If 4 fair dice are thrown simulatenously, what is the probability of getting at least one pair?
(I was able to compute the solution using Permutations, but I wanted to try calculating one pair and two pair probabilities)
Method1 -
One pair => there are 12/2 = 6 arrangements for a pair in XXYZ
(6*1*5*4) 6
----------- [First two dice form a pair, the other two don't]
(6*6*6*6)
=20/36 = 10/18
Similarly, Two different pair => there are 3 arrangements for a pair in XXYY
(6*1*5*1)*3
-----------
6*6*6*6
=5/72
Two same pair (i.e. four dice with the same number) => 6/(6*6*6*6) = 1/6*6*6
Total = 10/18 + 5/72 + 1/6*6*6 != 13/18, which is the correct answer. CRASHED
Please help
Method 2 (using perm) => 1-(6*5*4*3/6*6*6*6) = 13/18
Thanks
(I was able to compute the solution using Permutations, but I wanted to try calculating one pair and two pair probabilities)
Method1 -
One pair => there are 12/2 = 6 arrangements for a pair in XXYZ
(6*1*5*4) 6
----------- [First two dice form a pair, the other two don't]
(6*6*6*6)
=20/36 = 10/18
Similarly, Two different pair => there are 3 arrangements for a pair in XXYY
(6*1*5*1)*3
-----------
6*6*6*6
=5/72
Two same pair (i.e. four dice with the same number) => 6/(6*6*6*6) = 1/6*6*6
Total = 10/18 + 5/72 + 1/6*6*6 != 13/18, which is the correct answer. CRASHED
Please help
Method 2 (using perm) => 1-(6*5*4*3/6*6*6*6) = 13/18
Thanks
