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How to use dice to randomly select from 10 choices? [duplicate]
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This question already has an answer here:
How to generate a random number between 1 and 10 with a six-sided die?
21 answers
Today I wanted to randomly select a number (0 to 9) using a six-faced dice.
Because a single dice roll would not be sufficient (7, 8, 9, 0 would be missed), I determined that more than one dice roll would be necessary:
First roll
1 2 3 4 5 6
+------------
S 1 | 1 2 3 4 5 6
e r 2 | 7 8 9 0 1 2
c o 3 | 3 4 5 6 7 8
o l 4 | 9 0 1 2 3 4
n l 5 | 5 6 7 8 9 0
d 6 | 1 2 3 4 5 6
The first dice roll determines the column, the second dice roll determines the row from which the random number would be taken.
Unfortunately, I encountered the same problem i.e. 7, 8, 9, 0 appear less frequently in the table. The number of dice rolls could be increased further, but the same problem would be encountered.
My question: what is the easiest way to select 1 item from 10 choices if I only have a six-sided dice?
probability random
asked Jan 19 at 10:25
FluxFlux
13016
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marked as duplicate by Chris Culter, drhab probability
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Jan 19 at 11:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
How to generate a random number between 1 and 10 with a six-sided die?
21 answers
Today I wanted to randomly select a number (0 to 9) using a six-faced dice.
Because a single dice roll would not be sufficient (7, 8, 9, 0 would be missed), I determined that more than one dice roll would be necessary:
First roll
1 2 3 4 5 6
+------------
S 1 | 1 2 3 4 5 6
e r 2 | 7 8 9 0 1 2
c o 3 | 3 4 5 6 7 8
o l 4 | 9 0 1 2 3 4
n l 5 | 5 6 7 8 9 0
d 6 | 1 2 3 4 5 6
The first dice roll determines the column, the second dice roll determines the row from which the random number would be taken.
Unfortunately, I encountered the same problem i.e. 7, 8, 9, 0 appear less frequently in the table. The number of dice rolls could be increased further, but the same problem would be encountered.
My question: what is the easiest way to select 1 item from 10 choices if I only have a six-sided dice?
probability random
asked Jan 19 at 10:25
FluxFlux
13016
$endgroup$
marked as duplicate by Chris Culter, drhab probability
Users with the probability badge can single-handedly close probability questions as duplicates and reopen them as needed.
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Jan 19 at 11:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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You could order the throws. Let the first throw decide fifty-fifty whether the number will be $S-1$ or $5+S-1$ where $S$ is the second score. Throw again if $S=6$ until a non-six has been throwed.
$endgroup$
– drhab
Jan 19 at 10:40
add a comment |
$begingroup$
This question already has an answer here:
How to generate a random number between 1 and 10 with a six-sided die?
21 answers
Today I wanted to randomly select a number (0 to 9) using a six-faced dice.
Because a single dice roll would not be sufficient (7, 8, 9, 0 would be missed), I determined that more than one dice roll would be necessary:
First roll
1 2 3 4 5 6
+------------
S 1 | 1 2 3 4 5 6
e r 2 | 7 8 9 0 1 2
c o 3 | 3 4 5 6 7 8
o l 4 | 9 0 1 2 3 4
n l 5 | 5 6 7 8 9 0
d 6 | 1 2 3 4 5 6
The first dice roll determines the column, the second dice roll determines the row from which the random number would be taken.
Unfortunately, I encountered the same problem i.e. 7, 8, 9, 0 appear less frequently in the table. The number of dice rolls could be increased further, but the same problem would be encountered.
My question: what is the easiest way to select 1 item from 10 choices if I only have a six-sided dice?
probability random
asked Jan 19 at 10:25
FluxFlux
13016
$endgroup$
This question already has an answer here:
How to generate a random number between 1 and 10 with a six-sided die?
21 answers
Today I wanted to randomly select a number (0 to 9) using a six-faced dice.
Because a single dice roll would not be sufficient (7, 8, 9, 0 would be missed), I determined that more than one dice roll would be necessary:
First roll
1 2 3 4 5 6
+------------
S 1 | 1 2 3 4 5 6
e r 2 | 7 8 9 0 1 2
c o 3 | 3 4 5 6 7 8
o l 4 | 9 0 1 2 3 4
n l 5 | 5 6 7 8 9 0
d 6 | 1 2 3 4 5 6
The first dice roll determines the column, the second dice roll determines the row from which the random number would be taken.
Unfortunately, I encountered the same problem i.e. 7, 8, 9, 0 appear less frequently in the table. The number of dice rolls could be increased further, but the same problem would be encountered.
My question: what is the easiest way to select 1 item from 10 choices if I only have a six-sided dice?
This question already has an answer here:
How to generate a random number between 1 and 10 with a six-sided die?
21 answers
probability random
probability random
asked Jan 19 at 10:25
FluxFlux
13016
asked Jan 19 at 10:25
FluxFlux
13016
asked Jan 19 at 10:25
FluxFlux
13016
asked Jan 19 at 10:25
FluxFlux
13016
asked Jan 19 at 10:25
FluxFlux
13016
13016
marked as duplicate by Chris Culter, drhab probability
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Jan 19 at 11:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Chris Culter, drhab probability
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Jan 19 at 11:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
You could order the throws. Let the first throw decide fifty-fifty whether the number will be $S-1$ or $5+S-1$ where $S$ is the second score. Throw again if $S=6$ until a non-six has been throwed.
$endgroup$
– drhab
Jan 19 at 10:40
add a comment |
$begingroup$
You could order the throws. Let the first throw decide fifty-fifty whether the number will be $S-1$ or $5+S-1$ where $S$ is the second score. Throw again if $S=6$ until a non-six has been throwed.
$endgroup$
– drhab
Jan 19 at 10:40
$begingroup$
You could order the throws. Let the first throw decide fifty-fifty whether the number will be $S-1$ or $5+S-1$ where $S$ is the second score. Throw again if $S=6$ until a non-six has been throwed.
$endgroup$
– drhab
Jan 19 at 10:40
$begingroup$
You could order the throws. Let the first throw decide fifty-fifty whether the number will be $S-1$ or $5+S-1$ where $S$ is the second score. Throw again if $S=6$ until a non-six has been throwed.
$endgroup$
– drhab
Jan 19 at 10:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Mark you last row with $*$ in place of a digit. If you get one of these results then throw again. Repeat as necessary until you have a valid result.
Slightly more simply, throw the first die once and then throw the second repeatedly until you get something other than $6$.
There is a chance that you might need to continue throwing the second die forever but the probability is $0$.
The problem is related to the problem of writing $1 / 3$ as a decimal. You cannot get it exact with a finite number of decimal places. Your problem is trying to write $1 / 10$ exactly in vase $6$.
answered Jan 19 at 10:39
badjohnbadjohn
4,3521720
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add a comment |
$begingroup$
Create a probability chart like the one you already made, but leave diagonal entries crossed out. So this means if you roll doubles, just roll again. This leaves 30 outcomes, so list each possible choice three times.
answered Jan 19 at 10:41
Ryan GreylingRyan Greyling
3499
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Mark you last row with $*$ in place of a digit. If you get one of these results then throw again. Repeat as necessary until you have a valid result.
Slightly more simply, throw the first die once and then throw the second repeatedly until you get something other than $6$.
There is a chance that you might need to continue throwing the second die forever but the probability is $0$.
The problem is related to the problem of writing $1 / 3$ as a decimal. You cannot get it exact with a finite number of decimal places. Your problem is trying to write $1 / 10$ exactly in vase $6$.
answered Jan 19 at 10:39
badjohnbadjohn
4,3521720
$endgroup$
add a comment |
$begingroup$
Mark you last row with $*$ in place of a digit. If you get one of these results then throw again. Repeat as necessary until you have a valid result.
Slightly more simply, throw the first die once and then throw the second repeatedly until you get something other than $6$.
There is a chance that you might need to continue throwing the second die forever but the probability is $0$.
The problem is related to the problem of writing $1 / 3$ as a decimal. You cannot get it exact with a finite number of decimal places. Your problem is trying to write $1 / 10$ exactly in vase $6$.
answered Jan 19 at 10:39
badjohnbadjohn
4,3521720
$endgroup$
add a comment |
$begingroup$
Mark you last row with $*$ in place of a digit. If you get one of these results then throw again. Repeat as necessary until you have a valid result.
Slightly more simply, throw the first die once and then throw the second repeatedly until you get something other than $6$.
There is a chance that you might need to continue throwing the second die forever but the probability is $0$.
The problem is related to the problem of writing $1 / 3$ as a decimal. You cannot get it exact with a finite number of decimal places. Your problem is trying to write $1 / 10$ exactly in vase $6$.
answered Jan 19 at 10:39
badjohnbadjohn
4,3521720
$endgroup$
Mark you last row with $*$ in place of a digit. If you get one of these results then throw again. Repeat as necessary until you have a valid result.
Slightly more simply, throw the first die once and then throw the second repeatedly until you get something other than $6$.
There is a chance that you might need to continue throwing the second die forever but the probability is $0$.
The problem is related to the problem of writing $1 / 3$ as a decimal. You cannot get it exact with a finite number of decimal places. Your problem is trying to write $1 / 10$ exactly in vase $6$.
answered Jan 19 at 10:39
badjohnbadjohn
4,3521720
answered Jan 19 at 10:39
badjohnbadjohn
4,3521720
answered Jan 19 at 10:39
badjohnbadjohn
4,3521720
answered Jan 19 at 10:39
badjohnbadjohn
4,3521720
4,3521720
add a comment |
add a comment |
$begingroup$
Create a probability chart like the one you already made, but leave diagonal entries crossed out. So this means if you roll doubles, just roll again. This leaves 30 outcomes, so list each possible choice three times.
answered Jan 19 at 10:41
Ryan GreylingRyan Greyling
3499
$endgroup$
add a comment |
$begingroup$
Create a probability chart like the one you already made, but leave diagonal entries crossed out. So this means if you roll doubles, just roll again. This leaves 30 outcomes, so list each possible choice three times.
answered Jan 19 at 10:41
Ryan GreylingRyan Greyling
3499
$endgroup$
add a comment |
$begingroup$
Create a probability chart like the one you already made, but leave diagonal entries crossed out. So this means if you roll doubles, just roll again. This leaves 30 outcomes, so list each possible choice three times.
answered Jan 19 at 10:41
Ryan GreylingRyan Greyling
3499
$endgroup$
Create a probability chart like the one you already made, but leave diagonal entries crossed out. So this means if you roll doubles, just roll again. This leaves 30 outcomes, so list each possible choice three times.
answered Jan 19 at 10:41
Ryan GreylingRyan Greyling
3499
answered Jan 19 at 10:41
Ryan GreylingRyan Greyling
3499
answered Jan 19 at 10:41
Ryan GreylingRyan Greyling
3499
answered Jan 19 at 10:41
Ryan GreylingRyan Greyling
3499
3499
add a comment |
add a comment |
$begingroup$
You could order the throws. Let the first throw decide fifty-fifty whether the number will be $S-1$ or $5+S-1$ where $S$ is the second score. Throw again if $S=6$ until a non-six has been throwed.
$endgroup$
– drhab
Jan 19 at 10:40