In a dice game, 2 players each simultaneously cast 4 fair 6-sided dice, to form a 'hand'.
Each player then:-
(i) discards the highest-scoring die from his hand, and
(ii) scores the value of the next-highest (or highest-equal) die in his hand.
If my opponent has cast {6,4,4,3}, discarding the 6 to score 4, what is the probability that my score exceeds his?
(a) The probability that a single die exceeds 4 is 1/3.
The probability that a single die does not exceed 4 is 2/3.
(b) The probability that none of 4 dice exceed 4 = (2/3)^4 = 16/81
(c) The probability that exactly one die exceeds 4 = 4*(1/3)*(2/3)^3 = 32/81
As (b) and (c) are mutually exclusive, and as these are the only way to lose or tie, the second player wins with probability
1 - 16/81 - 32/81 = 33/81 = 11/27
This result agrees with the previous two answers
Edited on February 18, 2017, 10:34 pm
Solution (spoiler)
