page_title: Fischer random chess starting position
section_title: Polyhedral dice method
hierarchical_section_title: Fischer random chess starting position / Methods for creating starting positions / Polyhedral dice method
caption_attribution_description: English: Platonic-solids set of five dice, (from left) tetrahedron (d4), cube (d6), octahedron (d8), dodecahedron (d12), and icosahedron (d20).
context_page_description: A Fischer random chess starting position is one of 960 possible initial game positions in the chess variant Fischer random chess. The special arrangement of pieces on the players' first ranks is selected randomly before play according to Fischer random chess rules, and can be generated either by a computer program, or using dice, coin, cards, etc.
context_section_description: With polyhedral dice shaped like each of the five Platonic solids, one never needs to reroll any dice. 
  Roll all the dice in one throw and place White's pieces as follows:
  Place a bishop on one of the eight squares (counting from the left, a through h ) as indicated by the octahedron (d8).
  Place the other bishop on one of the four squares of opposite color as indicated by the tetrahedron (d4).
  Place the queen on one of the remaining six squares as indicated by the cube (d6).
  Take the value of the icosahedron (d20), divide by four (round up), and let x = the quotient, and y = the remainder + 1. Place a knight on the xth empty square. Then place the other knight on the yth remaining empty square. In other words, see the d20 as a d5 for the first knight: 1-4, 5-8, 9-12, 13-16 and 17-20. Then for the second knight, look within the group to get a d4. For example, a 20 is in the fifth group and the fourth spot in that group, so place the knights on the fifth square and the fourth square. An 11 is in the third group and the third spot.You can also use just a d10 since there are only ten unique placements of the knights once the bishops and queen has been placed. Hold one knight on the leftmost square and count one, two, three, four with the other knight on the empty square, then when it loops, move the leftmost knight one square to the right, five, six, seven, then it loops again, eight, nine, and finally with ten both knights are as far right as they go. For example, with a six the knight would be placed on the second of the five empty squares, then the second knight would be place on the second of the three squares that are empty to the right of the knight. Using a d10 in this way after two different colored d4:s and a d6 is a minimal one-roll way since 4×4×6×10 is exactly 960. (And, by subtracting one from each die and multiplying with 1, 4, 16 and 96 respectively, then adding those numbers together, you find the number in the Fischer Random Chess numbering scheme.) The d8, d4, d6, d20 still give equal chance for all 960 positions, but with every position being represented in four different ways.Or alternatively (using an additional die and different calculations): Place the first knight according to the value of the d20 die, by counting the five empty squares and looping back to the left whenever reaching the rightmost empty square. Then with four empty squares remaining, do the same for the other knight using the dodecahedron (d12) die. With this method, every position is represented in 48 different ways.
  Place the king between the rooks on the remaining three squares.
  Place the white pawns and mirror the position for Black.
caption: A set of Platonic solids dice, from left: tetrahedron (d4), cube (d6), octahedron (d8), dodecahedron (d12), icosahedron (d20)
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