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 Custom Dice (Posted on 2006-11-16)
You are given blank six-sided dice, with equal weight on each side (fair dice). You're allowed to write on each side of each die a number between 0 and 6. Numbers may repeat themselves on a single die, so for example, one die can have five 4s and one 0.

1) Can you create two such dice which, if they are rolled and their results added, will give equal odds for every number from 1 to 12?

2) Can you create three such dice which, if they are rolled and their results added, will give equal odds for every number from 1 to 18?

(NOTE: Don't get smart and make all the dice all 0s. The probability of getting 0 should be 0)

 No Solution Yet Submitted by TamTam Rating: 4.0000 (1 votes)

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 Solution (including 0-6 limit) | Comment 3 of 5 |
1)   0,0,0,6,6,6  and 1,2,3,4,5,6

2)  no you cannot.

Proof of 2:  Assume it can be done

all three dice must have 6s (to get 18)
one die must not have 0s (to get 0 prob for 0)
The other two must have 0s and the special one must have ones (to get 1s)
call special die (with 1s but not 0s) a and the others b and c,  and the number of xs on y xy  a6 = (# of 6s on a) for example so
a1>0 a6>0, b0>0 b6>0, c0>0 c6>0

a1*b0*c0 = 12
a6*b6*c6 = 12

a1*(b0*c6+b6*c0) <= 12    #all make 7
a6*(b0*c6+b6*c0) <= 12    #all make 12

a1*(b0*c6+b6*c0) <= a1*b0*c0
a6*(b0*c6+b6*c0) <= a6*b6*c6

b0*c6+b6*c0 <= b0*c0    # divide both sides by same postive
b0*c6+b6*c0 <= b6*c6    # divide both sides by same postive

b0*c6 < b0*c0                 # term removed positive
b6*c0 < b6*c6                 # term removed positive

c6 < c0
c0 < c6

Contradition.  So, it can't be done.

Edited on November 16, 2006, 10:22 am
 Posted by Joel on 2006-11-16 10:19:49

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