Math puzzles!

"Since each face has a one in six chance of coming up, when three dice are tossed there are therefore 3 in 6 chances that the number I bet on will appear... and if my number comes up more than once... I win even more! So it must be a good game to play."

ex: you bet $1 on 6.
Triple sixes occur. you get your dollar back and win 3 dollars on top of that.
double sixes occur. you get your dollar back and win 2 dollars on top of that.
single six occurs. you get your dollar back and win 1 dollar on top of that.
no sixes occur. You lose your dollar.


Is it? Your job is to work out the true odds of this game!

Going back to this one, cause it bothered me that nobody solved it this way...
For each correct die, you win $1.
so 1/6*$1 + 1/6*$1 + 1/6*$1 = $0.50 per game
But: If none of the dice are right, you lose $1
so 5/6*5/6*5/6*$-1 = 125/216*$-1 = $-0.5787 per game
Add them up: $-0.0787 per game, or $0.9213 return on $1 investment.