Combine Bets in Craps, But You Can't Cancel the House's Edge

he charisma of craps hinges heavily on the hodgepodge of bets resolved by rolling the dice. Players can back different ways the dice will land, and also select wagers with wide ranges of returns for equal equity at risk. These factors have seduced certain statistically-unschooled solid citizens into concocting combination bets they perceive pave the path to profits.

"Hedges" are especially popular. These are small bets with high payout on one side of the fence to "protect" larger wagers on the other. An example is $1 any craps, which pays $7, to protect $10 on the pass line against a two, three, or 12 during the "come-out" roll. Hedging is expensive, costing much more than it returns based on the probabilities of the game.

But schemes combining opposing bets go well beyond blowing a few bucks on bogus insurance. Some players think they practically guarantee success by cleverly exploiting loopholes nobody else ever figured out before.

I'll illustrate using one of my favorite horrible examples. Bet $12 on the don't pass line during the come-out roll. If the "point" thrown is a six or eight, place the same number for $12. If the point is four, five, nine, or 10, place it for $10.

The shooter hits the point. The don't pass bet loses $12. But the place bet wins $18 on four or 10 for a net of plus $6, and wins $14 on the other numbers for a net of plus $2.

The shooter "misses-out." The don't pass bet wins $12. But the place bet loses $12 on six or eight for a break-even, and loses $10 on the other numbers for a net of plus $2.

At worst, you break even; otherwise, you make $2 or $6 on every bet. Right? Wrong! This logic ignores the come-out roll when, of the 36 ways a pair of dice can land, 8 are instant losers, 3 are instant winners, 1 yields no decision, and 24 represent points for subsequent rolls. The 8 ways to lose the whole $12 versus only 3 ways to win $12 on the come-out are too high a price to pay for the sure but small returns generated thereafter.

I'll be more precise. Say you play through 3960 come-out rolls over a long period, just betting $12 on the don't pass line. With the statistically-correct distribution of results, you'd lose $648, as the following table shows:

outcome	win(+)/lose(-)
you'd win 330 $12 don't pass bets on the come-out	+ $ 3,960
you'd lose 880 $12 don't pass bets on the come-out	- $10,560
660 don't pass bets become points of four or 10 you'd win $12 on 440 sevens you'd lose $12 on 220 fours/10s	+ $ 5,280 - $ 2,640
880 don't pass bets become points of five or nine you'd win $12 on 528 sevens you'd lose $12 on 352 fives/nines	+ $ 6,336 - $ 4,224
1,100 don't pass bets become points of six or eight you'd win $12 on 600 sevens You'd lose $12 on 500 sixes/eights	+ $ 7,200 - $ 6,000
net	- $ 648

Here's the same game with $12 don't pass and $10 or $12 placed on the point. The loss rises to $1,640, as the next table shows:

outcome	win(+)/lose(-)
you'd win 330 $12 don't pass bets on the come-out	+ $ 3,960
you'd lose 880 $12 don't pass bets on the come-out	- $10,560
660 don't pass bets become points of four or 10 you'd win $2 on 440 sevens you'd win $6 on 220 fours/10s	+ $ 880 + $ 1,320
880 don't pass bets become points of five or nine you'd win $2 on 880, regardless of result	+ $ 1,760
1,100 don't pass bets become points of six or eight You'd break even on 600 sevens you'd win $2 on 500 sixes/eights	$ 0 + $ 1,000
net	- $ 1,640

Systems based on multiple opposing bets all fail because payoffs aren't arbitrary. They're the odds of bets actually winning, shaved to give the house an edge. Combining complementary bets averages out the edge. Combining opposing bets compounds it.

As Sumner A Ingmark, the percentage players' pundit, poetically put it:

Don't bet both sides, and that's my best advice,
You'll win just once, but pay commission twice.