Lunar New Year Winning (?)

I gambled for the first time in my life at the age of 5. I also learned that I was a sore loser. I cried watching my money get snatched away. It wasn't even my money. It was my dad's. 

I was playing the classic Vietnamese game bầu cua cá cọp. Commonplace during the Lunar New Years, this dice-game has 6 squares, each containing an image of one of the following animals: stag, crab, fish, prawn, rooster, and a fruit (calabash squash)! Place money, any amount, on any square, as many squares are you want. 3 dice are rolled. If one die faces up with an image that matches a square you have placed money on, you win the amount you've bet. If 2 dice face up with an image that matches a square you have placed money on, you win 2x the amount you've bet. If all 3 dice face up with an image that matches a square you have placed money on, you win 3x the amount you've bet! Any money on losing squares, squares who image does not appear face up on any of the dice, get gathered up by the "dealer" (the person running the game, most likely the owner of the board and die, and the bowl and plate it's shaken in).

So let's say I put $1 on the squash.

If one dice comes up as "squash", I receive $1 from the dealer.

The probability of this happening is the chance one die faces up with the squash and the others do not face up with the squash.

Scenario 1

    Dice 1 = "squash" AND Dice 2 = "NOT squash" AND Dice 3 = "NOT squash"

    P(dice 1 is squash) * P(dice 2 is not squash) * P(dice 3 is not squash)

    1/6 * 5/6 * 5/6 = 25/216

But Dice 2 could face up with squash, while the others do not, and you can still win $1!

Scenario 2

    P(dice 1 is not squash) * P(dice 2 is squash) * P(dice 3 is not squash)

    5/6 * 1/6 * 5/6 = 25/216

And don't forget Dice 3!

Scenario 3

    P(dice 1 is not squash) * P(dice 2 is not squash) * P(dice 3 is squash)

    5/6 * 5/6 * 1/6 = 25/216

So the probability that you win $1 at all is if Scenario 1 OR Scenario 2 OR Scenario 3.

    P(Scenario 1) + P(Scenario 2) + P(Scenario 3)

    25/216 + 25/216 + 25/216 = 75/216

What if we get lucky? Suppose the chances of having 2 dice come up with the squash.

    Dice 1 = "squash" AND Dice 2 = "squash" AND Dice 3 = "NOT squash"

    P(dice 1 is squash) * P(dice 2 is squash) * P(dice 3 is not squash)

    1/6 * 1/6 * 5/6 = 5/216

And don't forget the other combinations of dice:

    P(dice 1 is squash) * P(dice 2 is not squash) * P(dice 3 is squash) = 1/6 * 5/6 * 1/6 = 5/216

    P(dice 1 is not squash) * P(dice 2 is squash) * P(dice 3 is squash) = 5/6 * 1/6 * 1/6 = 5/216

The probability that you win 2x the amount you bet is the sum of the 3 combinations:

    P(2 dice face up with your image) = 5/216 + 5/216 + 5/216 = 15/216.

What if we got REALLY lucky? To win 3x the amount you bet, all 3 dice have to face up with your square's image:

    P(dice 1 is squash) * P(dice 2 is squash) * P(dice 3 is squash) = 1/6 * 1/6 * 1/6 = 1/216.

Winning 3x the amount you bet is rare. 1/216 means there is less than half of a percent chance to win! 

What are your chances that AT LEAST one die come up with your square? That's the probably that one die come up OR 2 dice come up OR 3 dice come up!

P(one die matches) + P(2 dice matches) + P(3 dice matches) = 75/216 + 15/216 + 1/216 = 91/216

That's a 42.13% chance that you win any money at all! That's pretty close to 50%! That's not bad! Now what is your average payout?

For that, we multiply each probability with the profit of each scenario:

($Profit for a 1 dice match)*P(1 dice matches) + ($Profit for a 2 dice match)*P(2 dice matches) + ($Profit for a 3 dice match)*P(3 dice matches) + ($Profit for no dice match)*P(no dice matches)

($1)*(75/216) + ($2)*(15/216) + ($3)*(1/216) + (-$1)*(125/216) = -$0.0787

On average, you lose 7 cents a game! That doesn't sound too bad in the long run. But what if you're the dealer? And there's multiple players? And they are betting multiple squares? That can really add up :)