A winning strategy?

It was suggested to me that, given enough patience, you'll end up ahead in roulette with the following strategy:

Always bet on 2/3 of the numbers.

If, like me, you know next to nothing about roulette, it may be hard to form an opinion about such a strategy. So, the following section provides some background for those of us who need it (or anyone who enjoys roulette fact checking). If you're already in the know, feel free to skip ahead.

Roulette background

roulette table

Using math

Returning to the strategy above, if you bet on 2/3 of the numbers (e.g. two columns or two dozens) and win, your payoff is 36/24-1 = .5 times your original bet.

If you always bet all your money and always won, your pot would increase by 50% each time. Not bad at all.

But, of course, you don't always win, and when you lose, instead of a 50% increase, you suffer a 100% decrease (i.e. it hurts to lose double how much it helps to win).

You're betting on 24 out of 38 of the numbers, so you should win ~63% percent of the time. Which means losing ~37% of the time. And if ~63 > 2*~37, you might be in business. But, unfortunately, ~63 < 2*~37, meaning, on average, you're going to be losing money.

Note - Even if there weren't the extra 0 and 00 (meaning you're betting on 24 out of 36 numbers), you still shouldn't come out ahead, since ~66 !> ~33 (in fact, 24/36 = 2*12/36, meaning you would, on average, break even).

Using repetition

Because I don't trust my math and don't mind a chance to use some python tricks I learned in CS 6.00x, I made a Monte Carlo simulation of this strategy (you can find my code here).

Some technical details:

First, I simulated a single person playing roulette. For me, this involved writing functions to

Here's an example of what might happen if you walked up to the table with $60 in your pocket and bet $15 total on 2/3 (minus 0 and 00) of the board, playing for 1000 rounds or until your cash runs out. (Remember, if you put $15 in and win, you get back $22.50. If you lose, you get back $0.) 

1 run, try 1

Okay, that's not going to make anyone fall in love with the strategy, but maybe you were just having a bad day. Let's try it again.

1 run, try 2

Yikes, it's looking like maybe our math wasn't too far off. Let's try one more time. Lucky number three...

1 run, try 3

Nope, still doesn't look like a winning strategy.

We could do this all day, but I promised you a Monte Carlo, so here's 5000 runs of walking up to the table with $60 in your pocket and betting $15 total on 2/3 (minus 0 and 00) of the board. Each time you play 1000 rounds or until your cash runs out.

5000 runs

You can see here that, averaging over all 5000 runs of this simulation, by around your 33rd bet, you've lost half your money, and it clearly doesn't get any better as you keep going.

Some technical details:

To do this, I simply wrote a function that would run my previous code for a single person playing roulette, but repeat it many times. For each round, I averaged the players' remaining money across all the runs so that I could graph is, again using pylab.


Two things: