Almost lost 10$ at the bar

#1
First off, I am kind of drunk, so excuse the simple mistakes. I was just given even odds on a bet on 1 roll of 5 dice. (I didn't take it) Meaning we each put $10 on the bar and whoever guesses what shows up on the 5 dice gets all of it. They bet that given 1 roll of a total of 5 dice, and given the fact that 1's are wild, that they could roll 3 of a kind. Meaning 1+1+2+3+4 is 3 of a kind 4's, and 4+4+4+2+3 is as well.

This is how I am trying to calculate it:

Over all goal is to calculate all possible wins / all possible combinations

a-f are all wins

(a) (odds of exactly 5 1s)
(1/6^5) +
(b) (odds of exactly 4 1s and anything for remaining 1 dice)
((1/6^4)*5) +
(c) (odds of exactly 3 1s and anything for remaining 1 dice)
((1/6^3)*10) +
(d) (odds of exactly 2 1s and anything for remaining 1 dice)
((1/6^2)*10) +
(e) (odds of exactly 1 1) times (odds of a pair (not 1's) out of remaining 4)
((1/6^1)*5)*(???????) +
(f) (odds of exactly 0 1's times (odds of three of a kind (not 1's) out of remaining 5) =
???????

/

all possible combinations is 6^5, or is it (6+5-1)! / ((5!)(6-1!)) ???

=

????

The guy claimed it was 1/3 but I really need to prove him wrong. Thank you so much for anyone who even read this!!!:wave: