Question

If you have 3 6-sided dice.

The first die has numbers 1-5; with 2 twice.

The second die has numbers 3-6, with numbers 4 three times,

The third die has only odd numbers from 1-6, with one twice and three thrice.

Die one is weighted to where the number 1 is 2% more likely to appear than the number 3 on die 2.

Die two is weighted to where 5 is 5% more likely to appear than 5 on die three.

Die three is weighted where 3 is 4% less likely to appear than the prob of 2 on die 1 and 4 on die 2.

What are the chances of 3 appearing on all 3 die?

Answer 1

Die I: 1, 2, 2, 3, 4, 5
Die II: 3, 4, 4, 4, 5, 6
Die III: 1, 1, 3, 3, 3, 5
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Die I Prop alpha (unchanged):
1: (1/6)
2: (2/6) = (1/3)
3: (1/6)
4: (1/6)
5: (1/6)
6: (0/6)
Die II Prop alpha (unchanged):
1: (0/6)
2: (0/6)
3: (1/6)
4: (3/6) = (1/2)
5: (1/6)
6: (1/6)
Die III Prop alpha (unchanged):
1: (2/6) = (1/3)
2: (0/6)
3: (3/6) = (1/2)
4: (0/6)
5: (1/6)
6: (0/6)
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Die I Prop. beta (altered, final)
1: 0.17
2: (2/6) = (1/3)
3: (1/6)
4: (1/6)
5: (1/6)
6: (0/6)
Die II Prop. beta (altered, final)
1: (0/6)
2: (0/6)
3: (1/6)
4: (3/6) = (1/2)
5: 0.175
6: (1/6)
Die III Prop. beta (altered, final)
1: (2/6) = (1/3)
2: (0/6)
3: 0.16
4: (0/6)
5: (1/6)
6: (0/6)
probability(2 on die 1 and 4 on die 2) = (1/3)(1/2) = (1/6)
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probability(3 on die 1, 3 on die 2, and 3 on die 3) = (1/6)(1/6)(0.16) = 0.004-bar ---------------------------------------------------------------
probability(odd on die 1, even on die 2, odd on die 3) = probability(1 or 3 or 5 on die 1)probability(2 or 4 or 6 on die 2) *probability(1 or 3 or 5 on die 3)

Answer 2

let probability(1 or 3 of 5 on die 1) = p(a)
let probability(2 or 4 or 6 on die 2) = p(b)
let probability(1 or 3 or 5 on die 3) = p(c)

p(a) = 0.17+(1/6)+(1/6) = 0.503-bar
p(b) = (0/6)+(1/2)+(1/6) = (2/3)
p(c) = (1/3) +0.16+(1/6) = 0.66

p(a) and p(b) and p(c) = (0.503-bar)(2/3)(0.66) = 0.22146

Answer 3

@satellite73, could you just see if the work seems right?