Asked by Anonymous
How do you calculate the probability of a sum with one fair die and one biased die?
P(1)=1/3
P(2)=P(3)=P(4)=P(5)=P(6)
P(1)=1/3
P(2)=P(3)=P(4)=P(5)=P(6)
Answers
Answered by
economyst
I presume you gave the probability of the biased die, where there is a 33.33% chance of rolling a 1 and a 13.33% chance of rolling a 2,3,4,5 or 6. For a normal die, the probability of rolling any given number is 1/6 = 16.67%.
There is one way to roll a 2, both dies are 1. so, Psum(2) = .1666 * .3333 = .0555
There are two ways to roll an 3, 1-2, and 2-1. So Psum(3) = (.16666*.3333)+(.16666*.13333) = .0777
Repeat for values 4 to 12.
There is one way to roll a 2, both dies are 1. so, Psum(2) = .1666 * .3333 = .0555
There are two ways to roll an 3, 1-2, and 2-1. So Psum(3) = (.16666*.3333)+(.16666*.13333) = .0777
Repeat for values 4 to 12.
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