Jane has a 3-sided die and Dick has an 8-sided die. They each roll independently until someone rolls a 1, at which point the other person does the dishes. The probability that Dick rolls a 1 before Jane is calculated as follows: The probability that Jane rolls a 1 on any given roll is 1/3, while the probability that Dick rolls a 1 is 1/8. The probability that Dick rolls a 1 before Jane is 2187/2099339.

1. Suppose Jane has a fair 3-sided die, and Dick has a fair 8-sided die. Each day, they roll their dice
(independently) until someone rolls a 1. (Then the person who did not roll a 1 does the dishes.)
Find the probability that
Dick rolls the first 1 before Jane does?
Solution
Jane has a 3 sided die, whereas Dick has an 8 sided die.
When Jane rolls the die, she has probability of 1/3 that she gets any one of the sides.
But Dick has an 8 sided die, and the probability of getting a particular face is 1/8.
So, lets calculate the probablity of Jane getting the particular combination in question.
In the first throw, the prob. is 1/3.
And the prob. in all the throws would be 1/3.
Thus, P( Jane gets the combination) = (1/3)^7 = 1/2187.
And prob that Dick gets the required combination = (1/8)^7 = 1/2097152/
Now, suppose one of them gets the combination.
P( any one of them getting the combination) = 1/2187 + 1/2097152
= (2187 + 2097152)/(2187*2097152)
When one of them gets the combination, the prob. that Dick gets it
= (1/2097152)/(1/2187 + 1/2097152)
= (2097152 * 2187)/[2097152*(2187+2097152)]
= 2187 / 2099339
This is the required probability.

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