a roulette paradox
May 5th, 2004 at 4:50 pmThis “paradox” is trivially solved with basic probability and expectancy calculations, but still it’s very cool:
Assume a casino roulette. In the US, a roulette has 38 numbers, 0 and 00, 13 black numbers and 13 red numbers. Let’s assume for simplicity that a roulette has 50 red, 50 black and 1 green (0) numbers.
Now, on one hand it’s a well known fact that the casino wins with the roulette. Say, for instance that I bet $100 on red. There’s a 50/101 chance for me to win, and 51/101 to lose. In the long run, I lose, the casino wins.
But lets consider the following strategy:
I bet $1 on red. If I win, I’ve won a dollar. If I lose, I double my bet - that is, bet $2 on red. If I win, I take my $2 and again start betting with $1, if I lose I double my bet - $4.
Think about it… eventually, a red will appear (the chances are a little smaller than 1/2 on each throw). When it will appear, I’ve won exactly $1, because my current bet is the sum of all previous bets + 1.
2^N - 1 = Sigma{n = 0 to N - 1} 2^n)
In more familiar terms, think of a binary tree. In a full (complete) tree the number of leaves is always the number of all non-leave nodes + 1.
In short, in each series of betting $1, then $2, then $4, etc, UNTIL I WIN, I will eventually earn $1. And hey, the chance that NO read will appear is very low even after 4-5 throws.
So, why can’t I rob the casino this way 
?
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August 21st, 2008 at 1:41 pm
There are two reasons why you won’t win.
Firstly, you actually win a tiny bit less than 50% of the time, because of the green 0. You’d need to increase your bet by more than double to allow for that. But putting that aside, the main reason you won’t rob the casino is that you run out of money long before they do. As you say, you keep doubling the bet: $1, $2, $4, … but after just ten doublings you have to bet $1024 just to break-even, and after twenty you would need more than a million dollars.
But it can be shown that the optimal strategy, that is the one with the least-worst expected return (the expected return is negative, that’s why casinos are richer than you) is to determine the maximum money you can afford to lose, then bet it all on one spin of the roulette wheel. If you lose, walk away. If you win, you’ve doubled your money. It might not be much fun, but mathematically it’s the best strategy.
August 21st, 2008 at 5:10 pm
How is that the best strategy?
1. For every 1$ you bet, your expected loss is 1/101$ (plus your initial 1$).
2. Therefore, the optimal strategy is not to play.
3. If you gain something else from playing besides money (like the thrill of gambling or free drinks), you might want to gamble as slowly as you can, to increase the number of minutes you spend in the casino.
Without any added incentive to bet, there is no optimal strategy besides “not playing”.