Easy OTB randomization of Chess960

Forget clumsy cards and rerolls of dices. With little money you can buy the right kind of dices, roll them all at once and set up the position.

Well, the easiest and cheepest way to randomize OTB chess today if you have a smartphone is to download chess960 app and it will show you a random position instantly. But if you don't want to use electronics, all you have to do is to buy the right kind of dices and it will be almost as simple.

Believe it or not, there actually exist 5-sided dices. They don't look well weighted, but they are.

So the easiest way is to buy three 4-sided dices of different colors, one 5-sided dice and one normal six sided dice. Throw all five dices at once and place out the pieces accordingly.

First you set up the bishops with the value of two of the three 4-sided dices. One for the squares ACEG and one for the squares BDFH (remember the Bishops always have to be on different colors).

Then read the normal six-sided dice to set up the Queen on one of the six available squares, count from A to H available squares only. Then read the 5-sided dice for one of the Knights. Then read the last 4-sided dice for the other Knight. Now there is three squares left. Place the King between the Rooks on those squares.

So this was the five dice approach to set up the board. I was thinking of how to reduce the amount of dices to a minimum and came up with a three dice system. The only problem was to translate the dice value to an understandable read of the setup. To do that, I had to buy two blank dices and paint values on them by hand. A lot of work, but I did it anyway and then I gave them to my local chess club so they would get used a lot.

So if you start by setting up the Bishops, there is actually 16 ways they can be set up, as they have to be on different colors. I bought a blank 16-sided dice and wrote all the combination of letters that they could be placed at (ie AB, BC, AH).

Now you set up the Queen. It's now six available squares for it, so just use a normal dice for that.

And after this, you have five available squares. You can put both Knights on these five squares in ten different ways. Since the Knights are identical to each other, it's not 5*4=20 combinations but (5*4)/2=10 combinations. For instance, the value 5/1 is the exact same setup as 1/5. I wrote all these combinations on a 10-sided dice with the highest value first. Count available squares from A to H).

Now there is only three available squares. Put King between Rooks and you're set.

The easiest way is of course to buy those five dices I first mentioned. There is no painting on dices needed, but I thought it was nice to have only three dices, and the local Chess club do like these dices I gave them

Why use any external equipment at all? You can use the chess set itself to decide the setup. Just write something at the bottom of the white Pawns, and allow black player to shuffle them before the white player sets them up. The basic idea is to write on the Pawns what piece should be behind it. Because of the restrictions on King, Rook and Bishop placement you need a little refinement. The rule that King should be between the Rooks is easily enforced: you just swap it for the nearest Rook when this would not be the case.

To get the Bishops on different square shades, without biasing the probabilities, is more tricky. This is done by distinguishing Pawns for equal piece types. Instead of R, B and N we write R0, R2, B, B*, N0 and N1. B* would indicate the 'relocatable Bishop', which will have to be moved when the B and B* Pawn inadvertantly end up on the same color. To decide where to move it, we look under the left-most (a-side) Rook Pawn and Knight Pawn, and add the accompanying numbers. E.g. if we find R2 and N0, we get 2+0=2. This means the relocatable Bishop goes to the 2nd square opposite shade, counted from the a-side, and we swap it for the piece there. We do that before putting the King between the Rooks.

So the procedure gets as follows:

1. Black shuffles the white Pawns, so white doesn't know what is on their bottom
2. White places his Pawns on the second rank. (Black can in the mean time place his Pawns.)
3. White looks at the bottom of the Pawns, placing the indicated pieces behind them
4. If the Bishops are now on different square shade, we skip the next three steps
5. We look under the Pawns in front of the a-side Rook and Knight, adding the numbers there to get N
6. We identify which Bishop has the B* Pawn in front of it
7. We swap that Bishop with the piece on the Nth square of opposite shade, counted from the a-side
8. If the King is not between the Rooks, we swap it with the nearest Rook
9. Black mirrors the white piece setup

You can streamline the procedure a bit by already noting the number when you place your first Rook and Knight, when you work from left to right, and refrain from placing the B* Bishop. After placing all pieces you are then left with a Bishop, and know whether this should go to the empty square, or a square of opposite shade indicated by the number you noted, the piece originally put there going to the square you left empty. After that you only have to correct the King placement, 2 out of 3 times.

HGMuller: That is a fascinating approach! Will all 960 positions be equally likely following this system?

I think so. The numbers are a way to generate a number 0-3 completely independent of how the pieces themselves are distributed. (Oh, I see I forgot that you have to start counting the squares at 0!) If you imagine the B/B* Pawns would be revealed first, the other pieces are still unknown, so that each other piece has equal probability to be swapped with the Bishop, and the distribution of the pieces other than Bishop over the non-Bishop squares will still be completely random.

But why not have R0 R1 N1 N3 generating a 1 - 4 Random number for which piece the B* should swap with?

An interesting check up I did just now to make sure it is possible that all 960 setups are equally likely. I noticed that all pawns have unique markings underneath. So the number of possible ways the pawns can be setup is easy to calculate. It is simply (8!). And 8!/960 is 42. Since these two numbers divided is an even number, it is very possible that all 960 positions are equally likely.

Indeed, that would be more intuitive. The only preparation that is needed is to write something on the bottom of 7 white Pawns. The B* needs not be written, to indicate that in step (3) we leave the square behind it empty. The shade of this square compared to that of the other Bishop then determines if we immediately place the second Bishop there, or if we move the piece indicated by the number there, and the 2nd Bishop to the square evacuated by that.

Note that there are other possible ways to decide what piece to swap for the 2nd Bishop, but that these would require more preparation. E.g. we could write on the bottom of the black Pawns AB, CD, EF and GH (each 2x), and then look under the Pawn on the same file as the empty (B*) square to decide on which file pair the second Bishop will be put. (The shade of the first Bishop determinig which square of the pair.)

In an old, slightly damaged set we could sometimes tell one individual pawn from another.

I did some more math and I got pretty much convinced that all setups are equally likely. We already know that the pawns can be setup in 42 times more different ways than legal positions. I also figured that all legal positions from the get go can be represented in 8 different setups of the pawns. The Knight pawns can swap places as well as Rook pawns and Bishop pawns (2*2*2). We can also calculate the chance to get a legal position from the get go. It is (1/3)(4/7). Multiply those odds with 42 and you get exactly 8.

@Evert: yes, on old sets with damage this would be a problem. But the etiquette to allow the non-owner of the set to place the white Pawns would largely solve that. The issue of damage would also occur when using cards to decide on the setup.

Life can be made a bit more difficult for the cheater by writing A to H on the bottom of the black Pawns, and preceed the whole procedure by looking at the bottom of all black Pawns (in order a-file to h-file) to determine with which Pawn the white Pawn originally set up in that file would have to be swapped before starting the piece setup. In that case a cheater would have to recognize both the white and the black Pawns. If just one of the white Pawns was different, and recognizable as (say) a Queen, you could not derive any advantage from it at all without knowing what is under a number of black Pawns. In fact the process of 8 swaps is so complex, that even when you would have the numbers on all black Pawns in plain view, you would still be clueless as to where you have to place a white Pawn to make it end up in a specific place. This does not only depend on the black Pawn in the same file, but also on other black Pawns that specify swaps with your first destination.

As to the likelihoods: I now see a quite simple way to prove that these are equal for all obtained positions. You can do this by simply counting the number of original permutations that would result in a give position. E.g. after the Bishop swap:

Each of the 960 positions can have originated from permutations:

• With B and B* on different shades in 8 ways, differing on which B is B/B*, which R is R0/R1 and which N is N1/N3. (Because in this case the numbers are not looked at).
• With B and B* on the same shade. In this case either of the Bishops could have been the B*, and it could have come from 3 different squares of opposite color not occupied by the other Bishop. That is 6 possibilities. In all these cases it is fixed (by the chosen original square) which numbering the R and N should have to select that square.

So each position could have resulted from 8 + 6 = 14 original permutations, and thus must all have the same likelihood, 14 times 1/8!. So after the Bishop swap we have 8!/14 = 2880 possible setups, each equally likely.

The King swap then just maps 3 of such positions with the Rooks and Kings distributed differently over their squares onto a single one, K-R-R and R-R-K being transformed to R-K-R. So in the end every setup can be obtained from 3*14 = 42 different permutations.

Couldn't we invent some purely mechanical device dedicated to generating a position? At least it would soon become a collector's item.

Well, it depends a bit on whether you also want it to indicate which of the three K&R squares holds the K, or that it just indicates three Rooks, where the user should understand the middle one is the King. I guess one could build such a device based on marbles. The marbles would have pictograms of Q, B, N and R on them. They would be thrown into a funnel, where they serialize to roll through a gutter that slightly slopes down. The B marbles would be magnetic, and a magnet along the track would deflect them to a side track, while the non-magnetic ones stay on the main track, which makes a much longer loop before the tracks recombine. This way the Bishops will be leading.

Then the track goes into a switch, which alternately steers the marbles in one track or the other, each passing marble flipping the switch to the other track. That way the B marbles will get separated, and each followed by 3 other marbles. Each of those tracks will dump the marbles into a funnel-like dish, in which they orbit until they lose enough energy by friction and collisions that they drop through the central hole in the bottom. This is supposed to randomize the order within each set of four. From the holes in the dishes they follow tracks that combine in a reverse switch, which alternately passes one from one track or the other, so that one set of four goes to the even positions, the other set to the odd positions. That should make them get out in a random order with unlike Bishops.

I like how this discussion is going. this mechanical thing would be awesome in big ches960 tournaments

i would leave out the three rooks in this mechanical device and just leave three positions empty.

I figured out a faster and easier setup without altering the chances. Scribble the following on the white pawns. 1 2 3 B* B N N Q 1. Place out pieces, leave 1,2 and 3 blank for now. 2. If Bishops are on different colors, place King between Rooks. 3. If Bishops are on the same color, place B* under 1 else 2 else 3 depending on which one first represents the opposite color. 4. place King between Rooks

I use the flipping of coins.  Much cheaper than buying D&D dice.

jonesmikechess, sounds tedious

I don't think so. For instance, a setup with all K&R on the same shade would originate from 24 permutations without swap (6 permutations of K&R, 2 of the N and B/B*). If it would have been created through a swap, it must have been the Bishop of opposite shade that has been swapped, (fixing the B/B* permutation) and this could have been swapped with any of the 123, which was the only 123 on the required shade, and thus not competing with any of the others for the lowest number. So each of the 12 K&R and N permutations in the final setup would be obtained from 3 original permutations through Bishop swapping. In total 36 + 24 = 60 original permutations would produce that same setup. This should have been 42 if all setups were equally likely. So positions with all K&R on the same shade are overrepresented.

(Somewhat more tediously you can show that each setup with K&R distributed over both shades will be obtained from 40 original permutations. Considering that 1/10 of the positions has all K&R on the same shade, this indeed averages to 1/10*60 + 9/10*40 = 6 + 36 = 42.)

As a moot addition, I also doubt that this procedure is simpler. You would have to search for the lowest number on the desired shade, as while first revealing the Pawn inscriptions you might not know yet which shade is desired. So if you want to do it by memory, you would need to remember where the lowest number on either shade is. Just adding the numbers on the first encountered R and N requires you only to remember a single number.

okay, how about three blank pawns, B* B N N Q. If Bishops are on same color, switch B* with piece to the right, exception if B* is on H then switch with piece on square A. then put King between Rooks.

The problem then is that setups with (cyclicly) adjacent Bishops can only be produced through a swap if the leftmost Bishop is the B* (i.e. from 12 original permutations of K&R and N), while those with non-adjacent Bishops could have come from either swapping left or right with the piece now left of it. Together with the 24 permutations for which you don't have to swap that makes 36 original permutations vs 48. (Since half the positions have (cycicly) adjacent Bishops, the average is (36 + 48)/2 = 84/2 = 42, as it should.) So adjacent Bishops would be under-represented.