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### Using a 6-sided Die to Generate a Random Number From 1 to 7

```Date: 12/23/2007 at 11:34:38
From: James
Subject: Generate random 1-7 number given a single six-sided die

During playing a card game I am asked to select a card at random from
my opponent's hand of 7 cards.  I have only a 6-sided die.  What
reasonable amount of time and number of steps can I take in order to
ensure a random selection of 1 card?  I would assign 1 to the left
most card (with respect to me) in my opponent's hand and continue to
the right assigning numbers 2-7.

If an opponent's hand has less than 6 cards, I can just re-roll on a
number that would not correspond to a card.  However I can not do such
a thing for 7 cards.  If I assign 1-5 for the first through fifth
cards in my opponent's hand I can't assign 6 to the last card and then
roll the die again to determine which to select (even for 6 and odd
for 7) since that would not yield a random result.

I thought about rolling the 6-sided die first and even and odd would
then tell me which half to select from.  For example roll 3:odd, take
the last four cards.  Then roll the die to determine which of the last
four to take from with 5-6 being re-rolls.  But I do not know if that
would give a skew.

```

```
Date: 12/24/2007 at 13:16:25
From: Doctor Tom
Subject: Re: Generate random 1-7 number given a single six-sided die

Hi James,

There are a bunch of options.  Here's an easy one:  roll the die
twice, keeping track of the first and second roll.  There are 36 outcomes:

(1,1), (1,2), (1,3), ..., (6,6)

(The first number is what you got on the first roll; the second is
what you got on the second.)

If you get a (6,6), just re-roll the die twice again until you get a
non-(6,6).

Now there are 35 equally-likely outcomes, so divide them into 7 groups
of 5 corresponding to the 7 choices among which you want to choose.

Here's an easy calculation that will do it.  Suppose you roll a (x, y)
with x = number on first roll and y = number on second roll.  First
calculate the following number:

N = 6(x-1) + y.

These possible values of N (all equally likely) will run from 1 to 35
(assuming you disallow a (6, 6)), including each value exactly once.

Then divide the number N by 7 and look only at the remainder.  For
example, if you rolled a (4, 5), the number N would be 6(4-1)+5 = 23.
Divide 7 into 23 and it goes in 3 times with a remainder of 2.  If
the number is exactly divisible by 7, the remainder is zero, so there
are 7 equally-likely possible outcomes for the remainder: 0, 1, 2, 3,
4, 5 and 6.

Note that this solution is not perfect in the sense that there is no
upper bound to the number of times you'd need to roll a pair of dice
to obtain a non-(6,6), but in the practical world, even getting two
(6,6) outcomes in a row will only happen less than one time in a
thousand.  Obviously you're willing to do something like that, since
you talked about getting one of 5 outcomes by rolling the die and
re-rolling if you get a 6.

- Doctor Tom, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Probability
High School Probability

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