Originally appeared Sep 16 2003

The Dinner Bet Will Put You on a Diet
By
Dale S. Yeazel

Every so often a gambler, either casual or serious, re-invents or reads about one of the mainstays of craps systems that I have heard described as the “dinner bet” or the “iron cross.” It is a hedging system that involves betting the field and placing the 5, 6 and 8.

Since the only numbers that cause the field to lose are 5, 6, 7 and 8 the bettor hopes to hedge the field bet by placing the 5, 6 and 8. This seems like a really groovy idea to those that don’t bother to examine this system too deeply. You either win even money, double (aces) or triple (boxcars) for your field bet or you net a small profit for your place bet minus the losing field bet. I mean if the seven doesn’t come up too often and wipe out your bets, you know? I’m sorry but when I deal to someone grinding it out on this system and after seeing the smugness in his expression after winning several bets in a row, I feel a sadistic joy when the seven appears five times in a row.

But neither the player’s smugness or my sadism or either one or our slightly skewed perception of reality is the basis for unbiased examination of the dinner bet as a means to financial freedom. Using mathematics to determine house percentage, not anecdotal musings, is the only reliable method of evaluating a method of playing craps (read: system).

I will examine the dinner bet using the amounts I have most often seen this played for: a $5 field bet, place the five for $5 and the six and eight for $6 each. Out of the thirty non-seven combinations, the bettor will win $15 on the one combination of twelve, $10 on the one combination of two, $70 on the fourteen combinations of 3, 4, 9, 10 and 11. The fourteen combinations of 5, 6 and 8 will win a net of $28 by virtue of winning seven dollars for the payoff for the place bet and losing five dollars on the field. The bettor will win a total of $123.

Of course “the devil” (seven) will cause bettor to lose $22 six times out of thirty-six for a total of $132. This brings the total to $132 - $123 = <$9>. If we add the winning and losing amounts we get a total of $255. If we then divide 9 by 255 we get .03529 or 3.529%. This is the house percentage against the dinner bet when played for the aforementioned amounts. The house percentage increases as the amount of the field bet is increased of the total amount played. In other words if you placed the five for $5 and the six and eight for $6 each but bet $10 in the field, you would suffer a worse HP. One would be better off playing the field alone (2.56%) or better yet, place the six and eight (1.51%).

For those of you that have a distain for numbers and prefer to look at things in a simple manner: you win a small amount 5/6 of the time before that 1/6 of the time when you lose an amount greater than your combined winnings. This of course when the dice behave as probability says they will and sadly enough, that is not the worse case scenario.