Dice Probability?

Question

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!

Basically, what are the chances of rolling 5 dice and getting one of the four combinations, a), b) c), or d)?

Answer 1

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)

Answer 2

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

Answer 3

a&b: 1 in 216
c: 1 in 36
d: 1 in 6