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It is possible to get by with fewer drawings if Mr Baljeu's suggeston of using marked pawns if followed. The pawns should carry the labels 1...8. At the first drawing, the numbers are associated with piece labels according to the piece setup in standard chess. ie. 1,8=R, 2,7=N, 3,6=B, 4=Q, 5=K. The white pieces should be put on the board, and, if the Bishops are on same colored squares, the pawns should be put back in the bag, and one should be drawn out. A number in the range 1...4 designates that the a-side Bishop should trade places with the appropriate piece on the opposite colored square. A number in the range 5...8 should be diminished by 4, and the h-side Bishop should trade places with the appropriate piece on the opposite colored square. This idea, a randomly chosen Bishop should move to a randomly chosen square of the opposite color, is due to David Wheeler. The same end could be achieved with 8 cards marked, on one side, with the numbers 1 through 8.
There is of course a numbering scheme for Chess960 (Fischer Random Chess). I invented it some years ago, and it is widely accepted. For details see e.g. in my (German) book on Chess960 http://www.chessbox.de/Compu/fullchess1_e.html or see at the two page document I gave (first page with table) to the Chess Tigers http://www.chesstigers.de/download/chess960_regeln.pdf . With best regards, Reinhard Scharnagl.
Here's a fair way of shuffling the pieces. All 960 positions are equally probable, and no extra equipment is needed. It works by splitting the placement into two phases for dark and light squares. 1. Randomly shuffle the 6 pieces not including the Bishops. Divide these pieces into two groups of 3 each. 2. Take the first 3 pieces and one bishop. Shuffle these 4 pieces. Place them on a1, c1, e1, and g1. 3. Take the remaining pieces and 1 bishop, and randomly shuffle them. Place these pieces on b1, d1, f1, and h1. 4. Correct the king position by swapping with a rook if needed. Here's the math: The first shuffle yields 6-choose-3 or 6*5*4/3/2/1 ways to divide the pieces. The second shuffle yields 4*3*2 permutations of 4 pieces. The third shuffle yields 4*3*2 permutations of 4 pieces. Divide by 2 because the knights are the same. Divide by 6 because of the king/rook unshuffle. End result = 5*4*4*4*3 = 960 positions. To make the shuffle totally fair, label the bottom of your eight pawns RRRBBNNQ and shuffle them instead of your pieces.
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I have a slight variation to David J. Coffin's approach that was outlined in the article. My change is to step 2:
And I also would like to quote Alan Baljeu's suggestion: