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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?

Flux
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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. – drhab Jan 19 '19 at 10:40

2 Answers2

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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 $6$.

badjohn
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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.

Ryan
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