----- Original Message -----
From: GEORGE PERKINS <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, June 29, 2000 2:07 PM
Subject: Dice Problem
> Recently a colleague came in the office with the following problem:
>
> Is there a way to 'load' two individual die so that all sums will be
equally likely?
>
> (I take it that they would like to load the die in such a way that the sum
of 2 is equally likely as a sum of 3, as a sum of 4...etc)
>
> Any ideas would be appreciated
> GP
If we specify that both dice shall be labelled 1,2,3,4,5,6, and that
they shall roll independently, then the answer is "no".
Proof: Let X,Y,S be the individual values and the sum.
P(X=6)P(Y=6) = P(S=12) > 0 so P(X=6) and P(Y=6) are positive.
Similarly P(X=1) and P(Y=1) are positive.
P(X=1)P(Y=1) = P(S=2) = P(S=7) >= P(X=1)P(Y=6) + P(X=6)P(Y=1)
> P(X=1)P(Y=6)
Thus P(Y=1) > P(Y=6); but interchanging values we also prove the
opposite inequality. Contradiction!
If the dice are labelled in a nonstandard way, we can get (eg) 36
equally probable outcomes: (1,2,3,4,5,6) and (0,6,12,18,24,30)
If the dice are in some way coordinated, so that the fall of one die
influences the fall of the other, we can have any probabilities we like.
-Robert Dawson
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