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Best of The Midnight Skulker

Gaming Guru

 

A Probable Misunderstanding

20 August 2000

A recent discussion on the Usenet craps newsgroup focused on a statement by Alan Krigman, respected author of many essays on gambling, and publisher of the newsletter Winning Ways. Mr. Krigman said that the probability of losing three consecutive lay bets against 4 is 3.7%. Another participant of the newsgroup posted that this was the Holy Grail all gamblers sought. All you had to do, this poster contended, was to observe two rolls of 4 with no intervening 7, then lay the 4. As Alan Krigman, respected author and newsletter publisher, had stated, you had a 96.3% chance of winning what was now the third decision of a no-4 lay bet without having actually lost any money on the first two. Why, it would be veritable child's play!

If only that were the way it worked. But alas, it is not. The poster received enlightenment from other newsgroup participants in short order, but I have seen and heard variations of this misunderstanding of probability so many times that I feel a detailed explanation of its fallacy is warranted.

The probability of losing the next three no-4 bets in a row is always 3.7%. Key word: "next." Probability refers to the future; past events may or may not alter that probability.

For example, what is the probability of losing a single no-4 bet? Answer: 33.33% on every decision. Now suppose that the dice are rolled separately instead of simultaneously, as all casinos require. Furthermore, let's assume that the first die rolled is not rolled again until the bet resolves (i.e., it is set aside, and only the other die is rolled). Before either die is rolled, the chances of rolling a total of 4 are 8.33% (3/36), and the chances of rolling a total of 7 are 16.67% (6/36). Because the chances of a 4 are half those of a 7, the probability of losing a no-4 bet is 33.33% (1/3). After the first die is rolled the chances of rolling a total of 7 remain 16.67% (1/6), but the chances of rolling a total of 4 change. They are 0% if the first die shows 4, 5, or 6; and 16.67% if it shows 1, 2, or 3. Obviously, the chances of losing a no-4 bet also change, to 0% and 50%, respectively. These are conditional probabilities, the chance of losing given the result of the first die's roll.

The same approach must be used in the original problem. What are the chances of losing three no-4 bets in a row? (I suppose I should mention we are back in a real casino, where the dice must be rolled simultaneously.) Before any of those decisions are known, there is a 3.7% chance of that happening, but as those bets are resolved the probability of losing those three bets will change. Obviously, if you win the first decision, your chances of losing those three decisions is nil (because you have already won one of them), but if you lose it your chances of losing those three decisions increases to 11.1%. On the other hand, a player starting a new series of three no-4 decisions has the original 3.7% chance of losing all three of them, because they are a different set of decisions.

The Midnight Skulker
The Midnight Skulker has been playing craps for over three decades and has played almost everywhere in the country. He is a computer expert and a frequent contributor to Internet newgroups, where his opinions and observations have earned him much respect.
The Midnight Skulker
The Midnight Skulker has been playing craps for over three decades and has played almost everywhere in the country. He is a computer expert and a frequent contributor to Internet newgroups, where his opinions and observations have earned him much respect.