How do Casinos make money at Chuck-a-Luck?

[IAm@Work.com]

I assume that the answer is that I don't understand the game, but here is what appears to happen:

The player can bet increments of $5 on one of the 6 possible outcomes of a roll of a die, 1-6. Say you put $5 on #1.
The casino rolls three die.
The player wins $5 for each one that appears and 10X or $50 if 3 ones appear.

My calculation says that the expected value of this game is $2.66 for a $5 bet. In other words, because of that 10x win for tripple ones the odds are in favor of the player.

What am I missing?

[urysohn]

I think your numbers are off. Or mine are - this is what I get...
# Odds Outcome Weighted
0 0.578703704 -5 -2.893518519
1 0.347222222 0 0
2 0.069444444 5 0.347222222
3 0.00462963 50 0.231481481

Expected value -2.314814815

[urysohn]

This slight modification let me match your numbers. You forgot to factor in that you lost your five dollar wager. But I think mine was off by a bit too.
# Odds Outcome Weighted
0 0.578703704 0 0
1 0.347222222 5 1.736111111
2 0.069444444 10 0.694444444
3 0.00462963 50 0.231481481

Expected value 2.662037037
What would the real net outcomes be?
0 --> -$5 (you lost your bet)
1 --> 0 or 5? (do you just break even or do you get your bet plus five more?)
2 --> 5 or 10?
3 --> 45 or 50?

[Sotally Tober]

This slight modification let me match your numbers. You forgot to factor in that you lost your five dollar wager. But I think mine was off by a bit too.
# Odds Outcome Weighted
0 0.578703704 0 0
1 0.347222222 5 1.736111111
2 0.069444444 10 0.694444444
3 0.00462963 50 0.231481481

Expected value 2.662037037
What would the real net outcomes be?
0 --> -$5 (you lost your bet)
1 --> 0 or 5? (do you just break even or do you get your bet plus five more?)
2 --> 5 or 10?
3 --> 45 or 50?

Exactly what I was wondering. If you break even with 1 of your number, the EV = -2.xxx, whatever that was. That's a HUGE house advantage. Best in the house.

[urysohn]

Geez, it's like post-padding heaven here.

If the outcomes are (-5, 5, 10, 50), then the EV = -$0.23 (sounds reasonable)

If the outcomes are (-5,0,5,45), then EV = -$2.34 (casino's making money hand over fist)

[Butters]

An actuary's website to help you with all of your gambling questions. (http://wizardofodds.com/games/chuckaluck.html)

[4sigma]

This is a deceptive game since it appears to be breakeven without the bonus. 1/6 of the tosses of each die come up "1", and there are 3 dice, right? So aren't you getting even money if you bet $5 and get $5 each time a "1" shows?

The game is more intuitive from the casino's point of view. Suppose you have one player betting each number each round.

If the dice show three different numbers (1,2,3), then the casino plays $5 to everyone who bet (1,2,3), but collects $5 from everyone who bet (4,5,6). So the casino breaks even.

But say two of the dice match. (1,1,2). Now the casino still pays out $15, but it collects from everyone who bet (3,4,5,6), thus collecting $20. Casino makes $5.

If all three dice show the same number (1,1,1), the casino pays out $50 but collects $25. A $25 loss for the house.

However, the odds of 2 dice matching are 15-1 compared to the odds of 3 dice matching. So the house makes $5 15 times as often as it loses $25.

[IAm@Work.com]

If the outcomes are (-5, 5, 10, 50), then the EV = -$0.23 (sounds reasonable)
These are the correct payouts, and yes, my original calc assumed that you got your bet back if you lost, kind of a silly outcome from the casino's standpoint.

Thanks. Sometimes the errors are glaring, but you still can't see them.