Playing It Smart: Hedging your bets increases the house advantage at craps

Craps players who are too smart by a half often use hedges, thinking they'll protect their primary wagers with small, high-payoff auxiliary bets. When a hedge pans out, they think they're onto one of those secrets the bosses don't want anyone to know.

Bet $10 on Pass and hedge it with $2 on Any Craps, for instance. On a seven or 11, you pick up $10 and lose $2 so you net $8; without the hedge, you'd pick up $10. The $8 still seems good, however. On a 2, 3, or 12, you pick up $14 and lose $10 for a gain of $4; without the hedge, you'd be out $10. Shrewd, eh?

On a four, five, six, eight, nine, or 10, Any Craps loses $2 right away. So, depending on the result of the point roll, the ultimate impact is that you're up $8 or down $12, rather than plus or minus $10. Further, by the time the coup is resolved, you've forgotten those trifling white chips the dealer grabbed. When you add it all up, though, the $2 you don't win when the shooter makes the point, and the extra $2 you lose when the seven pops aren't as insignificant as certain solid citizens assert.

A simplified secondary bet can help illustrate the idea. To set the stage, think initially about dropping $1 on either the three or 11, and nothing else. In 36 statistically-correct rolls, either of these will win $15 twice and lose $1 34 times. That's a net loss of $34 - $30 or $4. Next, Say you bet $10 on Pass, by itself. You'll win $10 eight times (on seven or 11) for $80 and lose $10 four times (on two, three, or 12) for $40. You'll neither win nor lose at this stage on any of the other 24 results. Your net will therefore be a gain of $80 - $40 = $40.

Alternately, bet $10 on Pass and $1 on the three. Again, consider only results of 36 statistically-correct come-outs. The hedge won't affect action on the point, assuming one is established.

A seven or 11 wins $10 from the line bet but loses the $1 on the three, a gain of eight times $9 or $72. A three will pop twice, yielding $15 for the three minus $10 for the money on the line; that's $5 x 2 + $10. A two or 12, together, will occur twice and cost you $11 each time $11 x 2 = $22. The 24 "other" rolls are no longer gimmes. Regardless of how they're later settled they'll set you back $1 x 24 = $24. The grand total is $72 + $10 - $22 - $24 or $36 profit as opposed to $40 on the line bet alone.

Instead, bet $10 on Pass and $1 on the 11. Now, a seven wins $10 and loses $1 six times; that's $9 x 6 = $54. An 11 will win $10 + $15 twice; that's $50. A two, three, or 12 will lose $11 four times; that's $44. The other 24 numbers will lose $1 each; that's $1 x 24 = 24. Net earnings are $54 + $50 - $44 - $24 = $36. The same as was the case with $1 on the three, notwithstanding the fact that the three opposed and the 11 reinforced the line bet.

These examples show that each wager gives the house an edge, independently of any other bet. On either the three or 11, for every 36 tries, edge is equivalent to an average loss of $4 per dollar bet. This factor would reduce the profit during come-outs for $10 Pass bets from $40 to $36. Guess what! It does.

I'll spare you the arithmetic, but similar reasoning applies to the harebrained scheme of betting equal amounts on Pass and Don't Pass, then taking Odds on Pass. The false logic is that the two flat bets are self-cancelling, leaving you with only the Odds on which the house has no advantage. Sound too good to be true?

Assume you do this with $10 bets. The flat bets cancel except when the shooter comes out with a 12. Then, Pass loses and Don't Pass pushes, a $10 net loss. This will happen once in every 36 statistically-correct trials. But (1/36) x $10 equals the house's theoretical combined earnings on bets of $10 Pass and $10 Don't Pass. So this strategy is equivalent to paying the casino a commission on $20 flat, then being able to take only half as much odds as would be allowed betting $20 on Pass. Once more showing what the perceptive poet, Sumner A Ingmark, knew when he noted: