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How to use dice to determine starting position in Fischerrandom?
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Using a set of Platonic dice (tetrahedron, cube, octahedron, dodecahedron, and icosahedron), what way of choosing the initial position in a game of Fischerrandom Chess (Chess 960) uses uses fewest dice throws?
I am looking for a way that
does not simply yield a number between 1 and 960 which you then look up on an ordered list of starting positions
allows pieces to be placed in sequence - for example, first the rooks, then the bishops, then the king, or in some other order
yields each of the 960 positions with equal probability
The rules state that for each player bishops must be on squares of opposite colour and the king must be between the rooks.
You do not have to use every die. If only say the octahedron need be used, that is fine!
fischer chess960
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h34
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up vote
4
down vote
favorite
Using a set of Platonic dice (tetrahedron, cube, octahedron, dodecahedron, and icosahedron), what way of choosing the initial position in a game of Fischerrandom Chess (Chess 960) uses uses fewest dice throws?
I am looking for a way that
does not simply yield a number between 1 and 960 which you then look up on an ordered list of starting positions
allows pieces to be placed in sequence - for example, first the rooks, then the bishops, then the king, or in some other order
yields each of the 960 positions with equal probability
The rules state that for each player bishops must be on squares of opposite colour and the king must be between the rooks.
You do not have to use every die. If only say the octahedron need be used, that is fine!
fischer chess960
asked 4 hours ago
h34
1314
New contributor
h34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Using a set of Platonic dice (tetrahedron, cube, octahedron, dodecahedron, and icosahedron), what way of choosing the initial position in a game of Fischerrandom Chess (Chess 960) uses uses fewest dice throws?
I am looking for a way that
does not simply yield a number between 1 and 960 which you then look up on an ordered list of starting positions
allows pieces to be placed in sequence - for example, first the rooks, then the bishops, then the king, or in some other order
yields each of the 960 positions with equal probability
The rules state that for each player bishops must be on squares of opposite colour and the king must be between the rooks.
You do not have to use every die. If only say the octahedron need be used, that is fine!
fischer chess960
asked 4 hours ago
h34
1314
New contributor
h34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Using a set of Platonic dice (tetrahedron, cube, octahedron, dodecahedron, and icosahedron), what way of choosing the initial position in a game of Fischerrandom Chess (Chess 960) uses uses fewest dice throws?
I am looking for a way that
does not simply yield a number between 1 and 960 which you then look up on an ordered list of starting positions
allows pieces to be placed in sequence - for example, first the rooks, then the bishops, then the king, or in some other order
yields each of the 960 positions with equal probability
The rules state that for each player bishops must be on squares of opposite colour and the king must be between the rooks.
You do not have to use every die. If only say the octahedron need be used, that is fine!
fischer chess960
fischer chess960
asked 4 hours ago
h34
1314
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h34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 4 hours ago
h34
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asked 4 hours ago
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asked 4 hours ago
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2 Answers
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Several methods to do so are described on Wikipedia. For example:
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 'x'-th empty square. Then place the other knight on the 'y'-th 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 multipying with 1, 4, 16 and 96 respectively, then adding those numbers together, you find the number in the Chess960 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.
It is stated (but not proven) that this method (and the others) "generat[e] random starting positions with equal probability".
answered 2 hours ago
Glorfindel
11.7k43355
Thanks. That's especially neat with the icosahedron. If we amend my method to use an icosahedron in that way to place the Ns, then we always need exactly 4 throws (d4 for each of the two Bs, d6 for Q, d20 for the two Ns together).
– h34
2 hours ago
add a comment |Â
up vote
1
down vote
Numbering from white's left, and assuming each die carries consecutive integers starting at 1, you can do it in the order BQN as follows:
- throw a tetrahedron; if n is up, place the white-squared B on the nth free white square
- do the same for the other B, except place it on the nth free black square
- throw a cube; if n is up, place the Q on the nth free square
- throw a cube; rethrow until a number n≠6 is up; place a N on the nth free square
- throw a tetrahedron; if n is up, place the other N on the nth free square
- now place RKR on the remaining free squares in that order
This method uses two dice: a tetrahedron and a cube. The tetrahedron is thrown 3 times; the cube a minimum of 2 times and a mean of 2.2 times.
Equivalently, use a single dodecahedron and interpret n base 4 or 6 according to whether you would otherwise use a tetrahedron or cube. You then need to throw at least 5 times and a mean of 5.2 times.
This is neat because all you need to remember is "BQN". But I do not know whether this yields all 960 positions with equal probability.
answered 2 hours ago
h34
1314
New contributor
h34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Several methods to do so are described on Wikipedia. For example:
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 'x'-th empty square. Then place the other knight on the 'y'-th 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 multipying with 1, 4, 16 and 96 respectively, then adding those numbers together, you find the number in the Chess960 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.
It is stated (but not proven) that this method (and the others) "generat[e] random starting positions with equal probability".
answered 2 hours ago
Glorfindel
11.7k43355
Thanks. That's especially neat with the icosahedron. If we amend my method to use an icosahedron in that way to place the Ns, then we always need exactly 4 throws (d4 for each of the two Bs, d6 for Q, d20 for the two Ns together).
– h34
2 hours ago
add a comment |Â
up vote
1
down vote
Several methods to do so are described on Wikipedia. For example:
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 'x'-th empty square. Then place the other knight on the 'y'-th 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 multipying with 1, 4, 16 and 96 respectively, then adding those numbers together, you find the number in the Chess960 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.
It is stated (but not proven) that this method (and the others) "generat[e] random starting positions with equal probability".
answered 2 hours ago
Glorfindel
11.7k43355
Thanks. That's especially neat with the icosahedron. If we amend my method to use an icosahedron in that way to place the Ns, then we always need exactly 4 throws (d4 for each of the two Bs, d6 for Q, d20 for the two Ns together).
– h34
2 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Several methods to do so are described on Wikipedia. For example:
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 'x'-th empty square. Then place the other knight on the 'y'-th 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 multipying with 1, 4, 16 and 96 respectively, then adding those numbers together, you find the number in the Chess960 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.
It is stated (but not proven) that this method (and the others) "generat[e] random starting positions with equal probability".
answered 2 hours ago
Glorfindel
11.7k43355
Several methods to do so are described on Wikipedia. For example:
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 'x'-th empty square. Then place the other knight on the 'y'-th 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 multipying with 1, 4, 16 and 96 respectively, then adding those numbers together, you find the number in the Chess960 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.
It is stated (but not proven) that this method (and the others) "generat[e] random starting positions with equal probability".
answered 2 hours ago
Glorfindel
11.7k43355
answered 2 hours ago
Glorfindel
11.7k43355
answered 2 hours ago
Glorfindel
11.7k43355
answered 2 hours ago
Glorfindel
11.7k43355
11.7k43355
Thanks. That's especially neat with the icosahedron. If we amend my method to use an icosahedron in that way to place the Ns, then we always need exactly 4 throws (d4 for each of the two Bs, d6 for Q, d20 for the two Ns together).
– h34
2 hours ago
add a comment |Â
Thanks. That's especially neat with the icosahedron. If we amend my method to use an icosahedron in that way to place the Ns, then we always need exactly 4 throws (d4 for each of the two Bs, d6 for Q, d20 for the two Ns together).
– h34
2 hours ago
Thanks. That's especially neat with the icosahedron. If we amend my method to use an icosahedron in that way to place the Ns, then we always need exactly 4 throws (d4 for each of the two Bs, d6 for Q, d20 for the two Ns together).
– h34
2 hours ago
Thanks. That's especially neat with the icosahedron. If we amend my method to use an icosahedron in that way to place the Ns, then we always need exactly 4 throws (d4 for each of the two Bs, d6 for Q, d20 for the two Ns together).
– h34
2 hours ago
add a comment |Â
up vote
1
down vote
Numbering from white's left, and assuming each die carries consecutive integers starting at 1, you can do it in the order BQN as follows:
- throw a tetrahedron; if n is up, place the white-squared B on the nth free white square
- do the same for the other B, except place it on the nth free black square
- throw a cube; if n is up, place the Q on the nth free square
- throw a cube; rethrow until a number n≠6 is up; place a N on the nth free square
- throw a tetrahedron; if n is up, place the other N on the nth free square
- now place RKR on the remaining free squares in that order
This method uses two dice: a tetrahedron and a cube. The tetrahedron is thrown 3 times; the cube a minimum of 2 times and a mean of 2.2 times.
Equivalently, use a single dodecahedron and interpret n base 4 or 6 according to whether you would otherwise use a tetrahedron or cube. You then need to throw at least 5 times and a mean of 5.2 times.
This is neat because all you need to remember is "BQN". But I do not know whether this yields all 960 positions with equal probability.
answered 2 hours ago
h34
1314
New contributor
h34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
Numbering from white's left, and assuming each die carries consecutive integers starting at 1, you can do it in the order BQN as follows:
- throw a tetrahedron; if n is up, place the white-squared B on the nth free white square
- do the same for the other B, except place it on the nth free black square
- throw a cube; if n is up, place the Q on the nth free square
- throw a cube; rethrow until a number n≠6 is up; place a N on the nth free square
- throw a tetrahedron; if n is up, place the other N on the nth free square
- now place RKR on the remaining free squares in that order
This method uses two dice: a tetrahedron and a cube. The tetrahedron is thrown 3 times; the cube a minimum of 2 times and a mean of 2.2 times.
Equivalently, use a single dodecahedron and interpret n base 4 or 6 according to whether you would otherwise use a tetrahedron or cube. You then need to throw at least 5 times and a mean of 5.2 times.
This is neat because all you need to remember is "BQN". But I do not know whether this yields all 960 positions with equal probability.
answered 2 hours ago
h34
1314
New contributor
h34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Numbering from white's left, and assuming each die carries consecutive integers starting at 1, you can do it in the order BQN as follows:
- throw a tetrahedron; if n is up, place the white-squared B on the nth free white square
- do the same for the other B, except place it on the nth free black square
- throw a cube; if n is up, place the Q on the nth free square
- throw a cube; rethrow until a number n≠6 is up; place a N on the nth free square
- throw a tetrahedron; if n is up, place the other N on the nth free square
- now place RKR on the remaining free squares in that order
This method uses two dice: a tetrahedron and a cube. The tetrahedron is thrown 3 times; the cube a minimum of 2 times and a mean of 2.2 times.
Equivalently, use a single dodecahedron and interpret n base 4 or 6 according to whether you would otherwise use a tetrahedron or cube. You then need to throw at least 5 times and a mean of 5.2 times.
This is neat because all you need to remember is "BQN". But I do not know whether this yields all 960 positions with equal probability.
answered 2 hours ago
h34
1314
New contributor
h34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Numbering from white's left, and assuming each die carries consecutive integers starting at 1, you can do it in the order BQN as follows:
- throw a tetrahedron; if n is up, place the white-squared B on the nth free white square
- do the same for the other B, except place it on the nth free black square
- throw a cube; if n is up, place the Q on the nth free square
- throw a cube; rethrow until a number n≠6 is up; place a N on the nth free square
- throw a tetrahedron; if n is up, place the other N on the nth free square
- now place RKR on the remaining free squares in that order
This method uses two dice: a tetrahedron and a cube. The tetrahedron is thrown 3 times; the cube a minimum of 2 times and a mean of 2.2 times.
Equivalently, use a single dodecahedron and interpret n base 4 or 6 according to whether you would otherwise use a tetrahedron or cube. You then need to throw at least 5 times and a mean of 5.2 times.
This is neat because all you need to remember is "BQN". But I do not know whether this yields all 960 positions with equal probability.
answered 2 hours ago
h34
1314
New contributor
h34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 2 hours ago
answered 2 hours ago
h34
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h34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 2 hours ago
h34
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answered 2 hours ago
h34
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h34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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h34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
h34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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h34 is a new contributor. Be nice, and check out our Code of Conduct.
h34 is a new contributor. Be nice, and check out our Code of Conduct.
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