Computing house edge of Lunar New Year dice game Here is an apt description of the game: https://medium.com/@Rob_TactVP/how-to-play-hoo-hey-how-812532d405a1
I’ll repeat the rules of the game here.

Traditionally, Hoo Hey How uses the
following pay table:

*

*1:1 for one matching symbol.

*2:1 for two matching symbols.

*3:1 for three matching symbols.

*30:1 for any triple.

For example: Betting \$5 on "Fish" and having one fish symbol turn up nets \$10 (your initial \$5, plus \$5 in winnings).
Betting \$5 on "Fish" and having it appear on two dice,
returns \$15 (your \$5 wager and \$10 in winnings). Three symbols pays
\$20 (your \$5 bet and \$15 profit). However, you're also allowed to
wager on a "Triple" option. This pays out at 30 to one, making it look
like an enticing option. It's not. The odds of all three dice
displaying the same symbol are very slim.

I wanted first to evaluate the expected value of betting on any of the 6 symbol choices, and ignore the Triple bet for now.
I came up with an expected value of \$0.875 for an investment of \$1, which is alarmingly low and significantly worse than the supposed 8% house advantage that I have been reading about (including from this article).
So let me walk you through the logic that I used to calculate this.
First, the probability of getting all 3 dice matching your chosen symbol:
$(1/6)^3 = 1/216$
Second, the probability of getting exactly 2 dice matching your chosen symbol:
$(1/6)^2 (5/6) = 5/216$
Third, the probability of getting exactly 1 die matching your chosen symbol which I derived by first finding the probability of at least 1 die matching, which should be:
$1 - (5/6)^3$
This is the inverse of the probability of all 3 dice not matching (each with a 5/6 chance of that).
Therefore the probability of exactly 1 die matching should be:
$1 - (\frac{5}{6})^3 - \frac{5}{216} - \frac{1}{216} = 1 - \frac{125 + 6}{216} = \frac{216 - 131}{216} = \frac{85}{216}$
And the expected value calculation:
\$1 invested yields $\$4\frac{1}{216} + \$3\frac{5}{216} + \$2\frac{85}{216} = \$\frac{4 + 15 + 170}{216} = \$\frac{189}{216} = \$\frac{7}{8} = \$0.875$
Have I neglected to consider something in this analysis? I avoided counting the cases out but it would not be too difficult to do this.
Based on this, a player should be losing around \$12.50 per \$100 spent, not \$8. I wonder if it’s just a simple mistranslation where a 1/8 loss got interpreted as an 8% loss.
Let’s revisit the 30:1 for triple. If I interpret this description literally, that’d be 31/36 * \$1 expected value, 0.861111, which is remarkably consistent with (and actually a smidge lower than) 0.875 which is a hint to me that it may actually be correct. This means the expected value is not going to change a whole lot whether the player bets more on individual symbols or on the triple.
 A: This showed up in my Google searches and I couldn't find where the 8% was originally calculated but hopefully the below will make sense.
Ignoring the triple bet, the expected value of this game can be calculated as follows.
Let us assume $1 is bet on fish. The expected value of the bet is:
$\text{EV = (Probability of Winning $\times$ Amount Won Per Bet) $-$ (Probability of Losing $\times$ Amount Lost Per Bet)}$
There are 4 outcomes possible when betting on fish:

*

*One fish appearing (\$1 Payout): $\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \times 3$


*Two fish appearing (\$2 Payout): $\frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} \times 3$


*Three fish appearing (\$3 Payout): $\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6}$


*No fish appearing (\$1 Loss): $\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}$
The expected value can be then calculated:
$\text{EV} = (\frac{25}{72} \times \$ 1 + \frac{5}{72} \times \$2 + \frac{1}{216} \times \$ 3) - (\frac{125}{216} \times \$ 1) = \ \sim-\$0.0787$
With that being said, don't take advantage of those around you with this information.
