## Visualizing Probability: Roulette

I wrote a post over on the Society of Information Risk Analysts blog and I was having so much fun, I just had to continue. I focused this work on the American version of Roulette, which has “0” and “00” (European version only has “0” producing odds less in favor of the house). The American versions also have “Five Numbers” option to bet, which the European version doesn’t have.

According to this site, the American version of roulette could do about 60-100 spins in an hour, I figured maybe 4 hours in the casino and being conservative, I decided to model 250 iterations of roulette. I then chose $5 bets, which isn’t significant, changing the bet would only change the scale on the left, not the visuals produced. I then ran 20,000 simulations of 250 roulette spins and recorded the loss or gains from the bets along the way. One way to think of this is like watching 20,000 people play 250 spins of roulette and recording and plotting the outcomes.

I present this as a way to understand the probabilities of the different betting options in roulette. I leveraged the names and payout information from fastodds.com. The main graphic represents the progression of the 20,000 players through the spins. Everyone starts at zero and either goes up or down depending on lady luck. The distribution at the end shows the relative frequency of the outcomes.

Enough talking, let’s get to the pictures.

### Betting on a single number

What’s interesting is the patterns forming from the slow steady march of losing money punctuated by large (35 to 1) wins. Notice there would be a few unlucky runs with no wins at all (the red line starts at zero and proceeds straight down to 1250). Also notice in the distribution on the right that just over half of the distribution occurs under zero (the horizontal line). The benefit will always go to the house.

### Betting on a pair of numbers

Same type of pattern, but we see the scale changing, the highs aren’t as high and the lows aren’t as low. None of the 20,000 simulations lost the whole time.

### Betting on Three Numbers

### Betting on Four Number Squares

### Betting on Five Numbers

### Betting on Six Numbers

### Betting on Dozen Numbers or a Column

### Betting on Even/Odd, Red/Green, High/Low

And by this point, when we have 1 to 1 payout odds, the pattern is gone along with the extreme highs and lows.

### Mixing it Up

Because it is possible to simulate most any pattern of betting, I decided to try random betting. During any individual round, the bet would be on any one of the eight possible bets, all for $5. The output isn’t really that surprising.

### Rolling it up into one

Pun intended. While these graphics help us understand the individual strategy, it doesn’t really help us compare between them. In order to do that I created a violin plot (the red line represents the mean across the strategies).

Looking at the red line, they all have about the same mean with the exception of Five Numbers (6-1). Meaning overtime, the gambler should average to just over a 5% loss (or a 7% loss with five number bets). We can see that larger odds stretch the range out, which smaller odds cluster much more around a slight loss. The “scatter” strategy does not improve the outcome and is just a combination of the other distributions. As mentioned, the 6-1 odds (Five Numbers) bet does stick out here as a slightly worse bet than the others.

Lastly, I want to turn back to a comment on the fastodds.com website:

While I may disagree that the only bets to avoid are limited to those (had to get that in), I also disagree with the blanket statement. Since they all lose more often than they win, trying to get less-sucky-odds seems a bit, well, counter-intuitive. I would argue that the bets to avoid are not the same for every gambler. The bets should align with the tolerance of the gambler. For example, if someone is risk-averse, staying with the 2-1 or 1-1 payouts would limit the exposure to loss, while those more risk-seeking, may go for the 17-1 or 35-1 payout – the bigger the risk, the bigger the reward. Another thing to consider is that the smaller odds win more often. If the thrill of winning is important, perhaps staying away from the bigger odds is a good strategy.

Now that you’re armed with this information, if you still have questions, the Roulette Guru is available to advise based on his years of experience.

Nicely done Jay!

Very cool analysis Jay!

I always argued that betting for two columns is a good strategy: probability of winning for each column is around 1/3. If you place a bet on two columns, probability of winning is 2/3.

If one of the two columns wins, you loose the bet on the other column but gain two times the bet on the first column, so overall you’re winning by one factor. Keep in mind that the probability of this happening is almost 2/3.

This is the same for dozens. I would love to see the result of this strategy on your statistical model and if you have any comments on the logic/math.

ANaimi: thanks for bringing that up. It actually doesn’t quite work out that way and I usually see it better when I write up the code to calculate the odds because I end up simplifying it down before coding it. I’ll try to walk through it. When a bet (say $5 to have real numbers) is placed on two columns, there are two possible outcomes: 1) when the third column or zeros come up, there is a loss of $10 with a probability of 14/38 or 36.8%. 2) win $5 with a probability of 24/38 or 63.2% because placing two bets at $5 each means one will always lose. So we’re $5 down but the other one wins 2-to-1, so we get $10 back, net gain is $5.

The result is going to look like the rest. Just slightly less than 2 out of 3 times the gambler would win $5, just slightly more than 1 out 3 times the gambler loses $10.

The odds are set up consistently across the board to pay out slightly less then what is gambled. The only winning strategy is based on random luck.

One clarification.

The net gain isn’t really $5. On average you lose 50 cents a spin on a dozens bet: $10 x 36.8% is lose $3.68, win $5*63.2 or $3.16.

Interesting visualization.

Of course you’re absolutely correct, I meant the net gain from the collection of two bets if one wins on a single spin.

I’ve tried to stay away from talking in terms of averages, because a player would never lose $3.68 or win $3.16 when betting $5 on roulette. It’s turning a discrete event into a continuous range and I’m not sure people understand the transition (can they see how it applies if they lose $10 on the first round?) But perhaps that’s a better message for some folks to say “on average you lose $3.68 or win $3.16″ and they can see the difference better.

Thanks!

Is there a way to model the double-x-double bet risk? $5 each spot, $20 per spin.

A person bets 2 of the columns, say first 12, second 12.

Then they also bet 2 horizontal columns, say top 12, middle 12.

8 numbers in the first 12, 8 numbers in the second 12… 16 are double payoffs. 42% to win $30 on a $20 total bet.

4 numbers in the first 12, 4 numbers in the second 12, 8 numbers in the third 12…16 total are single payoffs, 42% to win $15 on a $20 total bet.

4 numbers in the third 12 pay nothing, 0 and 00 pay nothing…6 total numbers are losers. 15% chance to lose $20 on $20 bet.

Is this just a more complicated model vs. betting a single column or a color?

There is an 84% chance you sort of win something and a 42% chance you win 50% of your bet.

Does this just create more play time?

How about modelling the only winning strategy, bet colour (1.05:1) and double down. So, bet 5$ on red. If you win, yay. If lose, bet 10$ on red… keep doubling until you win. Start again.

John, After a little searching, what you’re describing is the “Martingale System” and has some challenges, first the bet increases exponentially, $5, $10, $20, $40, $80, $160, $320, $640 and there is (conceivably) a limit to available pocket cash to wager, but more importantly there are table limits in place. From what I can tell, common table limits are $5 min to $1000 max, meaning that the gambler would have to lose 8 times in a row (starting at min bet) before hitting the table limit throwing off the system. My guess is the min and max are correlated and would negate the effectiveness of the martingale system.

I’d like to try and simulate this, but I’m not sure what to do when the table limit on wagers is hit, any idea on what to do at that point? Stop at the point? Surrender that round and start over? Keep betting at max until a win?

Well done! Would be interesting to perform the same simulation with the European Roulette (without 00) and compare the two.

There is no winning strategy. Let’s get that straight. No bet here wins, and no strategy, like John’s, is going to win in the long run, in part because of what Jay said. Some bets aren’t as bad as others.

It’s easy to see this, and make quick calcuations if you understand expected value. For any bet, you can expect to “win”: ($ paid on win)*(prob of winning) – ($ lost on loss)*(prob of losing).

For a $1 bet on a single number, that is :

35*(1/38) – 1*(37/38) = -2/38 = $-.0526

So, you can expect to lose about $13 on 250 $1 bets. Your plot out to the right is simply the normal curve approximating the binomial distribution (wiki it). What’s a little interesting about this is that most people would have just plotted the possible “states” (your diagonal lines), but you filled it in and it shows the density of intermediate states.

Replot it using only 5000 or 10000 people but do about 1000 reps each. You’re going to see a “tighter” looking curve. Do it with dollar bets, too. It’s easier to mentally scale.

I’ve been recently reading up on Hidden Markov Models as a method to detect a hidden state given an observed sequence. One of the classic examples is detecting an occassionally dishonest casino. This would be a casino were every once in a while, unfair odds were introduced into individual rounds of play. In roulette each side of a die normally has a 1/6 chance getting rolled. But if you had some weighted die, maybe the odds were shifted to something like like (1/2) *(1/6) for the weighted side and a corresponding increase to the odds for the other sides. From a modeling perspective you could give odds to transitioning between fair and unfair die. I’m curious given your visualization how quickly this unfairness would manifest itself graphically.