You play poker by rolling 5 ordinary die. Numbers on the die correspond to ranks on the cards. For example, two pairs means you rolled the numbers $aabbc$ in any order, where $a,b,c$ are different numbers. Calculate:
a) P(full house), ie the pattern $aabbb$
Attempt: By counting, we have $6.1.5.1.1. \frac{5!}{2! 3!}$ possible rearrangements. Dividing this by the sample space, $|s| = 6^5$ gives approx $0.04,$ I believe this is the right answer
b) P(one pair), ie the pattern $aabcd$
Attempt: So there are $6.1.5.4.3\frac{5!}{2!}$ possible rearrangements. Divide by $|s|$ to get a number > 1. I am not sure where I went wrong here. My reasoning for the $5!/2!$ was that for any one combination of 5 numbers, you can order them 5! ways. However, 2 of the numbers are the same, so we divide by 2!.
c) P({$aabbc$}) = $6.1.5.1.4.\frac{5!}{2!.2!}$ all divided by $6^5$ gives 0.46, exactly twice the suggested answer.
I also wonder now why the sample space would not be $(6^5 . 5!)$?
d) How many rolls you need for a full house on average? Attempt: So to get a full house, you need $aabbb$. Roll a die. Roll a die again.The expected number of rolls needed to get a matching number on the first roll is 6. I tried to extend this arguement to find how many rolls you need to roll a different number other than the one already got, but didn't get very far. I suppose this method also neglects the fact that we don't care about the order of the full house. Could I have a hint on this one (not the answer)?