Roulette is no match for geeks with a basic knowledge of the fundamentals of statistics
The quote above popped up on my Twitter timeline a few days ago. It’s not my purpose to embarrass the person who wrote it; their statement simply reflects several very common misconceptions around gambling probability, concepts that I’ve been meaning to write about for a while. Probability is often unintuitive, and when applied to gambling people can quickly get unstuck.
I think the most fundamental error here, more observational than mathematical, is the idea that casinos could somehow have a poor grasp of probability theory, and therefore roulette can be beaten with a “system”. Notwithstanding that casinos must have, over centuries, evolved a powerful grip on probability theory else quickly perish, whenever I confront people with this notion I find their strategy boils down to a variation of the debunked Martingale System. One of the simplest Martingale strategies, as it applies to a roulette player betting on red or black for example, is to wait for a run of a colour (let’s say 5 to 7 reds in a row) and then place a wager on the opposite outcome (i.e. black). This strategy lends itself to the Gamblers Fallacy, a mistaken belief that a future outcome in a random independent process is somehow “more likely” based on past behaviour. And while it’s true, given infinite time, that the roulette ball in a fair game must fall as often on black as it does on red, you’ll go broke long before then for reasons I hope to make clear below.
The casino’s strategy, on the other hand, is simple, fullproof and ruthless: keep you playing. Like a predator stalking its prey to exhaustion, the casino can wear you down to broke using a statistical concept known as Expected Value. Put simply, the expected value, as it applies to gambling, is the average payoff over time.
Take, for example, the roulette wheels at Adelaide’s “SkyCity” casino. These tables have 37 numbers consisting of 18 red, 18 black, and one “0” slot. Even money bets such as black/red pay even money, meaning that if you bet $1 and win (with probability of 18/37) you gain an additional $1. Conversely, there’s a 19/37 chance you will lose your $1 stake. So the punters’ average payoff over time is:
Expected Value [$1 even-money bet] = Prob(loss) x Price of losing + Prob(win) x Price of winning
= (19/37 x -$1) + (18/37 x $1)
Therefore you can expect to lose, on average, about 2.7 cents for every $1 even-money bet over the long term.
In other words, the house doesn’t always win, but it will ultimately prevail.
Here’s a clip showing how it can all go horribly, horribly wrong. Note how the player starts with a wager of just €1 and watch the losses increase exponentially [warning: some language].