This is post is the finale of our study of probability by examining the popular casino game, roulette. In case you missed either of our first two installments, here is our introduction to probability in roulette and here is our analysis of your expected winnings in playing roulette.
As we have mentioned over the last two posts, gambling strategies are popular in casinos to “skew the odd” in a gambler’s favor. A popular strategy for roulette, that is also widely used in currency trading, is The Martingale. And if you are like us and thought that was also a popular bird, we are afraid that you are actually thinking of that feathery troubadour, the Nightingale.
For those who are not familiar with it, the basic idea (and this works with any binary game) is to start your betting at a reasonable level and double it each time you lose. This increases your chance of winning in the short term. In other words, you bet $5 your first spin. If you win, you walk away with your $10 and are happy. If you lose, you bet $10 on the next spin. If you lose again, then you bet $20. And this goes on until you win. The reason why the casinos LOVE these "gambling strategies" is because they have enough money to weather an anomalous streak....while most gamblers do not.
The Martingale strategy is interesting because even though it seems to provide better odds of winning, it actually returns a WORSE expected value.
For instance, say we are trying to win $1 using the Martingale strategy, by betting on let us say Red. We set our stop loss at $127 which is 7 losses in a row (1 + 2 + 4 + 8 + 16 + 32 + 64 = 127).
The probability of me losing my $127 is (20/38)^7 which is approximately .01. Additionally, the probability of me winning my $1 is 1 - (20/38)^7 which is about .99. The odds look good right? Let us calculate the expected value.
E= .99 * 1 - .01 * 127 = .99 - 1.27 = -.28, or an expected loss of 28 cents. So before you all run out to Vegas, this suggests that even though you may win your $1 most of the time, you should expect to lose big eventually. The graph below charts the monetary goal between $1 - $100 against the probability of actually winning it using a stop loss of $127. The black line is the regression line, also known as the line of best fit.
As you can see, winning $50 is still over a 50% chance and a $100 over 25%!
Ultimately however, if you want to win more than $1, be prepared to increase your stop loss or initial bet. For example, if you want to win $10 using the same initial bet and stop loss, the probability of winning is simply (.99)^10 which equals .904, as you can tell from the graph. Consequently, to actually have a .99 probability of winning $10, the chance of winning $1 needs to be the 10th root of .99, or .99999.
We can now solve for our max losses. 1 - (20/38)^n = .99999 → .00001 = (20/38)^n which we can solve using WolframAlpha. The result is n = 18, or 18 max losses. This is equivalent to a loss of (1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048 + 4096 + 8192 + 16384 + 32768 + 65536 + 131072 = $262,143) 1% of the time!
By changing our initial bet to $5, but keeping our stop loss, our max losses drops to 4. Our chance of winning $10 then becomes (1 - (20/38)^4)^2 = .85. Not quite as good odds as above, but at least we would lose less money!
Our simulation of this is written in both R and Python and you can change any or all of your stop loss, target winnings, and initial bet to determine the probability of winning. Running this simulation across 100,000 games actually proves all of these probabilities discussed here, so check it out if you are interested! Naturally they all reside on our GitHub.
While it seems a success is incredibly likely, personally, we still wouldn't potentially sacrifice $127 for the chance of making just one more dollar! Let us know what you think in the comments!
The SaberSmart Team