Nonuniformity of dice totals. Consider dice whose six faces are labeled 1,2,3,4,

Question

Nonuniformity of dice totals. Consider dice whose six faces are labeled 1,2,3,4,5,6 but are not necessarily equally likely. Show that there does not exist a die with the property that, when two identical copies of it are tossed, the 11 possible totals, 2-12, are equally likely. Hint: let pi denote the probability that the die lands with face i up. Let X denote the sum of two dice. Relate P[X=2], P[X=3], P[X=4] to p1,p2,p3 and likewise relate P[X=12], P[X=11], P[X=10] to p4,p5,p6 and show there is a contradiction if such a die exists.

Explanation / Answer

P(X =2) = p1*p1

P(X = 3) = 2 p1 * p2

P(X = 4) = 2*p1*p3 + p2 ^2

P[X=12] = p6^2

P[X=11]= 2 *p5*p6

P[X=10]= p5^2 + 2 p4*p6

P(x4) = P(X5) = P(X6) = P(X12) = P(X11) = P(x10)

p1*p1 = 2 p1 * p2 =2*p1*p3 + p2 ^2 = p6^2=2 *p5*p6= p5^2 + 2 p4*p6

hence

p1 = p6

p2 = p5

p3 = p4

p1 = 2 p2

2 p1 * p2 =2*p1*p3 + p2 ^2   so 4 p2^2 = 4 p2*p3 + p2^2

3 p2^2 = 4 p2* p3

p3 = 3 p2 / 4

p1 + p2 + p3 + p4 + p5 + p6 = 1

p1 + p2 + p3 = 1/2

2 p2 + p2 + 3 p2/4 = 1/2

p2 ( 2 + 1 + 3/4) = 1/2

p2 (15/4) = 1/2

p2 = 2/15 = p5

p1 = 4/15 = p6

p3 = 1/10 = p4

now

P(X=2) = P(X=3) =..P(X = 12) = p

hence

P(X=2) = P(X=3). = 1/11 { all are equally likely)

but

P(X = 2) = p1^2 = (4/15)^2 = 16/225

which is not equal to 1/11

hence there is contradiction

hence such a die does not exist

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