Game Volatility at Baccarat
The purpose of this article is to explain in a probabilistic sense the degree of volatility that can be expected. By Andrew MacDonald Executive General Manager, Gaming Risk Management, Crown Ltd 1998 |
Introduction | Mathematics | Variable Betting | Conclusion |
The purpose of this paper is therefore to establish and verify a formula to determine the number of hands played which would result in a specified tolerance level of departure from the mean win % at the game of Baccarat for a given confidence interval.
Determination based upon the above of a level of turnover which should generate results replicating the theoretical win percentage at the game. The number of decisions (n) at which the win percentage will deviate by a given fraction from the mean may be calculated using the following formula: |
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n = | 1 / [(tolerance level/confidence interval)^2] | x Variance of the game | |||||||||||||||||||
where n | = resolved decisions (hands) | ||||||||||||||||||||
tolerance level (T) | = dispersion from the mean in %. | ||||||||||||||||||||
= the z score relative to the desired confidence interval of a normal distribution curve (z = 2.33 for a 98% confidence interval) | |||||||||||||||||||||
confidence interval (C) | |||||||||||||||||||||
= the variance of the game in question (ie the average squared result) | |||||||||||||||||||||
Variance (V) | |||||||||||||||||||||
Therefore: n
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= 1 / [( T / C )^2] x V | ||||||||||||||||||||
A desirable level of tolerance for which to calculate may be one where breakeven is achieved for the specified confidence limit.
Breakeven may be calculated as follows: |
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Costs as a percent of turnover | = E % | ||||||||||||||||||||
Gaming Tax | = G % | ||||||||||||||||||||
Breakeven | = E % / (1 – G %) | ||||||||||||||||||||
To consider this issue further it may also be useful to consider game bet distribution patterns and actual results derived over time.
The number of hands (resolved decisions) required based upon the above may be calculated as follows: |
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n = | 1 / [(A% – (E% / (1 – G%) / 2.33)^2] | x 0.975 | |||||||||||||||||||
The number of decisions played for the theoretical win % to be achieved within 0.04% on the same basis is astronomical. | |||||||||||||||||||||
n = | 1 / [( 0.04% / 2.33)^2] | x 0.975 | |||||||||||||||||||
n = 33,082,359 for the theoretical win % to fall within 1.225% and 1.305% at a 98% confidence limit.
If flat bets of $200,000 were considered, the volume of turnover required to generate such results would be: |
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For n = 33,082,359 | turnover = | $6,616,471,800,000 (ie. $6.62 tn) |