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PART I
Mathematical Expectation
The Mathematics of Gambling by Dr. E. O. Thorp

I have already made reference to the concept of mathematical expectation. The principle is central to an understanding of the chapters to follow.

Imagine for a moment a coin toss game with an unbiased coin (a coin we assume will produce 50% heads and 50% tails). Suppose also that we are offered an opportunity to bet that the next flip will be heads and the payoff will be even money when we win (we received a $1 profit in addition to the return of wager). Our mathematical expectation in this example is:

(.5)(1) + (.5)(-1) = 0

The mathematical expectation of any bet in any game is computed by multiplying each possible gain or loss by the probability of that gain or loss, then adding the two figures. In the preceding example, we expect to gain nothing from playing this game. This is known as a fair game, one in which the player has no advantage or disadvantage.

Now suppose the payoff was changed to 3/2 (a gain of $1.50 in addition to our $1 bet). Our expectation would change to:

(.5)(1.5) + (.5)(-1) = +.25

Playing this game 100 times would give us a positive expectation of $25.

The two examples presented thus far are admittedly simple, but often this type of analysis is all that is needed to evaluate a proposition. Consider the "dozens" bet in Roulette. Our expectation for a $1 bet is:

(12/38)(2) + (26/38)(-1) = -.0526

As another example, suppose that on the first hand of four-deck Blackjack the player bets $12, he is dealt 6, 5, and the dealer then shows an ace up. The dealer asks the player if he wants insurance. This is a separate $6 bet. It pays $12 if the dealer's hole card is a ten value. It pays -$6 otherwise. A full four-deck pack has 64 tens and 144 non tens. Assuming the deck is "randomly" shuffled (this means that all orderings of the cards are equally probably), the chances are equally likely that each of the 205 unseen cards is the dealer's hole card. Thus the player's expectation is:

(64/205)(12) + (141/205)(-6) = -78/205

Or about -$.38. The player should not take insurance.

Different betting amounts have different expectations. But the player's expectation as a percent of the amount bet is always the same number. In the case of betting on the Red in Roulette, this is 18/38-20/38 = -2/38 = -1/19 or about -5.26%. Thus the expectation of any size bet on Red at American double zero Roulette is -1/19 or about -5.265 of the total amount bet. So to get the expectation for any size bet on Red, just multiply by -5.26%. With one exception, the other American double zero Roulette bets also have this expectation per unit bet. The player's expectation per unit is often simply called the players disadvantage. What the player loses, the house wins, so the house advantage, house percentage, or house expectation per unit bet by the player is +5.26%.

A useful basic fact about the player's expectation is this. The expectation for a series of bets is the total of the expectations for the individual bets. For instance, if you bet $1 on Red, then $2, then $4, your expectations are -$2/38, -$4/38, and -$8/38. Your total expectation is -$14/38, or (a loss of) about -$.37. Thus, if your expectation on each of a series of bets is -5.26% of the amount bet, then the expectation on the whole series is -5.26% of the total of all bets. This is one of the fundamental reasons why "staking systems" don't work: a series of negative expectation bets must have negative expectation.
 
 
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