The Mathematician’s WastebasketVolume 1, Issue 4

Stephen Devereaux April 28, 2013

Beating Roulette?An analysis with probability and statistics.

Every time I watch the film 21, I feel like I’ve made the right decision with my life bygoing into math. The movie is about college students who learn how to count cards in black-jack then make a killing at Los Vegas casinos. There’s something alluring about being ableto use mathematics to beat the system of gambling and make a lot of money. Unfortunately,counting cards is much more difficult and complicated in real life than it appears in themovie, and the system isn’t flawless. But if you can’t use mathematics to cheat at gambling,then what’s the point?

Naturally, I spend a lot of my free time wondering how I could apply mathematics to theworld of gambling. After all, it’s not really gambling if you know you’re going to win.

My attention turned to Roulette. The game is so simple, there just has be a way to beat it.

For those of you who live outside the gambling world (much like myself), allow me toexplain how the game works.1 The game involves a spinning wheel, and the casino employeerolls a ball around the outside of the wheel. Inside the wheel are thirty-eight notches/holes

1The explanation that follows is for American Roulette. European Roulette is slightly different, thoughvery much alike. The main difference is that European Roulette has only 0 and not 00.

that the ball can fall into when it loses momentum. Thirty-six of the holes are numbered1 through 36 (not in order) and alternate in black and white colors. Two of the holes arelabelled 0 and 00 and are colored green. The gamblers then bet on where the ball will land.There are many ways to do this, but I’m only going to focus on two.

One way to bet is by dozens; you try to predict if the ball will land in 1-12, 13-24, or25-36. You won’t quite win a third of the time, because 0 and 00 are mucking up the works.But the odds are still twelve in thirty-eight, which ends up being about 31.58%, and thepayout is two-to-one. Now, here was the big idea I came up with in the shower: what if Ialways bet on the last dozen?

No, Steve, that’s a terrible idea, you’re thinking. Right. But suppose I bet one chip onthe first spin. If I win, great; I’ll bet one chip on the next spin. If I lose, no problem, I’ll bettwo chips on the next spin. If I lose that, I’ll bet four chips on the next spin. Then eight.Then sixteen. Then thirty-two. And so on. I’ll keep doubling the previous bet until I win.As soon as I win, I make a profit, and since the payout is two-to-one, the profit will be atleast a little worth my while.

Bet Chips If I Win Chips Lost in Previous Bets Total Profit1 2 0 22 4 1 34 8 3 58 16 7 916 32 15 1732 64 31 3364 128 63 65128 256 127 129256 512 255 257

In a fascinating twist, it seems I make a profit faster if I lose a few times before I win. Seemssound, right? Seems foolproof, even. I’m bound to win sooner or later, so why not do this?In fact, why isn’t everyone doing this? Why aren’t casinos out of business?

It turns out I’m not the first person to come up with this. I was almost finished writingthis article when I discovered that this strategy, called the Martingale System, became pop-ular in eighteenth-century France. The method is bold, but ultimately fails. Why?

One of the problems is that many Roulette tables have a rule governing the maximumbet that is allowed. Suppose for a moment that we’re at a table whose maximum bet is$300. That means I can place nine bets before I’m not allowed to double my previous bet.So if I strike out nine times in a row, I’m going to have to suck it up, count my losses, andstart again with betting only one chip. Yes, I’ve lost 511 chips, but big deal, right? I’m go-ing to make that up again. And besides, what are the chances I strike out nine times in a row?

Let’s answer that question, actually. This is an easy question to answer because spinson the Roulette wheel are what we call independent events in the study of probability. Inother words, the result I get from the current spin is in absolutely no way dependent on the

previous spin. The odds are always the same.2 Since the events are independent, we canreason how likely we are to lose n consecutive times. First, we note that since the probabilityof winning a single spin is 0.3152, the probability of losing on a spin is 1− 0.3152 = .6842,or about 68.42%. Thus, the probability that I lose n times in a row is

(.6842) · (.6842) · (.6842) . . . (.6842)︸ ︷︷ ︸n

= (.6842)n

Since 0.6842 < 1, higher powers result in lower numbers, so the probability of me losingn times in a row decreases with higher values of n.3 Therefore, it seems as though it is highlyunlikely that the player should lose many times in a row; as long as he or she keeps bettingon the same dozen every time, the probability of the ball landing in his or her dozen getsbetter and better.

n Approximate Probability of Losing n Consecutive Times1 68.42%2 46.81%3 32.03%4 21.92%5 15.00%6 10.26%7 7.02%8 4.80%9 3.29% (cut-off for our example)

10 2.25%15 0.34%20 0.05%

Major casinos in Vegas will sometimes have a maximum bet of $1000 allowed. The catchis that the minimum bet required is $10 rather than the $1 we’ve been assuming. But sup-pose I start with $300 and am allowed to bet $1 with a maximum bet of $1000. I createda simulation to test my hypothesis (before I learned of its history) and was ready to watchthe cash flow in.

I have never been so wrong.

2This is in stark contrast to what is known as the Gambler’s Fallacy–the belief that if I’ve lost severaltimes in a row, the chances of me winning this time are higher. This isn’t true; the chances are exactly thesame as they’ve always been.

3While at first this might appear to violate the Gambler’s Fallacy, it does not. The Fallacy concerns thesingle, current bet, whereas our observation here involves several bets in a row. The mathematical truth inour observation is likely the cause of the popularity of the fallacy in the first place; it is tempting to treatthe single event and the multiple event interchangeably, which is not mathematically valid.

Occasionally, I made money–for a little while. But as time goes on, the story is alwaysthe same.

Assuming I could keep borrowing enough money to play a thousand spins, what wouldbe the end result? To answer that question, I played one thousand games of one thousand

spins. On average, I would lose $1005. And that number actually seems generous—in somesimulations, I lost over thirteen thousand dollars!

Min Q1 Median Mean Q3 MaxMoney Made −13260 −2721 −987 −1005 838.2 6064

Spins Until Bankrupt 9 32 75 161.8 200.8 1000Max Money 299 395.8 704 1165 1483.5 6829

Games Without Bankruptcy 40

Here, the top row of the table gives the amount (in dollars) that I would be left withafter one thousand spins. These values assume that if I ever run out of money, I can keepborrowing money for the next spin (that way, even if I’m thousands of dollars in the hole, Ican try to chip away at that debt). The second rows gives the number of spins before I hadto start making new friends to lend me some cash.4 The third row tells me the maximumamount of money that I ever had at any point in the game. Finally, the last row tells mehow many games (out of a thousand) I had in which I never went bankrupt.

Now you’ll notice that in some cases, I actually did make a profit. Sometimes, I evenmade a rather hefty one. And yet, it just doesn’t quite seem worth the risk knowing thatI’ll still go bankrupt nine hundred and sixty times out of a thousand games. On average,I get around one hundred sixty-two spins before I’m out of money. And looking over thesimulation, the same thing always happens to make me go bankrupt.

I lose nine times in a row.

How can this be? The chances seemed so small! Maybe we can redeem this. What if Icould double my bet up to twenty times? Then the probability of me losing twenty timesin a row is a (.6802)20 = 0.05% chance. My fantasy casino here is allowing a $1 minimumbet and a maximum bet of $1,050,000.5 Of course, I would need to be able to start withenough money to be able to double up to twenty times, so let’s say I start with a milliondollars (this isn’t quite enough to cover twenty consecutive losses, but it is enough to covernineteen. No need to be greedy yet, right?) The good news is that I was unable to run thefirst simulation again because I made so much money that the computer ran out of memoryto hold all of the digits. Don’t you wish that would happen in real life? For the secondsimulation, we had eight hundred sixty-three out of one thousand games that never wentbankrupt—a significant improvement, to be sure.

But what about those other one hundred thirty-seven times? Even in this fantasy casino,I’m still going bankrupt from time to time. I’m still having days when I walk in with amillion dollars and walk out with empty pockets. In fact, in this last simulation, I went

4This statistic only applies to games in which I actually went bankrupt. The forty games in which I neverwent bankrupt were not included with the data.

5We would need to be able to bet 220 = 1048576, so we’ll round up to make an already unbelievablesituation a little easier to swallow.

bankrupt in as few as thirty-four spins. What gives?

The thing to remember here is that we’re playing for a really long time. A typicalRoulette spin takes about fifteen seconds; if we suppose it takes another fifteen seconds formoney and chips to change hands, a thousand spins alone would take over eight hours. Thekey thing to remember with probability is that, given enough time, even an event with a0.05% chance is bound to turn up once in a while. With our more believable case of startingwith $300 and doubling up to nine times, losing nine times in a row seemed unlikely (a littleover a 3% chance), but after a thousand spins, it’s very likely that it’ll happen at least once.And it only takes one time for it to bankrupt my account because we’re not making moneyfast enough to cover that kind of expense.

Another way to bet would be to bet on whether the number will be odd or even. Here,the payout is only even money6, so we make money much slower. But the silver lining is thatwe face a 0.31% probability of striking out nine times in a row. Of course, as we’ve alreadyseen, if we play a thousand times, this is still going to happen sooner or later, but it won’thappen as fast or as often as it was before.

Let’s run the simulation again and see how things differ. We avoided bankruptcy 163times.

Min Q1 Median Mean Q3 MaxMoney Made −4659 −567 −537 −330.3 472 519

Spins Until Bankrupt 9 136 291 356.5 503 1046Max Money 299 398 528 575.6 789.5 846

The potential profit isn’t as glamorous, but the potential pitfall isn’t as devastating. Westill end up losing around three hundred dollars each time we play.

The moral of the story is this: don’t play Roulette. The house always wins. Sit down atthe poker table instead and give yourself a chance. ♠

c©2013 by Stephen Devereaux.Stephen Devereaux is a graduate student at Western Michigan University.

6In other words, a one-to-one payout. Whatever you bet, that’s what you win.