Dice wheel edge

I'm trying to figure out the house edge to this 54 spot dice wheel. I believe each number appears equally, just pared down to 1/4th of the normal 216 ways that 3 dice can come up.

I specifically went through the 6's and found that of the 54 spots:
1 six: 7 times
2 sixes: 4 times
3 sixes: 4 times

You win 1x on a single, 2x on a double, and 3x on a triple.

15 out of 54 spots win. You win 27.77% of the time.
39 out of 54 spots you lose your unit
The other spots work out to winning 27 units. (7x1) + (4x2) + (4x3)

Do you divide 27/39 to get 69.2% and thus a 30.8% house edge?



I suck at formatting tables on here. Using your numbers,

# of 6's. ways......pays......probability
one 6..... 7.......;... 1.......... 0.12962963
two 6..... 4.......... 2.......... 0.148148148
three 6... 4.......... 3.......... 0.222222222
no 6....... 39......... -1....... -0.722222222
total ........54................... -0.222222222 (House Edge)

It'd be 27 - 39 = -12 / 54 units = 22.2% house edge

The distribution isn't equivalent to a standard roll of 3 dice, though -- there's only one way for any given triple to appear and a lot more ways for a singleton. This wheel version is significantly worse than normal sic bo or chuck-a-luck.

Thank you for the fast reply and yes not sure what I was thinking as there are 4 triples of each number on this wheel. The 54 spots being 1/4 of 216 got me down the wrong path.

Again, thank you...