Dice and probability

Playing dice recently we got into a conversation about dice odds. When I first played I figured if the odds of rolling a 1 or a 5 (a scoring number) was 1 in 3 for one die, it had to be about even for three dice. Looking closer, that's not exactly the case.

The odds for one die: 4 of 6 losing combinations, meaning 2 of 6 wins. 1 in 3

The odds for two dice: 4/6 x 4/6 = 16/36 = .444 or 44 of 100 losses, meaning 55.6 of 100 wins. 11 in 20 (roughly)

The odds for three dice: 4/6 x 4/6 x 4/6 = 64/216 = .296 or 30 of 100 losses, meaning 70 of 100 wins. For three dice, we must subtract 4 other scoring combinations of triple digits (222, 333, 444, and 666, as 1s and 5s are already accounted for). So instead of 64/216, the number really is 60/216 = .277 or 28 of 100 losses, meaning 72 of 100 wins. 7.2 in 10

For four dice, it gets more complicated because we must count triple digit scoring rolls (111, 222, 333, etc.) as well as any 1s or 5s...without double-counting 1 and 5 triples. My math ran out of gas at this point. (Anyone good with probability math who can explain the formula to me for figuring scoring probabilities for 4 through 8 dice?)

Maybe it's enough to say that, if my math is correct, the odds look something like this:

# of dice ----- odds of a roll that scores
1 die ----- 33 in 100
2 dice ----- 56 in 100
3 dice ----- 72 in 100


[I described the dice game in the first comment.]