# Why All Roulette Combination Bets Are the Same (Explained)

Do combination bets in Roulette increase your chances of winning? Does it help, to cover more numbers with different bets? When studying and simulating various Roulette betting systems and strategies, I found the following insight, when I asked myself: What is the best Roulette combination bet?

All combination bets in Roulette are exactly the same in terms of the expected average profit per round, which always computes to the house edge. This can be shown explicitly by analysis of probabilities and profits of the individual bets in a combination, and then combining them correctly.

But how does this work in detail? Read on, and I’ll show you everything you need to know in order to understand why and how this counterintuitive result comes about.

## What is the Best Roulette Combination Bet?

This is the underlying question fo all of this, right? For me, it was a process of figuring things out step after step, when I was trying to find an answer to that very question.

For example, the James Bond Roulette strategy is just a combination bet. Analyzing that in detail, I found various nice independences of a combination bet. In summary, the expected average profit per round of a combination bet in Roulette does not depend:

• on the number of individual bets involved
• on the size of the individual bets
• on whether or not the individual bets cover some numbers twice or even several times or not

These may sound surprising, but they are factual. How all this works, will be the subject of the following sections.

So, the short and simple answer to the question, what the best combination bet in Roulette is, is really that they are all the same. There is no best one.

## The Key Metric for Roulette and Other Games of Chance: The Expected Average Profit Per Round

The key to understanding how all combination bets in Roulette are the same, is to see how the potential for success of a particular type of bet is tied to the expected average profit per round of playing. This expected average profit per round is based on the following idea:

Imagine that a player has large enough funds to allow for a large number of rounds of playing. The player plays the same kind of bet very many times in a row. How much money does the player win (or lose) over time? And how much is that, averaged per round?

Simulating this, we would record the simulated player’s earnings and in the end, divide the difference in his funds (the funds at the end minus the funds at the beginning) by the number of rounds, i.e., the number of times that particular bet was played. The result of this division is the average profit that a player can expect per round in this particular game.

Playing a game of chance , each player at each round can win or lose, which is based on luck. But for many players over many rounds, we can compute rather precisely, how probable a loss or a win are, and how much profit can be made on average.

## The Expected Average Profit Per Round for Roulette and the House Edge in a Casino

It should not come as a surprise that the expected average profit per round is negative for the player in a casino. This is called the house edge , and it provides the basis of the casino’s business model. On average, the casinos can calculate rather precisely, how much money the’ll make for each playing customer.

The house edge for Roulette essentially depends on the number of zero-like or green numbers on the Roulette wheel. For a European Roulette wheel, there is a single green number, zero, for which the house edge is universally quoted as 2.7 percent. We’ll see soon, why this house edge is, in fact, rather universally valid across all kinds of bets.

For American Roulette, there are two green numbers, the 0 and the 00. This leads to a house edge of 5.26 percent. There are even Roulette wheels with three zeros on them, thus raising the house edge further to a whopping 7.69 percent.

Why is the house edge an issue for the player? Because that is the percentage of your bets that you will lose over time on average per round. To see this in action, here is a table that shows you what remains of a 1000 \$ starting amount of cash, just by subtracting the house edge a number of times.

RoundEuropean house edge of 2.7 %American house edge of 5,26 %Triple-zero house edge of 7.69 %
01000.01000.01000.0
1972.97947.37923.08
2946.68897.51852.07
3921.09850.27786.53
4896.2805.52726.02
5871.97763.12670.18
6848.41722.96618.62
7825.48684.91571.04
8803.17648.86527.11
9781.46614.71486.57
10760.34582.36449.14
11739.79551.71414.59
12719.8522.67382.7
13700.34495.16353.26
14681.41469.1326.08
15663.0444.41301.0
16645.08421.02277.85
17627.64398.86256.47
18610.68377.87236.75
19594.18357.98218.53
20578.12339.14201.72
21562.49321.29186.21
22547.29304.38171.88
23532.5288.36158.66
24518.11273.18146.46
25504.1258.8135.19
26490.48245.18124.79
27477.22232.28115.19
28464.32220.05106.33
29451.78208.4798.15
30439.57197.590.6

We can easily see the funds disappearing over time. After 30 rounds of playing, on average, the European Roulette house edge leaves you with less than half of what you started with. For the American Roulette house edge, it is less than one fifth (less than 20 %). And for the triple-zero Roulette house edge, we land below 10 % of the starting funds.

Bummer. Keep in mind that these are the averages. On the other hand, you can nicely see that it is good to be the casino. Given enough players such that the law of large numbers takes care of making statistical predictions a reality, you are earning money non-stop (on average).

Now, I have just given you a number (the house edge), and told you that this is what players can expect to lose per round, on average, at a Roulette table. But I have not explained why or given you any kind of proof. We’ll remedy that in the next couple of sections.

## Calculating the Expected Average Profit Per Round of a Simple Bet in Roulette

The origin of the house edge in Roulette lies in the combination of the probability for a win for each type of bet and the corresponding payouts. In the following sections I have prepared a couple of examples that show you both the effect of different Roulette wheels and different probabilities and payouts on the expected average profit per round.

### Calculating the Expected Average Profit Per Round of a Single Bet in European Roulette

Take a bet on a Single number for example: for a bet of one dollar, the payout in case of a win is 35 to 1, i.e., you make a profit of 35 dollars. In case of a loss, the profit is -1 dollar, i.e., you lose you bet.

The probability for a win is 1/37 (for a European Roulette wheel), because there are 37 numbers on the wheel and only one wins. The probability for a loss is therefore 36/37 (all other numbers lose).

These profits and probabilities for all possible outcomes must be combined to calculate the expected average profit as follows: for all possible outcomes, the probability and profit are multiplied, and all of these terms are summed up. For our example, the simple bet, this yields

$\frac{1}{37} \cdot 35 + \frac{36}{37} \cdot (-1)$

In this expression, we find the common factor 1/37. If we extract that, and look at the rest, we find

$\frac{1}{37} (35 + 36 \cdot (-1)) = \frac{1}{37} (35 – 36) = \frac{1}{37} (-1) = \frac{-1}{37}$

Evaluating this, we get -0.027, which is the expected average profit for a one dollar bet. But this result is more general. Let’s say that we bet a dollars, where the variable a can be set to any amount that we would like to use as a bet. Then we have

$\frac{1}{37} \cdot 35 a + \frac{36}{37} \cdot (-a)$

and the result is (check it)

$\frac{1}{37} (-a) = a \frac{-1}{37}$

Now we understand that, on average per round, we lose 0.027 times our bet, which is exactly what a percentage means.

### Calculating the Expected Average Profit Per Round of a Single Bet in American Roulette

The Single bet in European Roulette has shown us, where the house edge comes from. In order to see the difference to American Roulette, we’ll do this calculation again, but with the additional number 00 on the wheel.

First of all, the payout in case of a win is the same: 35 to 1, i.e., you still make a profit of 35 dollars and a loss of -1 dollar, i.e., you lose you bet, depending on what the outcome is.

The probability for a win is 1/38 for an American Roulette wheel, because there are now 38 numbers on the wheel, but still only one wins. The probability for a loss is thus 37/38 (all other numbers lose).

Combining profits and probabilities for all possible outcomes again analogously to the previous calculation, we find

$\frac{1}{38} \cdot 35 + \frac{37}{38} \cdot (-1)$

In this expression, we now find the common factor 1/38 instead of 1/37. If we extract that, and look at the rest, we find

$\frac{1}{38} (35 + 37 \cdot (-1)) = \frac{1}{38} (35 – 37) = \frac{1}{38} (-2) = \frac{-2}{38}$

Evaluating this, we get -0.0526, which is the expected average profit for a one dollar bet on a Single number in American Roulette. Again, we can generalize this to

$\frac{1}{38} \cdot 35 a + \frac{37}{38} \cdot (-a)$

for a bet of the size a, and the result is (feel free to check it once again)

$\frac{1}{38} (-2 a) = a \frac{-2}{38}$

So, what happened here? Since the probabilities shifted, but the payout stayed the same, we arrived at a number almost twice as large as for the European house edge. The house edge in American Roulette is 5.26 percent.

### Calculating the Expected Average Profit Per Round of a Single Bet in Triple-Zero Roulette

To make things even better for the casino, clever strategists came up with a Triple-Zero wheel, which has the usual numbers 1 to 36, but in addition the 0, the 00, and the 000. This shifts the probabilities even more, but the payouts are kept the same as before, which tilts the risk-reward balance even further.

Let us quickly evaluate the house edge for such a wheel, and then move on to other types of bets. The payouts, i.e. winning and losing profits, remain the same for the Single bet: For a bet of the size a you win 35 a or you lose a.

The probability for a win is now 1/39, and the probability for a loss is 38/39. Combining these to get the average expected profit per round, we get

$\frac{1}{39} \cdot 35 a + \frac{38}{39} \cdot (-a)$

which computes to

$\frac{1}{39} (-3 a) = a \frac{-3}{39}$

How big a percentage is this? The result is 0.0769 or 7.69 percent. And we have already seen in the table above, what kind of damage this can do. But now, on to an interesting question.

### Calculating the Expected Average Profit Per Round of a Other Types of Bets in European Roulette

Alright, you might say, now we know where the house edge comes from and how to compute the average expected profit per round for a single bet, but what about all the other types of bets? Well, the calculation we need to perform is always the same. It is just that we have to know the probability and the payout.

While the payout is a matter of definition (someone decides how much the player should win for any given type of bet), the probability can be easily computed by counting the favorable versus all possible outcomes. So, for example, for a Six Line (betting on two consecutive rows, containing three numbers each) the payout is 5 to 1.

What is the probability for a win? Six out of all numbers win, so if we stick to American Roulette for simplicity (you can do the same calculation for European Roulette), we have six winning outcomes versus 38 possible outcomes. The winning probability is thus 6/38, the losing probability is 32/38. Combining that with the payout, we get the expected average profit per round as

$\frac{6}{38} \cdot 5 a + \frac{32}{38} \cdot (-a)$

where a stands for the bet again. We take out the factor 1/38 as always and combine the two terms to get

$\frac{1}{38} (-2 a) = a \frac{-2}{38}$

The result, it appears, is again the house edge for American Roulette. As it turns out, this works out for all other types of bets as well, because the payouts are always such that the profit plus the bet would give you a fair return with 36 numbers on the wheel. The additional zeros then shift the sum by exactly their number into the negative, so -2 for two zeros, for example, divided by the total number of numbers on the wheel.

As a result, we can go to the next section with the general assumption under our belt that the expected average profit per round is the same for all types of bets.

## Calculating the Expected Average Profit Per Round of a Combination Bet in Roulette

In the previous sections, we have shown that we can assume the same amount of average expected profit per round for different types of bets in Roulette. Why is that important or even interesting? Because it lays the foundation for the calculation of the expected average profit per round of any combination of such simple bets.

This is what I will show you here. In particular, we’ll see that the number of combined bets doesn’t matter. Furthermore, it does not matter how the total bet is distributed on the parts, i.e., the individual bets contained in the combination bet. Sounds hard to believe at first, I know, but let me show you:

### Necessary Restrictions on Combination Bets in Roulette

Assume that we have a total sum of some amount of dollars to bet, once again. We’ll make life easier for us and replace that sum by the usual variable a. This time, we split this amount up into smaller amounts $$a_1$$, $$a_2$$, $$a_3$$, and so on, which are restricted only by the requirement that their sum must be a:

$a_1 + a_2 + a_3 + \ldots = a$

We could expect that also the ratios of these parts need to be restricted somehow, but as it turns out, they don’t. It is important to explain here that I already know that they don’t, because I have already done the calculation and seen the result.

But one particular aspect of the beauty of math is that we don’t need to impose any restrictions, until they really become necessary. What that means is that we can go ahead with the calculation without further restrictions and see, whether or not we get away without imposing more. If more restrictions were actually necessary, the calculation would reflect this. But not only that, the calculation would also show us, which the necessary additional restrictions would be.

So, we just leave it at the one restriction that we know must be there, and keep in mind that we’ll see what happens during the calculation anyway.

### Defining and Finding Parts of a Combination Bet in Roulette

The first thing we need for computing the average expected profit per round for a combination bet in Roulette is an overview of the parts that are needed. For that, we need to understand what we must do. The task is to sum over all possible outcomes (in terms of wins and losses). In order to do that, we’ll divide the entire set of possible outcomes (numbers on the Roulette wheel) into suitable parts.

Now, what is a suitable part? These parts are best defined and visualized by the types of bets we are placing and how they overlap. Yes, we can also deal with overlapping bets. If we make a bet on the High numbers (19 – 36), for example, and in addition, we make a second bet on the Six Line with the numbers 31 – 36, we need to split the numbers on the Roulette wheel up as follows:

• The numbers from 0 to 18 all lose (because we haven’t placed any bet on them).
• The numbers from 19 to 30 win via the High-number bet, but lose with respect to the bet on our Six Line.
• The numbers from 31 to 36 win via the High-number bet and the Six-Line bet.

So the rule is to divide the numbers up into groups such that within a group all numbers share the same … what, actually? They share the same profit(s) – yes, there can be more than one profit for a group, if it is covered by two or more bets. So we divide all possible numbers up into groups according to profits.

Then, what’s left to find are the probabilities for a number from each group to occur. More precisely, we need to count the numbers in the group and compare them to all possible outcomes (numbers), which is a straightforward task.

### General Formulation of a Combination Bet in Roulette

So, in analogy to the list in the previous section, we need to create a list of all groups of numbers that have the same profits, get their probabilities of occurrence sorted out, and add all of that to the list as well. After that, we can sum up the terms in the list.

To make our life easier, we will use variables in the calculation – sounds somewhat counterintuitive, i know. But you’ll see soon that it really helps to see what is going on independently of whatever bets we want to place on various different possibilities.

So let’s take the above example and run with it. We have the variables a for the total bet, and $$a_1$$ and $$a_2$$ for the individual bets on our two parts. Let’s say we bet $$a_1$$ on the High numbers and $$a_2$$ on the Six Line. Then, we have the following list of possible different outcomes:

• The numbers 0 to 18: we lose all our bets, i.e., the profit here is $$-a$$. The probability for this to happen is 19/37 (European) or 20/38 (American) – just counting the losing numbers and dividing the result by the total number of numbers on the wheel.
• The numbers 19 to 30: we lose the Six Line but win the High numbers (payout 1 to 1). Thus, our profit is $$a_1 – a_2$$. The probability for this to happen is 12/37 (European) or 12/38 (American).
• The numbers 31 to 36: we win both the Six Line (payout 5 to 1) and the High numbers (payout 1 to 1). Thus, our profit is $$a_1 + 5 a_2$$. The probability for this to happen is 6/37 (European) or 6/38 (American).

Now, in order to calculate the average expected profit for the combination bet, we need to add up all products of probabilities and profits, like this (for American Roulette):

$\frac{20}{38}\cdot (-a) + \frac{12}{38}\cdot (a_1 – a_2) + \frac{6}{38}\cdot (a_1 + 5 a_2)$

As you can see, I just put the terms from the list together, multiplied each fraction (the probability) with the corresponding profit, and put plus signs in between them. Now, let’s evaluate this sum. As usual, it helps to collect the factor 1/38, so we’ll do that:

$\frac{1}{38}\cdot\left( -20 a + 12 (a_1 – a_2) + 6 (a_1 + 5 a_2) \right)$

Evaluating everything inside the large round brackets, we get

$\frac{1}{38}\cdot\left( -20 a + 18 a_1 + 18 a_2 \right) = \frac{1}{38}\cdot\left( -20 a + 18 (a_1 + a_2) \right)$

And now, we use the only thing we have restricted the partial bets to do, namely, to add up, by replacing $$a_1 + a_2 = a$$. This yields:

$\frac{1}{38}\cdot\left( -20 a + 18 a \right) = \frac{1}{38}\cdot\left( -2 a \right)$

As usual, we can rewrite this to give us the expression above which led us to understand the house edge:

$a \frac{-2}{38}$

So what does that mean? It means that for the combination bet the expected average profit per round is exactly the same as for each of the individual bets in Roulette. This is a result of the premise that that expected profit is also the same among all different individual bets to begin with, but that is the way Roulette is set up.

Yes, I admit, this was not the most general calculation of a combination bet in Roulette, just one example. But we can be more general quite easily, so let’s do that as well, shall we?

### Even More General Formulation of a Combination Bet in Roulette

If you dislike the use of variables, you may want to skip this. On the other hand, the use of variables has incredible advantages, so why not stay, read on, and see why? Well, it’s up to you.

The general case is described by a certain number $$n$$ of individual bets that are combined into one combination bet. We are able to divide all numbers on the Roulette wheel into $$k \ge n$$ separate groups of numbers, for which the profit is the same for all numbers inside the group (but can be different from the profit in other groups).

So far, so good? Why is $$k \ge n$$ and not just equal? Because we can overlap the bets, as we have just seen in the example above. In particular, for partially overlapping bets, we have more possibilities to account for than there are bets. Unclear? Example!

Assume that in our previous example we had chosen the Six Line such that one of its Lines (three numbers) had been 16 to 18, and the second Line had been 19 to 21. Then we have two individual bets, but three possible winning outcomes, plus the one additional outcome where we lose. The three winning outcomes (groups of numbers) in this case are:

• 16 to 18: Only the Six Line wins
• 19 to 21: Both the Six Line and the High numbers win
• 22 to 36: Only the High numbers win.

Ok, and since I mentioned the loss for those numbers where we didn’t place any bet, there is another one that we need to account for. In total, we have $$k$$ different outcomes, for which we can write down profits and probabilities. We also have our $$n$$ individual bet amounts set up such that

$a = a_1 + a_2 + \ldots + a_n$

But how can we write down any possible combination of bets with correctly counted probabilities $$p_1, p_2, \ldots, p_k$$ and profits $$c_1, c_2, \ldots, c_k$$ for all possible outcome groups of numbers? The way we have done it so far, yields the sum

$p_1\cdot c_1 +p_2\cdot c_2 + \ldots +p_k\cdot c_k$

We know that we can in principle calculate this by counting the numbers in each group $$i$$ to get the probability $$p_i$$, and we can compute the profit from the bets and payouts of the winning bets and all losing bets, but that is not an easy task to generalize. We need to go at this from a different angle.

So, what we’ll be getting at is that we can disentangle the individual bets and sum them up individually as well, no matter whether or not they overlap. Doesn’t sound convincing? I know, but let’s use an example again to see how that works. In fact, let’s use the same example that we just saw with the six line and the high numbers partially overlapping.

First, we’ll compute the expected profit according to the general formula you just saw. The probabilities and profits sum up the following list of possible different outcomes:

• The numbers 0 to 15: we lose both our bets, i.e., the profit here is $$-a$$. The probability for this to happen is 16/37 (European) or 17/38 (American) – just counting the losing numbers and dividing the result by the total number of numbers on the wheel.
• The numbers 16 to 18: we win the Six Line (payout 5 to 1) but lose the High numbers. Thus, our profit is $$5 a_1 – a_2$$. The probability for this to happen is 3/37 (European) or 3/38 (American).
• The numbers 19 to 21: we win both the Six Line (payout 5 to 1) and the High numbers (payout 1 to 1). Thus, our profit is $$5 a_1 + a_2$$. The probability for this to happen is 3/37 (European) or 3/38 (American).
• The numbers 22 to 36: we lose the Six Line but win the High numbers (payout 1 to 1). Thus, our profit is $$a_2 – a_1$$. The probability for this to happen is 15/37 (European) or 15/38 (American).

Now, if we go ahead and sum these for terms up, we find (for American Roulette)

$\frac{17}{38}\cdot (-a) + \frac{3}{38}\cdot (5 a_1 – a_2) + \frac{3}{38}\cdot (5 a_1 + a_2) + \frac{15}{38}\cdot (a_2 – a_1)$

At this point, we could just go ahead and sum everything up in order to find the house edge at the end once again. However, instead we’ll do something different. We are going to try and rearrange these terms such that we get a sum of all the individual bets (in terms of probabilities and profits) as if they were played independently (which of course, in fact, they are).

What do I mean by that? Well, first of all, we’ll take apart the losing term at the beginning, which has a profit of $$-a$$ such that it shows all the terms that sum up to $$a$$ instead:

$\frac{17}{38}\cdot (-a_1 – a_2) + \frac{3}{38}\cdot (5 a_1 – a_2) + \frac{3}{38}\cdot (5 a_1 + a_2) + \frac{15}{38}\cdot (a_2 – a_1)$

Now we can group all terms with $$a_1$$ and $$a_2$$, respectively and keep the corresponding fractions around to get

$a_1 \frac{-17 + 3\cdot 5+ 3\cdot 5 – 15}{38} + a_2 \frac{-17 – 3 + 3 +15}{38}$

At this point, again, we are tempted to just sum everything up in the numerators of those fractions, but we won’t. Why? Because we can see what is going on, if we only combine all positive and negative terms in each numerator, like this:

$a_1 \frac{32\cdot (-1) + 6\cdot 5}{38} + a_2 \frac{20\cdot (-1) +18}{38}$

Pretty nice! Not only can we see that, if we sum the terms up in each numerator, we indeed get the house edge. We can also see that for each term, the positive and negative terms correspond exactly to the probabilities and profits for that bet alone.

That is the reason why I wrote those $$-1$$ factors separately in the parentheses: $$-1$$ is the profit for each individual loss, and it is multiplied by the probability in terms of the count of numbers that lead to a loss (32 for the Six Line and 20 for the High numbers). So, in effect, any combination bet can be viewed and calculated as the sum of the individual bets, in particular for the computation of the expected average profit.

This is the most important lesson here. With this in mind, we can now easily answer a few questions, just like follow-ups:

## Is There a Roulette Combination Bet That Always Wins?

The short answer to this is no. In order to always win, the player needs to cover all numbers with a bet or a combination thereof. Since any combination bet has the same expected profit than the individual ones, that means, however, that the player is guaranteed to lose by covering all numbers.

In fact, somewhat counterintuitively to what a beginner’s viewpoint may be, covering all numbers on the Roulette table is the only way you are guaranteed to lose. As soon as you leave some numbers uncovered, you have the potential to win.

## Probabilities for Roulette Combination Bets

The probabilities of combination bets in Roulette are based on the probabilities of the individual bets in the combination. As always in Roulette, winning probabilities are calculated by counting the numbers with a bet on them and divide the result the total number of numbers on the Roulette wheel.

## Payouts for Roulette Combination Bets

The payout of a combination bet in Roulette is calculated from the individual parts. Furthermore, it has to be viewed on a case-by-case basis, since the bets can be overlapping, not overlapping, covering various areas of the numbers, etc.

The simplest method for doing such a calculation is taking the combination bets parts and looking at them individually.

## Systems and Techniques for Roulette Combination Bets

Since all combination bets in Roulette are the same in terms of the expected average profit per round, there is no optimal way to combine individual bets in order to maximize profit or winning probability. While some combination bets have a higher probability of winning as opposed to losing, the amount of money won or lost always leads to the same expected profit on average.

However, since a combination bet is the same as an individual bet in this regard, it means that betting strategies or systems can be applied to combination bets in much the same way as to simple bets. If you are interested in such strategies, be sure to check out my analyses of, e.g., the famous Martingale Roulette strategy or the Reverse Martingale Roulette strategy.

## Tips and Tricks for Roulette Combination Bets

As for tips and tricks for Roulette combination bets, I will say this: know, what you are doing and what you can expect. That said, you can use combination bets to impress people, play Roulette like James Bond, or you can choose to educate others playing combination bets, and make a friend or two.

What all this boils down to, however, is just that playing combination bets at the Roulette table helps you to lose your money faster on average, if you bet more in one round on a combination bet than you would have bet on anything simple.

Finding a system that helps you winning more money at Roulette will be unsuccessful due to the facts yielded by a proper statistical analysis . If you don’t mind how much you win, but you’d like to win more often (even smaller amounts), then the Martingale strategy may be the thing for you.

This isn’t just a cheap throwaway line, but it is based on statistical analysis as well. Check out my article Martingale Roulette Strategy: Full Outcome Analysis with Charts for details.

Andreas

I am a nerd and trained in theoretical physics (with a PhD and everything ...). My interests are life as a human being and how computing can teach us valuable things about that as well as about nature in general. Read more about me.