Stats question regarding the game of Craps?

Question

I will quickly run through the bets involved so that any stats experts who don't know the game can follow along.

This question will be dealing with 2 types of bets: Place bets and the Field bet.

Place bets are when you bet on a specific number to come up, and every time it hits you win, and if a 7 shows you lose. The 3 I will be dealing with are 5, 6, and 8. Betting on 5 pays outs $7 for every $5 bet, and the bets on 6 and 8 each pay out $7 for every $6 bet.

Any amount can bet bet on the field. If a 2 or 12 shows, double your bet is paid (i.e., you win $10 if you bet $5). If 3, 4, 9, 10, or 11 show, you win even money. If any other number shows, you lose.

So, if you were to bet $10 on 5, $12 each on 6 and 8, and $6 on the field, you'd win $8 if the 5, 6, or 8 shows ($14 minus the $6 field bet that is lost), $12 if the 2 or 12 shows, and $6 if the 3, 4, 9, 10, or 11 show.

My actual questions is regarding house edge. "House edge" is defined as the amount of money you will lose...

Answer 1

The house edge or vigorish "vig" is the result of one of the c**p numbers (either 2 or 12) being barred. Not every combo can be covered. If you get into a floating craps game, always bet "wrong". The edge is the amount that the house will earn based on the amount paid out to the amount bet. A vig of 1.47% means that the house will return $98.53 for each $100.00 wagered. E-mail me if you want.

Answer 2

Good question! It had me scratching my head for a while.

The explanation is fairly simple. The house edge for the place bets you quoted are based on only the rolls where those bets win or lose, excluding the rolls where nothing happens. However, when you have the combination of bets something happens on every roll so those rolls are included.

For example, consider the place bet on a 5.

Only 10 of the 36 combinations have an impact on that bet. So, using a $100 bet so I don't have to convert to a percentage later, I get

$140 x (4/10) = $56

-$100 x (6/10) = -$60

$56 - $60 = -$4 = 4% house edge

However, if we use all 36 possible rolls, we get

$140 x (4/36) = $15.555555555...

-$100 x (6/36) = -$16.666666666...

$15.555555555... - $16.666666666... = -$1.11111111... = 1.1111111...% house edge.

So, while I agree your 1.39% house edge is correct, it is higher than the house edge on the place bets if you calculate the place bets edge on all 36 dice combinations, not just the combinations that win or lose.