A first choice die
$begingroup$
You and your friend have met to play a game of dice.
This is a very simple game where both players roll a die at the same time, and the player with the lower score pays a coin to the other player.
In case of tie, no money is exchanged and the turn is repeated.
You propose your friend to make the game more interesting by replacing the usual playing dice with three special dice you have designed.
Your dice have a number of dots between 1 and 6 on each face like regular dice, but some numbers are repeated on different faces and some numbers are missing, and no two dice are identical.
Apart from this twist, those are pretty normal playing dice, not unbalanced towards any side.
At each turn, you propose, first your friend will pick a die and then you will pick one of the remaining two dice.
Your friends accepts, thinking that, if at all, he's being offered an advantage.
If one of your dice were stronger than the other two, your friend could use that die at every turn because he can choose first, and that would make his winning chances better.
But you know this isn't going to be the case, and in fact, no matter which die your friend chooses, you will be always able to pick a stronger one.
How did you design your dice such that your chances to win are always better than your friend's?
logical-deduction mathematics probability dice
asked Nov 13 '14 at 13:25
GOTO 0GOTO 0
9,66654089
$endgroup$
|
show 6 more comments
$begingroup$
You and your friend have met to play a game of dice.
This is a very simple game where both players roll a die at the same time, and the player with the lower score pays a coin to the other player.
In case of tie, no money is exchanged and the turn is repeated.
You propose your friend to make the game more interesting by replacing the usual playing dice with three special dice you have designed.
Your dice have a number of dots between 1 and 6 on each face like regular dice, but some numbers are repeated on different faces and some numbers are missing, and no two dice are identical.
Apart from this twist, those are pretty normal playing dice, not unbalanced towards any side.
At each turn, you propose, first your friend will pick a die and then you will pick one of the remaining two dice.
Your friends accepts, thinking that, if at all, he's being offered an advantage.
If one of your dice were stronger than the other two, your friend could use that die at every turn because he can choose first, and that would make his winning chances better.
But you know this isn't going to be the case, and in fact, no matter which die your friend chooses, you will be always able to pick a stronger one.
How did you design your dice such that your chances to win are always better than your friend's?
logical-deduction mathematics probability dice
asked Nov 13 '14 at 13:25
GOTO 0GOTO 0
9,66654089
$endgroup$
$begingroup$
This is a fairly well known puzzle that I feel deserves a place in Puzzling.SE. I will accept the best answer.
$endgroup$
– GOTO 0
Nov 13 '14 at 13:25
$begingroup$
oh my god thats a lot of work :) hopefully someone will try to solve it originally here
$endgroup$
– skv
Nov 13 '14 at 13:33
1
$begingroup$
@Victor afaik, die is the singular name form, while dice can be both singular or plural. I don't have any particular feelings about the word choice. You're welcome to edit the question if you feel my wording is not clear enough.
$endgroup$
– GOTO 0
Nov 13 '14 at 13:48
1
$begingroup$
@GOTO0 Not to be a stickler, but dice is only plural, strictly speaking. People tend to use it as singular but it is incorrect.
$endgroup$
– Neil
Nov 13 '14 at 14:18
2
$begingroup$
@GOTO0 english.stackexchange.com/questions/167104/singular-of-dice/…
$endgroup$
– Neil
Nov 13 '14 at 14:29
|
show 6 more comments
$begingroup$
You and your friend have met to play a game of dice.
This is a very simple game where both players roll a die at the same time, and the player with the lower score pays a coin to the other player.
In case of tie, no money is exchanged and the turn is repeated.
You propose your friend to make the game more interesting by replacing the usual playing dice with three special dice you have designed.
Your dice have a number of dots between 1 and 6 on each face like regular dice, but some numbers are repeated on different faces and some numbers are missing, and no two dice are identical.
Apart from this twist, those are pretty normal playing dice, not unbalanced towards any side.
At each turn, you propose, first your friend will pick a die and then you will pick one of the remaining two dice.
Your friends accepts, thinking that, if at all, he's being offered an advantage.
If one of your dice were stronger than the other two, your friend could use that die at every turn because he can choose first, and that would make his winning chances better.
But you know this isn't going to be the case, and in fact, no matter which die your friend chooses, you will be always able to pick a stronger one.
How did you design your dice such that your chances to win are always better than your friend's?
logical-deduction mathematics probability dice
asked Nov 13 '14 at 13:25
GOTO 0GOTO 0
9,66654089
$endgroup$
You and your friend have met to play a game of dice.
This is a very simple game where both players roll a die at the same time, and the player with the lower score pays a coin to the other player.
In case of tie, no money is exchanged and the turn is repeated.
You propose your friend to make the game more interesting by replacing the usual playing dice with three special dice you have designed.
Your dice have a number of dots between 1 and 6 on each face like regular dice, but some numbers are repeated on different faces and some numbers are missing, and no two dice are identical.
Apart from this twist, those are pretty normal playing dice, not unbalanced towards any side.
At each turn, you propose, first your friend will pick a die and then you will pick one of the remaining two dice.
Your friends accepts, thinking that, if at all, he's being offered an advantage.
If one of your dice were stronger than the other two, your friend could use that die at every turn because he can choose first, and that would make his winning chances better.
But you know this isn't going to be the case, and in fact, no matter which die your friend chooses, you will be always able to pick a stronger one.
How did you design your dice such that your chances to win are always better than your friend's?
logical-deduction mathematics probability dice
logical-deduction mathematics probability dice
asked Nov 13 '14 at 13:25
GOTO 0GOTO 0
9,66654089
asked Nov 13 '14 at 13:25
GOTO 0GOTO 0
9,66654089
asked Nov 13 '14 at 13:25
GOTO 0GOTO 0
9,66654089
asked Nov 13 '14 at 13:25
GOTO 0GOTO 0
9,66654089
asked Nov 13 '14 at 13:25
GOTO 0GOTO 0
9,66654089
9,66654089
$begingroup$
This is a fairly well known puzzle that I feel deserves a place in Puzzling.SE. I will accept the best answer.
$endgroup$
– GOTO 0
Nov 13 '14 at 13:25
$begingroup$
oh my god thats a lot of work :) hopefully someone will try to solve it originally here
$endgroup$
– skv
Nov 13 '14 at 13:33
1
$begingroup$
@Victor afaik, die is the singular name form, while dice can be both singular or plural. I don't have any particular feelings about the word choice. You're welcome to edit the question if you feel my wording is not clear enough.
$endgroup$
– GOTO 0
Nov 13 '14 at 13:48
1
$begingroup$
@GOTO0 Not to be a stickler, but dice is only plural, strictly speaking. People tend to use it as singular but it is incorrect.
$endgroup$
– Neil
Nov 13 '14 at 14:18
2
$begingroup$
@GOTO0 english.stackexchange.com/questions/167104/singular-of-dice/…
$endgroup$
– Neil
Nov 13 '14 at 14:29
|
show 6 more comments
$begingroup$
This is a fairly well known puzzle that I feel deserves a place in Puzzling.SE. I will accept the best answer.
$endgroup$
– GOTO 0
Nov 13 '14 at 13:25
$begingroup$
oh my god thats a lot of work :) hopefully someone will try to solve it originally here
$endgroup$
– skv
Nov 13 '14 at 13:33
1
$begingroup$
@Victor afaik, die is the singular name form, while dice can be both singular or plural. I don't have any particular feelings about the word choice. You're welcome to edit the question if you feel my wording is not clear enough.
$endgroup$
– GOTO 0
Nov 13 '14 at 13:48
1
$begingroup$
@GOTO0 Not to be a stickler, but dice is only plural, strictly speaking. People tend to use it as singular but it is incorrect.
$endgroup$
– Neil
Nov 13 '14 at 14:18
2
$begingroup$
@GOTO0 english.stackexchange.com/questions/167104/singular-of-dice/…
$endgroup$
– Neil
Nov 13 '14 at 14:29
$begingroup$
This is a fairly well known puzzle that I feel deserves a place in Puzzling.SE. I will accept the best answer.
$endgroup$
– GOTO 0
Nov 13 '14 at 13:25
$begingroup$
This is a fairly well known puzzle that I feel deserves a place in Puzzling.SE. I will accept the best answer.
$endgroup$
– GOTO 0
Nov 13 '14 at 13:25
$begingroup$
oh my god thats a lot of work :) hopefully someone will try to solve it originally here
$endgroup$
– skv
Nov 13 '14 at 13:33
$begingroup$
oh my god thats a lot of work :) hopefully someone will try to solve it originally here
$endgroup$
– skv
Nov 13 '14 at 13:33
1
1
$begingroup$
@Victor afaik, die is the singular name form, while dice can be both singular or plural. I don't have any particular feelings about the word choice. You're welcome to edit the question if you feel my wording is not clear enough.
$endgroup$
– GOTO 0
Nov 13 '14 at 13:48
$begingroup$
@Victor afaik, die is the singular name form, while dice can be both singular or plural. I don't have any particular feelings about the word choice. You're welcome to edit the question if you feel my wording is not clear enough.
$endgroup$
– GOTO 0
Nov 13 '14 at 13:48
1
1
$begingroup$
@GOTO0 Not to be a stickler, but dice is only plural, strictly speaking. People tend to use it as singular but it is incorrect.
$endgroup$
– Neil
Nov 13 '14 at 14:18
$begingroup$
@GOTO0 Not to be a stickler, but dice is only plural, strictly speaking. People tend to use it as singular but it is incorrect.
$endgroup$
– Neil
Nov 13 '14 at 14:18
2
2
$begingroup$
@GOTO0 english.stackexchange.com/questions/167104/singular-of-dice/…
$endgroup$
– Neil
Nov 13 '14 at 14:29
$begingroup$
@GOTO0 english.stackexchange.com/questions/167104/singular-of-dice/…
$endgroup$
– Neil
Nov 13 '14 at 14:29
|
show 6 more comments
3 Answers
3
active
oldest
votes
$begingroup$
There are many examples. This is one:
A: 1, 1, 3, 5, 5, 6
B: 2, 3, 3, 4, 4, 5
C: 1, 2, 2, 4, 6, 6
The probability of A winning versus B (or B vs. C, C vs. A) is...
17/36. The probability of a draw is 4/36, so that only 15 out of 36 rolls lose. So the overall winning expectation is higher.
Source: http://en.wikipedia.org/wiki/Nontransitive_dice
edited 4 mins ago
North
2,3081735
answered Nov 13 '14 at 13:42
Ivo BeckersIvo Beckers
9,90923051
$endgroup$
add a comment |
$begingroup$
I solved this without looking at any of the previous answers and was surprised that there are many solutions!
I wrote a C program to brute-force the solution here: https://gist.github.com/timmontague/4244afcde2750f6f9c67
One of the "best" solutions that I found is the following set of dice:
d1: 1 1 4 4 4 4
d2: 3 3 3 3 3 3
d3: 2 2 2 2 5 5
I defined "best" as the set of dice with the largest average win percentage. In the example above there are no draws and the win percentages are:
d1 beats d2 24 out of 36 times
d2 beats d3 24 out of 36 times
d3 beats d1 20 out of 36 times
answered Nov 18 '14 at 22:44
TimTim
1412
$endgroup$
add a comment |
$begingroup$
One (of many) solutions:
Die A has sides 2, 2, 4, 4, 9, 9.
Die B has sides 1, 1, 6, 6, 8, 8.
Die C has sides 3, 3, 5, 5, 7, 7
The probability that die A rolls a higher number than B, the probability that B rolls higher than C, and the probability that C rolls higher than A are all 5/9, so this set of dice is "non-transitive". In fact, it has the even stronger property that, for each die in the set, there is another die that rolls a higher number than it more than half the time.
The game whith such dice is biased in favor of the second player, because the second player can always find a die that will beat the first player's die with probability 5/9.
So your friend should return the "courtesy" and allow you the first choice! :-)
answered Nov 13 '14 at 13:37
George R.George R.
42735
$endgroup$
1
$begingroup$
the description say that a side only can have numbers between 1 and 6
$endgroup$
– Ivo Beckers
Nov 13 '14 at 13:39
$begingroup$
@ Ivo Beckers: Oops! "Read fast,answer wrong!" :-) The link from Wiki you provided covers the matter thoroughly ,I think.
$endgroup$
– George R.
Nov 13 '14 at 16:41
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are many examples. This is one:
A: 1, 1, 3, 5, 5, 6
B: 2, 3, 3, 4, 4, 5
C: 1, 2, 2, 4, 6, 6
The probability of A winning versus B (or B vs. C, C vs. A) is...
17/36. The probability of a draw is 4/36, so that only 15 out of 36 rolls lose. So the overall winning expectation is higher.
Source: http://en.wikipedia.org/wiki/Nontransitive_dice
edited 4 mins ago
North
2,3081735
answered Nov 13 '14 at 13:42
Ivo BeckersIvo Beckers
9,90923051
$endgroup$
add a comment |
$begingroup$
There are many examples. This is one:
A: 1, 1, 3, 5, 5, 6
B: 2, 3, 3, 4, 4, 5
C: 1, 2, 2, 4, 6, 6
The probability of A winning versus B (or B vs. C, C vs. A) is...
17/36. The probability of a draw is 4/36, so that only 15 out of 36 rolls lose. So the overall winning expectation is higher.
Source: http://en.wikipedia.org/wiki/Nontransitive_dice
edited 4 mins ago
North
2,3081735
answered Nov 13 '14 at 13:42
Ivo BeckersIvo Beckers
9,90923051
$endgroup$
add a comment |
$begingroup$
There are many examples. This is one:
A: 1, 1, 3, 5, 5, 6
B: 2, 3, 3, 4, 4, 5
C: 1, 2, 2, 4, 6, 6
The probability of A winning versus B (or B vs. C, C vs. A) is...
17/36. The probability of a draw is 4/36, so that only 15 out of 36 rolls lose. So the overall winning expectation is higher.
Source: http://en.wikipedia.org/wiki/Nontransitive_dice
edited 4 mins ago
North
2,3081735
answered Nov 13 '14 at 13:42
Ivo BeckersIvo Beckers
9,90923051
$endgroup$
There are many examples. This is one:
A: 1, 1, 3, 5, 5, 6
B: 2, 3, 3, 4, 4, 5
C: 1, 2, 2, 4, 6, 6
The probability of A winning versus B (or B vs. C, C vs. A) is...
17/36. The probability of a draw is 4/36, so that only 15 out of 36 rolls lose. So the overall winning expectation is higher.
Source: http://en.wikipedia.org/wiki/Nontransitive_dice
edited 4 mins ago
North
2,3081735
answered Nov 13 '14 at 13:42
Ivo BeckersIvo Beckers
9,90923051
edited 4 mins ago
North
2,3081735
edited 4 mins ago
North
2,3081735
edited 4 mins ago
North
2,3081735
2,3081735
answered Nov 13 '14 at 13:42
Ivo BeckersIvo Beckers
9,90923051
answered Nov 13 '14 at 13:42
Ivo BeckersIvo Beckers
9,90923051
answered Nov 13 '14 at 13:42
Ivo BeckersIvo Beckers
9,90923051
9,90923051
add a comment |
add a comment |
$begingroup$
I solved this without looking at any of the previous answers and was surprised that there are many solutions!
I wrote a C program to brute-force the solution here: https://gist.github.com/timmontague/4244afcde2750f6f9c67
One of the "best" solutions that I found is the following set of dice:
d1: 1 1 4 4 4 4
d2: 3 3 3 3 3 3
d3: 2 2 2 2 5 5
I defined "best" as the set of dice with the largest average win percentage. In the example above there are no draws and the win percentages are:
d1 beats d2 24 out of 36 times
d2 beats d3 24 out of 36 times
d3 beats d1 20 out of 36 times
answered Nov 18 '14 at 22:44
TimTim
1412
$endgroup$
add a comment |
$begingroup$
I solved this without looking at any of the previous answers and was surprised that there are many solutions!
I wrote a C program to brute-force the solution here: https://gist.github.com/timmontague/4244afcde2750f6f9c67
One of the "best" solutions that I found is the following set of dice:
d1: 1 1 4 4 4 4
d2: 3 3 3 3 3 3
d3: 2 2 2 2 5 5
I defined "best" as the set of dice with the largest average win percentage. In the example above there are no draws and the win percentages are:
d1 beats d2 24 out of 36 times
d2 beats d3 24 out of 36 times
d3 beats d1 20 out of 36 times
answered Nov 18 '14 at 22:44
TimTim
1412
$endgroup$
add a comment |
$begingroup$
I solved this without looking at any of the previous answers and was surprised that there are many solutions!
I wrote a C program to brute-force the solution here: https://gist.github.com/timmontague/4244afcde2750f6f9c67
One of the "best" solutions that I found is the following set of dice:
d1: 1 1 4 4 4 4
d2: 3 3 3 3 3 3
d3: 2 2 2 2 5 5
I defined "best" as the set of dice with the largest average win percentage. In the example above there are no draws and the win percentages are:
d1 beats d2 24 out of 36 times
d2 beats d3 24 out of 36 times
d3 beats d1 20 out of 36 times
answered Nov 18 '14 at 22:44
TimTim
1412
$endgroup$
I solved this without looking at any of the previous answers and was surprised that there are many solutions!
I wrote a C program to brute-force the solution here: https://gist.github.com/timmontague/4244afcde2750f6f9c67
One of the "best" solutions that I found is the following set of dice:
d1: 1 1 4 4 4 4
d2: 3 3 3 3 3 3
d3: 2 2 2 2 5 5
I defined "best" as the set of dice with the largest average win percentage. In the example above there are no draws and the win percentages are:
d1 beats d2 24 out of 36 times
d2 beats d3 24 out of 36 times
d3 beats d1 20 out of 36 times
answered Nov 18 '14 at 22:44
TimTim
1412
answered Nov 18 '14 at 22:44
TimTim
1412
answered Nov 18 '14 at 22:44
TimTim
1412
answered Nov 18 '14 at 22:44
TimTim
1412
1412
add a comment |
add a comment |
$begingroup$
One (of many) solutions:
Die A has sides 2, 2, 4, 4, 9, 9.
Die B has sides 1, 1, 6, 6, 8, 8.
Die C has sides 3, 3, 5, 5, 7, 7
The probability that die A rolls a higher number than B, the probability that B rolls higher than C, and the probability that C rolls higher than A are all 5/9, so this set of dice is "non-transitive". In fact, it has the even stronger property that, for each die in the set, there is another die that rolls a higher number than it more than half the time.
The game whith such dice is biased in favor of the second player, because the second player can always find a die that will beat the first player's die with probability 5/9.
So your friend should return the "courtesy" and allow you the first choice! :-)
answered Nov 13 '14 at 13:37
George R.George R.
42735
$endgroup$
1
$begingroup$
the description say that a side only can have numbers between 1 and 6
$endgroup$
– Ivo Beckers
Nov 13 '14 at 13:39
$begingroup$
@ Ivo Beckers: Oops! "Read fast,answer wrong!" :-) The link from Wiki you provided covers the matter thoroughly ,I think.
$endgroup$
– George R.
Nov 13 '14 at 16:41
add a comment |
$begingroup$
One (of many) solutions:
Die A has sides 2, 2, 4, 4, 9, 9.
Die B has sides 1, 1, 6, 6, 8, 8.
Die C has sides 3, 3, 5, 5, 7, 7
The probability that die A rolls a higher number than B, the probability that B rolls higher than C, and the probability that C rolls higher than A are all 5/9, so this set of dice is "non-transitive". In fact, it has the even stronger property that, for each die in the set, there is another die that rolls a higher number than it more than half the time.
The game whith such dice is biased in favor of the second player, because the second player can always find a die that will beat the first player's die with probability 5/9.
So your friend should return the "courtesy" and allow you the first choice! :-)
answered Nov 13 '14 at 13:37
George R.George R.
42735
$endgroup$
1
$begingroup$
the description say that a side only can have numbers between 1 and 6
$endgroup$
– Ivo Beckers
Nov 13 '14 at 13:39
$begingroup$
@ Ivo Beckers: Oops! "Read fast,answer wrong!" :-) The link from Wiki you provided covers the matter thoroughly ,I think.
$endgroup$
– George R.
Nov 13 '14 at 16:41
add a comment |
$begingroup$
One (of many) solutions:
Die A has sides 2, 2, 4, 4, 9, 9.
Die B has sides 1, 1, 6, 6, 8, 8.
Die C has sides 3, 3, 5, 5, 7, 7
The probability that die A rolls a higher number than B, the probability that B rolls higher than C, and the probability that C rolls higher than A are all 5/9, so this set of dice is "non-transitive". In fact, it has the even stronger property that, for each die in the set, there is another die that rolls a higher number than it more than half the time.
The game whith such dice is biased in favor of the second player, because the second player can always find a die that will beat the first player's die with probability 5/9.
So your friend should return the "courtesy" and allow you the first choice! :-)
answered Nov 13 '14 at 13:37
George R.George R.
42735
$endgroup$
One (of many) solutions:
Die A has sides 2, 2, 4, 4, 9, 9.
Die B has sides 1, 1, 6, 6, 8, 8.
Die C has sides 3, 3, 5, 5, 7, 7
The probability that die A rolls a higher number than B, the probability that B rolls higher than C, and the probability that C rolls higher than A are all 5/9, so this set of dice is "non-transitive". In fact, it has the even stronger property that, for each die in the set, there is another die that rolls a higher number than it more than half the time.
The game whith such dice is biased in favor of the second player, because the second player can always find a die that will beat the first player's die with probability 5/9.
So your friend should return the "courtesy" and allow you the first choice! :-)
answered Nov 13 '14 at 13:37
George R.George R.
42735
answered Nov 13 '14 at 13:37
George R.George R.
42735
answered Nov 13 '14 at 13:37
George R.George R.
42735
answered Nov 13 '14 at 13:37
George R.George R.
42735
42735
1
$begingroup$
the description say that a side only can have numbers between 1 and 6
$endgroup$
– Ivo Beckers
Nov 13 '14 at 13:39
$begingroup$
@ Ivo Beckers: Oops! "Read fast,answer wrong!" :-) The link from Wiki you provided covers the matter thoroughly ,I think.
$endgroup$
– George R.
Nov 13 '14 at 16:41
add a comment |
1
$begingroup$
the description say that a side only can have numbers between 1 and 6
$endgroup$
– Ivo Beckers
Nov 13 '14 at 13:39
$begingroup$
@ Ivo Beckers: Oops! "Read fast,answer wrong!" :-) The link from Wiki you provided covers the matter thoroughly ,I think.
$endgroup$
– George R.
Nov 13 '14 at 16:41
1
1
$begingroup$
the description say that a side only can have numbers between 1 and 6
$endgroup$
– Ivo Beckers
Nov 13 '14 at 13:39
$begingroup$
the description say that a side only can have numbers between 1 and 6
$endgroup$
– Ivo Beckers
Nov 13 '14 at 13:39
$begingroup$
@ Ivo Beckers: Oops! "Read fast,answer wrong!" :-) The link from Wiki you provided covers the matter thoroughly ,I think.
$endgroup$
– George R.
Nov 13 '14 at 16:41
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@ Ivo Beckers: Oops! "Read fast,answer wrong!" :-) The link from Wiki you provided covers the matter thoroughly ,I think.
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– George R.
Nov 13 '14 at 16:41
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This is a fairly well known puzzle that I feel deserves a place in Puzzling.SE. I will accept the best answer.
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– GOTO 0
Nov 13 '14 at 13:25
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oh my god thats a lot of work :) hopefully someone will try to solve it originally here
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– skv
Nov 13 '14 at 13:33
1
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@Victor afaik, die is the singular name form, while dice can be both singular or plural. I don't have any particular feelings about the word choice. You're welcome to edit the question if you feel my wording is not clear enough.
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– GOTO 0
Nov 13 '14 at 13:48
1
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@GOTO0 Not to be a stickler, but dice is only plural, strictly speaking. People tend to use it as singular but it is incorrect.
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– Neil
Nov 13 '14 at 14:18
2
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@GOTO0 english.stackexchange.com/questions/167104/singular-of-dice/…
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– Neil
Nov 13 '14 at 14:29