Chuck-a-Luck: A fine game involving probability

I stumbled onto this game while reading a fantastic book, 'Innumeracy' by John Allen Paulos. The rules of the game is simple: you play against a host. The  host has three dices with number 1-6 on them. Then you would be asked to choose a number in that range. Let's say you have chosen number 4. The host would then roll the three dices at the same time. Now there are three outcomes:

1. If all three dices have 4 on them, you get 3 dollars.
2. If two of the dices come up with 4, you get 2 dollars.
3. If only dice has 4 on it, you get 1 dollar.
4. Else(none of the dices has came up with 4) then you pay the host 1 dollar.

Now, what do you think about your chances are in this game? Obviously, as it is played in casinos, the casino owners won't have allowed this game if there were no profit for them. So even if this game seems quite winnable, you should re-calculate your chances.

Here is the expected outcome of this game:

expected outcome(EO) = sum(all the ways you can win * associated rewards) - sum(all the ways you can lose * associated cost)

EO = sum(P(all three dices has 4) * 3 + P(any two dices has 4) * 2 + P(any dices has 4) * 1) - P(no dice has 4) * 1

= (3C3 * (1/6)^3 * (5/6)^0)*3 + (3C2 * (1/6)^2 * (5/6)^1)*2 + (3C1 * (1/6)^1 * (5/6)^2)*1) - (3C3 * (5/6)^3 * (1/6)^0)*1

= 3/216 + 10/72 + 25/72 - 125/216 = -.08

As you can see, the outcome is quite counter intuitive. It seems that you would lose .08 dollars every time you play the game.

Following is a simulation of the game play. It shows how many times you can play before you lose all of your money in the cold-hearted probabilistic eventuality:

'''
code is also hosted at https://github.com/hasanIqbalAnik/probabilities/blob/master/monty_hall_simul.py
'''
import random
def dice_roll():
    return random.randint(1, 6)

balance = 10 #you start with 10 dollars
count_games = 0
choice = random.randint(1,6)

while(balance > 0):
    outcomes = [dice_roll(), dice_roll(), dice_roll()]
    if(outcomes.count(choice) == 3):
        balance = balance + 3
    elif(outcomes.count(choice) == 2):
        balance = balance + 2
    elif(outcomes.count(choice) == 1):
        balance = balance + 1
    else:
        balance = balance - 1
    count_games = count_games + 1

print balance # 0 :(
print count_games # sample output, 74 times

This also matches the probability calculation that you would be able to play around 80 times with 10 dollars, losing .08 dollars on average each time.