For the Same Totals, Odds at Craps Beat Flat Bets

The rationale for taking or laying odds on Pass or Don't Pass bets eludes craps veterans and neophytes alike. Almost everyone who enjoys the game has read or been told that betting the odds reduces house advantage on the line. But house advantage doesn't smack shooters in the face like a bill presented by a pit boss at the end of a game. More, benefits and risks of betting odds are masked by changes in what's won or lost under various conditions.

A helpful way to view the question of betting the odds at craps uses the "even-money equivalents" of different wagers as standards of comparison. Even-money equivalents are theoretical bets paying exactly 1-to-1, adjusted so their sizes and chances of winning yield the same edge and bankroll swing characteristics as the actual wagers.

Picture dropping $10 on Pass or Don't Pass, with no odds. The even-money equivalent of the Pass bet is $10 with a 49.29 percent chance of winning. In this case, actual amounts and probabilities equal their even-money equivalents because flat bets on Pass, in fact, either win or lose on every come-out cycle and pay 1-to-1. The even-money equivalent of $10 on Don't Pass with no odds, in contrast, is $9.86 with a 49.31 percent chance of winning. These differ from the nominal values because bets on Don't Pass can push rather than win or lose during a come-out cycle.

Betting $30 flat triples the even-money equivalent amounts -- to $30 for Pass and $29.58 for Don't Pass. But effective chances of winning remain 49.29 and 49.31 percent, respectively.

Say that, instead of tripling your flat Pass bet, you go to the $30 level with $10 on the line and $20 odds. You're up to win $10 on the come-out and $34 on points of six or eight, $40 on five or nine, and $50 on four or 10. The even-money equivalent, $28.58, has 49.75 percent chance of winning. What most solid citizens think is the same $30, spread this way rather than flat, is like betting $1.42 less and winning an extra 46 times in 10,000 tries.

For Don't Pass, $10 on the line with double odds positions you to win $10 on the come-out and a total of $30 thereafter by laying $24 on six or eight, $30 on five or nine, and $40 on four or 10. The even-money equivalent is $28.53 with a 49.76 percent chance of winning. Relative to $30 flat, this is like betting $1.05 less and winning an additional 45 times in every 10,000 tries.

More odds for the same nominal total bet further heighten the effect. Consider $5 on Pass with five-times odds: $25 on four, six, eight, and 10; $30 on five and nine. The even-money equivalent is $30.87 with 49.89 percent chance of winning. Likewise for $5 on Don't Pass with quintuple odds: $30 on six and eight, $45 on five and nine, and $50 on four and 10; this has a $30.86 even-money equivalent with 49.89 percent chance to win.

Simply adding odds to flat bets increases bankroll swings. Effectively, you'll average a $28.58 or $28.53 step on each come-out cycle with $10 on the line and double odds, compared with $10 or $9.86 by betting $10 flat. If the extra money means you're overbetting your bankroll, you've too great a risk of tapping-out early -- notwithstanding the reduced house advantage.

Proportioning roughly the same money with the least flat and the most odds is another matter altogether. Now, betting the odds moderates swings relative to the flat wager, as you can see by comparing the amounts of the alternate even-money equivalents. And the lower edge shows up in the improved the chance you'll win, as is evident from the probabilities of the equivalents.

The differences are small. Do they matter? Decide for yourself. Assume 1,000 players start with $1,500 each. Betting $10 with double odds instead of $30 flat means that between 10 and 20 more will still be in the game, if not ahead, after 500 come-out cycles -- a reasonable day's play. The writer of airs, Sumner A Ingmark, had this to say about such splitting of hairs: