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

What are the changes of an odd number appearing on die 1 and 3, and an even number appearing on dice 2?

Answer 2

1st create your sample space, allocate probabilities to each without the weights.

When you compute the probability, use the weights of the 3's by adjusting the probabilities you gave them above according to their relative weights.

The weights just introduce bias. For example, if we have a coin and say that heads is 5% heavier than tails, we say rather than 50/50 of heads to tails, we say prob of heads is 0.5*1.05 and prob of tails is 0.5 * 0.95 so prob of heads is 1.05/(1.05+0.95) = 0.525 rather than the original, 0.5.

This will be rather long process but that is the basic idea.

Answer 3

Die I: 1, 2, 2, 3, 4, 5
Die II: 3, 4, 4, 4, 5, 6
Die III: 1, 1, 3, 3, 3, 5
--------------------------------------------------
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
--------------------------------------------------
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)

--------------------------------------------------------
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)