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Multiple-Bet Games Have No Loopholes

6 July 1998

Craps and roulette fans spend countless hours trying to dope out combinations of wagers and amounts that can't lose, regardless of what number shows. Sorry, folks. Odds and payoffs associated with individual wagers are structured to make such a bet impossible.

Unfortunately, this statement is mathematically unprovable. And the absence of proof falsely encourages hopeful solid citizens.

A basic tenet of logic, not restricted to gambling, holds that impossibilities can't be formally proved. People who say it's possible to live on Venus could show it's true by doing it. How could they certify such a feat as impossible? By failing to find anyone who succeeded? That doesn't mean nobody did - or could.

This philosophical tangent brings me to my old friend Olaf. Olaf has been looking for loopholes in craps since the day, in the back room of Darwin's Deli, we discovered something better to play with dice than Monopoly.

He won't believe me when I insist he's wasting his time. He keeps demanding proof. And, since it's not forthcoming, Olaf is forever trying to find the anomaly, sure his efforts will be rewarded.

Olaf usually begins by building up combinations of bets that cover more and more numbers. He adjusts and readjusts the amounts, trying for a way to win on at least one number and do no worse than break even on the rest. Here's the general concept.

Start with a plain vanilla $5 Field bet. Seven numbers are winners and four are losers:

Roll
Result
2, 12
win $10
3, 4, 9, 10, 11
win $5
5, 6, 7, 8
lose $5

 

Add $2 on Any Seven. Alone, this wins $8 on a seven and loses $2 on anything else. In combination with the $5 field bet, eight numbers are winners while three are losers:

Roll
Result
2, 12
win $8
3, 4, 7, 9, 10, 11
win $3
5, 6, 8
lose $7

 

Add a place bet on the five for $5. Alone, this pays $7 on a five and loses $5 on a seven. With the amalgamated bet, seven numbers are winners, one breaks even, and three are losers:

Roll
Result
2, 12
win $8
3, 4, 9, 10, 11
win $3
5
push
7
lose $2
6, 8
lose $7

 

Raise the bet on Any Seven to $3 and cut that on the Field to $3. Now, four numbers are winners, five break even, and only two are losers:

Roll
Result
7
win $4
2, 12
win $3
5
win $1
3, 4, 9, 10, 11
push
6, 8
lose $6

 

You get the drift. Olaf will next try to cover the six, adjusting everything else so eight is the only loser. If he gets this to work, he'll jump on the eight.

I made a mistake one day and told him the technical term - iteration - for the procedure he uses. "A lot of computer programs use a similar approach," I casually added.

Olaf apparently took my idle chatter to mean that science had legitimized his theory. He phoned the next day to propose a partnership. He'd supply the ideas - like Field, Any Seven, and Place combos; sequences of rising Come and Any Craps bets; and whatnot. I'd dash off some quick computer programs. And we'd divvy the dividends. "I could get a college kid for the computer stuff," he said, "but you proved it could be done with whatcha callems, itinerations, so you deserve to share in the shekels."

By now, I'm inured to saying one thing and having gamblers hear whatever they want to think anyway. So I reminded Olaf of the taxonomic text by the epistemological poet, Sumner A Ingmark:


Hokum with a fancy name,
Hokum in a gilded frame,
Still is hokum, all the same.


ASK AL


I'm just learning to play craps and love the game. The "Field" looks good to me because you win on the two, three, four, nine, 10, 11, and 12, and only lose on the five, six, seven, and eight. But a friend told me it was a "sucker bet." Is she right?

Your chances don't depend on how many final numbers win or lose. The dice form 11 numbers, two through 12, but can land 36 ways. So some numbers occur more than others. For instance, two is formed one way (1-1), three can be formed two ways (1-2 & 2-1), four can be formed three ways (1-3, 2-2, & 3-1), and so forth.

In all, winning numbers on the Field can be formed 16 ways while losing numbers can be formed 20 ways. So, you can expect to win this bet only 16 times for every 20 you lose. You may have noticed that the Field pays double when the shooter rolls a two or 12, which offsets the disadvantage somewhat.

There's another factor. Placing a number like nine wins if the shooter rolls nine and loses if the result is seven. Nothing else counts. There are an average of 10 decisions on the nine every 36 rolls. The Field, a one-roll bet, wins or loses whenever the dice are thrown. This means 36 decisions every 36 rolls. So, unless you're very lucky on the Field, the odds against winning and the rate of decisions can quickly erode your bankroll.

Alan Krigman

Alan Krigman was a weekly syndicated newspaper gaming columnist and Editor & Publisher of Winning Ways, a monthly newsletter for casino aficionados. His columns focused on gambling probability and statistics. He passed away in October, 2013.
Alan Krigman
Alan Krigman was a weekly syndicated newspaper gaming columnist and Editor & Publisher of Winning Ways, a monthly newsletter for casino aficionados. His columns focused on gambling probability and statistics. He passed away in October, 2013.