# Concepts 12: Casino Gaming: Theoretical Win, Expected Value and Standard Deviation

Theoretical win and the standard deviations of expected value have been widely covered in numerous articles.  Unfortunately for me, the discussion on the topic has seldom been comprehensive enough to be understood by the layperson (meaning me).

This post attempts to explain the issue in the simplest possible way.

Why is this important?

Theoretical win is derived from the probabilities built into any casino game.  As all casino games are designed (in theory) to guarantee a return to the casino, the theoretical win (winnings for the player) is always negative while the expected value, also known as expectation (winnings for the casino) is always positive.

However, like with all probabilities, an element of randomness exists.  The standard deviation of theoretical win thus provides a threshold for casino managers to decide if play has passed the limit where it becomes suspect.

Theoretical Win and Expected Value

The average of results is determined by the theoretical win formula.  Hence, 50% of all results will achieve this result.

For example:

A split wager on single-zero roulette pays 17:1.

The probability of winning a split bet is 2/37 (2 numbers wagered on from a total of 37 numbers)

The probability of losing the bet is 37-2 / 37 = 35/37 (for all the other numbers not wagered on)

The theoretical win formula is:

(Probability of winning x payout in terms of units of wager) – (Probability of losing x wager)

(2/37 x 17) – (35/37 x 1) = -0.02703

This means that 50% of all play on split wagers in single-zero roulette will (in the long-term) achieve a loss of \$0.02703 per dollar wagered by the player.

Standard Deviation

The majority of results will fall within 3 standard deviations from the average. (Source: http://schools-wikipedia.org/)

In order to obtain the standard deviation, we turn to the binomial distribution, which is a distribution where only 1 of 2 outcomes is possible.  In our case, the outcomes being a win or a loss.

Games with only 2 Outcomes

The formula for a binomial standard deviation is:

Square Root (number of games x probability of winning x probability of losing) x average wager x (unit of wager + payout in terms of units of wager)

Using our current split wager example, with an average wager of \$2 over 3 games, the formula would be as follows:

Square Root(3 x 2/37 x 35/37) x \$2 x (1+17)

=Square Root(3 x 2/37 x 35/37) x \$2 x 18

= 0.391659 x \$2 x 18

= \$14.09972 (standard deviation of theoretical win)

Games with 3 or more Outcomes

How about games with more than 2 possible outcomes, perhaps a win, loss or tie? Baccarat fits this description and for this we turn to the multinomial distribution:

This table describes the possible outcomes for a Player wager. A win on Player would give +1 and a loss on Player would give -1. A tie pays 0.

For a Banker wager, a win would pay 0.95, a loss -1 and a tie, again 0.

Player Wager

 Wager Probability Outcome Player 0.446246609 1 Banker 0.458597423 -1 Tie 0.095155968 0

The expectation of the Player wager is just the product of the probability of each result by the outcome, the sum of which is -0.012350813:

 Wager Probability Outcome Expectation Player 0.446246609 1 0.446246609 Banker 0.458597423 -1 -0.458597423 Tie 0.095155968 0 0 Total Probability = 1 Total Exp. -0.012350813

The standard deviation of expectation for this is worked out below.

First, you get the variance for each wager:

 Wager Probability Outcome Expectation Variance Player 0.44625 1 0.44625 0.44625 x (1- -0.01235)^2 = 0.45734 Banker 0.45860 -1 -0.45860 0.45860 x (-1- -0.01235)^2 = 0.44734 Tie 0.09516 0 0.00000 0.09516 x (0- -0.01235)^2 = 0.00001

Notice that the working is really:

Probability x (Outcome – Expectation)^2 = Variance

The standard deviation for this is then:

SQRT(Sum of all Variances) = Standard Deviation of Expectation

 SD. Of Exp. SQRT( 0.45734+ 0.44734+ 0.00001 ) = 0.95115

This means that the standard deviation of expectation of a \$1 Player wager for 1 coup is \$0.95115. For a \$5 wager over 10 coups, it would be \$0.95115 x \$5 x SQRT(10).

The Result

To find out if the win of a player is improbable, we use the Z-table and the following formula:

Win – Average Win (Theoretical Win)

Standard Deviation

Our example will yield the following result should the player have won \$15.  The player played 3 games with an average wager of \$2, so the total wager would be \$6.

15 – (-0.02703 x \$6)

14.09972

=1.075352

The result is 1.075352, meaning that the result of the player’s gaming is 1.075352 standard deviations from the average.  This means that the player’s gaming results are within the top 15% of all results.  Cause for some concern.

There you have it.  Good luck!

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