As I've said before, probability theory was invented to solve problems in games of chance. Supposedly the first nontrivial probability problem was the problem of points. Two people are flipping a coin, one bets on heads and the other bets on tails. They agree flip until either heads has occurred ten times or tails has occurred ten times. But they have to quit when the score is seven to five; how should they split up the money?

Anyway, the Forbes article claims: "If your goal is to nab the best risk-adjusted return (as opposed to playing for hours on end), place fewer, smarter and larger bets." I'm not entirely sure what this means, because "risk-adjusted return" is a vague phrase. If you bet, say, $10 each on 100 spins of the roulette wheel, or $100 each on 10 spins of the roulette wheel, either way you expect to lose about fifty bucks -- $52.63 to be exact. This is what we mean when we say that the "house edge" in roulette is 5.26% -- it means that the house will, on average, take 5.26% of your bet. (If all your bets are on "red" or "black", then roulette is basically like flipping a coin -- a very complicated coin -- except that there are certain special outcomes where both "red" and "black" lose.)

But the variance of your winnings on a single spin at $10 is 100 "square dollars" or so, so the variance of your winnings on 100 spins is 10,000 "square dollars". The standard deviation is about $100. For a single spin at $100, the variance is 10,000 "square dollars", and so the variance on 10 spins at $100 each is 100,000 "square dollars"; the standard deviation is about $316. What does this mean? It means that if you make a few big bets then the random fluctuations won't cancel out. So what you do have is a greater probability of coming out ahead. Maybe this is what they mean by "risk-adjusted return".

To make the statement about "random fluctuations" more precise, I could use the central limit theorem. The central limit theorem says, essentially, that when you add up a bunch of random things you get something that's approximately normally distributed. But it's not really fair to do that here. Why? Because the normal distribution is a

*continuous*distribution -- that is, it can take whatever value we like. The distribution of the amount of money you have after ten spins is quite "lumpy" -- it can be, say, $0 (if we win on five spins and lose on five spins), or $200 (if we win on six and lose on four), but not something in between. I'm afraid of losing something in the holes between $0 and $200, as it were.

(By the way, there are roulette bets that let you bet that any of 1, 2, 3, 4, 5, 6, 12, or 18 numbers come up. These all have the same house edge -- except for the 5. Don't take that one.)

Also, rumor has it that there are devices that you can sneak into a casino which will watch the roulette wheel and tell you where it's more likely to come up. I'm not sure if they're small and unobtrusive enough -- or sufficiently good at prediction -- to be good for anything.

They also seem to claim that it's possible to win at blackjack --

**in expectation**without counting cards; that's not so! The house edge in blackjack under most rules is about 0.5%. But even if you could win in expectation, I'm not sure if it's worth the trouble. If I'm going to have to pay attention to what I'm doing, I don't want to have a high chance of losing money.

## 4 comments:

You should try the no zero game over at betfair (think this is limited to UK players), or at least play online roulette which only has one zero (reducing the house edge to 2.7%). I can't believe people still play American Roulette with the double zero online. It just doesn't make sense when single zero games are everywhere these days.

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