this method to introduce my brother in law to a Chess960 game this weekend since we will be out in the country away from internet access.

The standard method that is used to create a starting position is explained here:

https://en.wikipedia.org/wiki/Chess960_starting_position

Hi, this is an old conversation, but like Jim, I am also a "Math Guy" and need to point out that his calculation is FINE. It doesn't matter what order you enumerate the positions for in combinatorics (as long as you follow the logic for each position at each step), thanks to the Rule of Product:

https://en.wikipedia.org/wiki/Rule_of_product

Now I know that name chess960 is not randomly named. It's have a specific reason to be named.

There is so much interesting information here.

Very good