Dice Probability?

I came a across this math problem and was looking for some math wiz's to help me solve it! Here it is:





Suppose you had 5 regular dice. What are the odds of rolling one of the four following combinations? (the order rolled does not matter)





a) 1,1,3,4,6


b) 3,3,4,5,6


c) 2,3,5,5, and a random number


d) 2,3,5, and two random numbers





Thanks in advance!

Dice Probability?
P(a) = 5! / 2! * (1/6)^5 = 5/648


P(b) = same as (a) = 5/648





Also note that a and b are mutually exclusive


c however is a subset of d, ie when you get c you automatically get d.





Working out P(d) is a little more complicated.


There are 5 cases to investigate:


Case 1: 2,3,5 and one of 2,3,5 appearing twice more


Case 2: 2,3,5 and one of 1,4,6 appearing twice


Case 3: 2,3,5 and two more of 2,3,5


Case 4: 2,3,5 and two of 1,4,6


Case 5: 2,3,5 and one of 2,3,5 and one of 1,4,6





1) Consider 2,3,5,2,2


There are 5!/3! ways of arranging that. The same is true for


2,3,5,3,3


2,3,5,5,5





2) Consider 2,3,5,1,1


2,3,5,4,4


2,3,5,6,6


For each there are 5! / 2! ways of arranging them.





3) Consider


2,3,5,1,4; 2,3,5,1,6; 2,3,5,4,6


For each there are 5! ways of arranging





4) Consider 2,3,5,2,3; 2,3,5,2,5; 2,3,5,3,5


There are 5!/2!/2! arrangements





5) Consider 2,3,5,1,2; 2,3,5,1,3; 2,3,5,1,5 etc


For each there are 5! / 2! arrangements





P(d) = 1/6^5 * [3*5!/3! + 3*5!/2! + 3*5! + 3*5!/2!/2! + 9*5!/2!]


= 205/1296





P(a or b or d) = 5/648 + 5/648 + 205/1296


= 25/144 = 0.1736





NOTE that P(d) already incorporates P(c)
Reply:a%26amp;b: 1 in 216


c: 1 in 36


d: 1 in 6
Reply:For (a) the probability is 5!/2x6^5=20/6^4=5/588


For (b) the probability is again 5/588


For (c)4!/2x6^4=


For (d)3!/6^3=1/36


Since any of (a), (b), (c), (d) is allowed add all probabilities,


(5/588)+(5/588)+( )+(1/36)


=calculate