Chapter 7
Roulette Revisited

In this chapter, we make use of some of the concepts we learned so far to address additional interesting questions about roulette. Although we concentrate on roulette, many of the ideas discussed here can be extended to other games based on independent rounds such as craps.

7.1 Gambling Systems

Gambling systems are strategies that increase or decrease the size of a bet according to whether the player is winning or losing. They are promoted as tools that allow players to beat the house advantage; plenty of books have been written on the topic; and many gamblers have rediscovered the same tactics over and over again. However, it is important to emphasize that for games that rely on independent rounds of play such as roulette or craps, no system can be devised to beat the house.

7.1.1 Martingale Doubling Systems

Martingale doubling systems are very simple. To play the system, you must keep betting until you win, doubling your bet every time you lose. More specifically, you start a betting cycle by making a small bet, such as $1. If you lose, you double your bet and gamble again; if you win, you take your winnings and start a new cycle by betting $1. Typically even bets, such as the color bet in roulette, are used. However, this is not a requirement.

In a world where you can keep making bets indefinitely, the martingale doubling system guaranties that you will make $1 at the end of a betting cycle no matter what the probability of winning is. Indeed, say that it takes individual bets to complete a cycle. Since the bet is even, at that point you win $2. Also, since you lost the previous wagers, you have lost a total of $ . Note that the amount you have lost is the sum of terms of a geometric series. Recall from Chapter 6 that

In this case, we have . Therefore,

and your profit for the cycle is

no matter what the value of is.

At the first sight, this calculation suggests that a doubling system should allow you to always make money. What can go wrong? The underlying assumption of this system is that you can keep playing indefinitely until you win. However, in real life, your bankroll is finite, and the bets you need to make to keep the system going grow very fast. So you might not be able to cover the next bet required by the system to keep going, at which point you will lose all your money.

To illustrate this, assume that you have $1000 and your initial bet is $1. How many losses in a row can you take before you run out of money to make the next bet? We just showed that the accumulated loss after rounds of the game is (see also Table 7.1). Hence, if we lose 9 times in a row, we will only have $489 left, which is not enough money to cover the 10th bet that the system requires (which would be for $512)! In general, the number of rounds that you can play is simply given by where is the amount of money in your bankroll and means “round the number down to the nearest integer” (in this case, , so ). This expression also makes it clear that doubling your bankroll (e.g., taking your initial money from $1000 to $2000) only buys you one additional round before you go bust!!

Table 7.1 Accumulated losses from playing a martingale doubling system with an initial bet of $1 and an initial bankroll of $1000

Round ()	Bet on this round	Accumulated loss	Money left
1	1	0	1000
2	2	1	999
3	4	3	997
4	8	7	993
5	16	15	985
6	32	31	969
7	64	63	937
8	128	127	873
9	256	255	745
10	512	511	489

Now, you may argue that losing 9 times in a row when making color bets in roulette is a very unlikely event. Because the spins of the roulette wheel are independent, the exact probability of this happening is

and therefore

Consequently, even if we start with $1000 and bet only $1 initially, we have that roughly every 300 cycles we will not be able to cover the next bet and the martingale system will fail (the exact number is cycles). In the meantime, we would have made a profit of about $300, but even if we reinvest the winnings we are bound to eventually run out of money to cover the next bet required by the system.

Some additional intuition can be obtained by simulating the running profit of playing a martingale doubling system with an initial bet $1 over 2000 spins of the roulette:

Figure 7.1 shows the result of one such simulation. The increasing trend in the cumulative profit suggests that, as advertised, the system makes money as long as we can keep playing it indefinitely. Note, however, that the positive trend is punctuated by sporadic big losses (of over $2000 in one case, even though our initial bet was only $1). It is these big sporadic loses that make the system fail in real life!

Figure 7.1 The solid line represents the running profits from a martingale doubling system with $1 initial wagers for an even bet in roulette. The dashed horizontal line indicates the zero-profit level.

7.1.2 The Labouchère System

To play the Labouchère system, you need to decide how much money you want to win and then write a list of positive numbers that add up to that quantity. For the sake of argument, say that you want to make $100, and you decide to use the numbers 15, 15, 20, 25, 20, 5 in your game. You always bet the sum of the first and last numbers in the list (if a single number remains, you use that number). If you win, the two numbers are removed from the list; if you lose, the amount of the losing bet is added at the end of the list. You stop playing once there are no more numbers in the list.

First, you need to convince yourself that the system, if completed, will indeed allow you to win the sum of the amounts in the list. To see this, assume first that it is your lucky day and you win all your bets straight. In our example, that means that the first time you bet (and win) $20, the second time you win $35, and the third time you win $45. So the total amount you win is $100 as expected. What if you lose your first bet but then win all others straight? In that case, your list now contains the numbers 15, 15, 20, 25, 20, 5, 20 and you are losing $20. But if you now win all your bets in a row you will be making $120, so your net profit will be $100 again. In general, by adding the amounts you lose to the end of the list you make up for any loses you might have incurred in the middle of the game before stopping, which ensures that you will make the desired amount of money. The following R code simulates the running profit of the Labouchère based on an initial list with 50 entries of $10 each used on an even roulette bet. The cumulative profit from the system can be seen in Figure 7.2.

Figure 7.2 Running profits from a Labouchère system with an initial list of $50 entries of $10 for an even bet in roulette. Note that the simulation stops when the cumulative profit is ; the number of spins necessary to reach this number will vary from simulation to simulation.

Just like the martingale doubling system, the Labouchère system would seem to ensure that you always make money when playing roulette. However, Labouchère systems share the same weaknesses as martingale doubling systems. If you hit a bad enough losing streak you might run out of money before you have the chance to recoup your previous loses or make any money. However, as the simulation suggests, since the size of the bets in a Labouchère system grow linearly rather than exponentially, the number of games you are able to play before going bankrupt tends to be larger.

7.1.3 D'Alembert Systems

The D'Alembert system is based on the idea that a win is less likely if you have just won and more likely if you have just lost. Hence, you should increase the amount of your bet after you lose and reduce it after you win. The recommended progression is typically linear, so that you add a fixed quantity (say, $1) to your bet when you lose, and subtract the same quantity every time you win, to the table minimum.

Whereas the martingale doubling systems and the Labouchère systems are based on mathematically sound principles (they do not work only because in real life we do not have an infinite amount of money in the bank), the D'Alembert system is based on an erroneous probabilistic argument. The spins of the roulette are independent from each other, which means that the probability of winning or losing does not depend on the past (the game is memoryless). It is true that, before you make any spin, the probability of getting 9 loses in a row when playing even bets in roulette is very small. However, after you have already seen 8 loses, the probability of getting the 9th is exactly the same as the probability of getting the first one.

The following R code simulates the cumulative profit from applying the D'Alembert system with an initial bet of $5, change in bets of $1, minimum bet of $1, and maximum bet of $20 to an even roulette bet. The clear decreasing trend and large negative values in Figure 7.3 corroborate our argument that the D'Alembert system does not work.

Figure 7.3 Running profits over 10,000 spins from a D'Alembert system with an initial bet of $5, change in bets of $1, minimum bet of $1 and maximum bet of $20 to an even roulette bet.

7.2 You are a Big Winner!

Even though the expected profit in roulette is negative, it is actually not uncommon for players to be able to get ahead for a while. Indeed, you can temporarily make a lot of money in roulette, but the law of large numbers implies that if you want to keep it, you need to stop playing and never do it again for the rest of your life!

For example, let's compute the probability of winning exactly 10 rounds out of 15 played when making $1 color bets (that would mean that you are ahead by $5 after playing 15 rounds). There are many ways in which this could happen; for example, you could win the first 10 rounds and lose the next 5,

or you could lose the 2nd, 3rd, 5th, 12th, and 13th,

Consider first the probability of each one of these sequences. Since the rounds are independent, all sequences of 15 spins that include 10 wins and 5 loses have the same probability,

(recall that the probability of winning a color bet is , while the probability of losing it is ). This means that

Now, to compute the total probability of winning 10 rounds out of 15, we need to sum the probabilities of all sequences that match the criteria. Since all of the different sequences have the same probability, this boils down to counting the number of sequences that match the criteria.

To compute the total number of ways in which you can get 10 wins in 15 spins, recall again the combinatorial numbers we discussed in Chapter 4. We need to pick 10 positions in the list out of 15, and the order in which the 10 positions are selected is of no consequence to us. Therefore, there are ways in which you can have 10 wins in 15 spins of a roulette. Hence,

More generally, consider a random variable that counts the number of wins out of rounds. The same argument we used before leads to

Note that if , then you made $k from the rounds you won, and lost $(n−k) from those you lost, making your profit from playing the game ) dollars.

Now let's put this result to good use. Say that you have been playing roulette all night. For the sake of the argument, say you have played 300 rounds (which means about 5 hours at a rate of 60 spins an hour) by betting $1 each time on color. After all this, you are ahead by $20 (this means you had 20 more wins than losses) and you feel very unlucky because you have made so little money. Are you justified?

One way to address this question is to compute the probability that somebody would win $20 or more after 300 games. Now, for you to be ahead by $20 or more, you would need to win at least 160 of the 300 rounds you have played, so we need to compute

Computing this quantity by hand is difficult, but you can use R to obtain the number (see Sidebar 7.1):

This means that, for every 100 players, only about 2 would have made $20 or more after playing for 5 hours... I would consider you quite lucky!

7.3 How Long will My Money Last?

We can use some of the tools we developed to study the martingale doubling system to answer other interesting questions about roulette. For example, suppose that you want to go out and play roulette tonight. Since the expected profit from this game is negative, you know for sure that you will eventually lose all your money. However, how long you play is a random variable whose distribution will depend on how much money you have and how much you bet each time.

To make things simple for now, say that you start with just $1, that you bet $1 each round, that you do not reinvest your winnings, and that you try to make your original $1 last for as long as possible by playing even bets such as a color bet. If you are a very unlucky individual, you might lose on the first spin, so that you might be able to play only one round. So, if you let

then we have .

Now, for you to be able to play exactly two rounds, you would need to win the first round and lose the second. Therefore, since spins are independent, we have . More generally, for you to play exactly rounds, you need to win the first rounds and lose in the th round, which happens with probability

where could be any integer number greater or equal than 1. Table 7.2 shows a graph of probability as a function of ; as you would expect, the longer the streak, the lower its probability.

Table 7.2 Probability that you play exactly rounds before you lose your first dollar for between 1 and 6

1	0.5263158
2	0.2493075
3	0.1180930
4	0.0559388
5	0.0264973
6	0.0125514

To compute the average length of one such streak, that is, , you would need to compute

which can be rewritten as

With a little bit of algebra, and using again the formula for the sum of the terms of a geometric series on each row, we get . Accordingly, on an average night, you would play for a little bit less than two rounds!

The previous scenario is probably too simple to be of practical use. For example, even if you decide not to reinvest your winnings, you would probably not go to the table with only $1. So, let's say that you start with $10, and you make $1 bets (but do not reinvest your winnings). There are a couple of ways in which you can work with the random variable

If you only care about the expectation, you can proceed in the following way. Since you make $1 bets, you can think about the gambling process as making 10 bets of $1, and riding each one until you lose the dollar. This means that you can write

where each corresponds to one independent realization of our original random variable. Therefore, we can easily see that

In other words, if you start with $10 and make $1 bets, you can expect to play for about 38 minutes on an average night (assuming about one spin every 2 minutes).

If you care about the whole distribution of , the following approach is a bit simpler than dealing with the sum of multiple random variables. For you to play exactly rounds before losing all your money, you need a sequence of wins and losses that satisfies two conditions: (1) there are exactly 10 losses, and (2) the th round (the last one) is a loss. In other words, you need a sequence such as

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Sidebar 7.1 The Binomial Distributions in R

R includes functions that allow you to compute probabilities associated with a host of well known random variables. For example, the functions dbinom() and pbinom() allow you to compute and either or when follows a binomial distribution. For example, for a binomial distribution with and

which can be computed in either of these two forms:

On the other hand, for

while is obtained as

Finally, the function rbinom() can be used to generate random numbers that follow the binomial distribution:

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Now, this sequence is of length 25 and (since rounds are taken together and are independent from each other) it has probability

However, note that this is not the only possible sequence that satisfies these criteria. As a matter of fact, there are such sequences (recall our discussion on combinatorial numbers from Chapter 4 and notice the last position has to be a loss, so we need to pick 9 positions for the remaining losses among 24 options). Since all of these sequences have the same probability:

More generally, the probability that you are able to play for rounds if you started with $ and you bet $1 per round and do not reinvest your winnings is

for any .

The case in which winnings are reinvested is a bit trickier and beyond the scope of this book. However, a simulation in R can provide you with some intuition:

Note that by reinvesting your winnings you can significantly prolong the amount of time your $10 will last. However, since the expected value of the game is negative, you are bound to eventually go bankrupt!

7.4 Is This Wheel Biased?

In Chapter 3, we discussed the use of Chebyshev's inequality to approximately determine the number of spins needed to detect bias in a wheel. We consider now the related question of whether a sample consisting of a given number of spins provides evidence of a biased wheel. For example, let's assume that you have collected the results of 10,000 roulette spins and you observe that the number 31 has appeared 270 times (recall you would have expected to see it just about times in this many spins). Does this suggest that the roulette is biased in favor of the number 31?

To answer this question, let's compute the probability that you observe the number 31 at least 270 times in 10,000 spins of the wheel if the roulette is not biased,

This leads to

As before, you can use R to compute this number

Since this number is relatively large, there is little reason to think that the wheel is biased (the difference between 263 and 270 is small enough for it to be likely due to randomness).

7.5 Bernoulli Trials

When you look at the outcomes of multiple rounds of roulette, you are looking at an example of a very particular type of experiment called Bernoulli trials. A set of Bernoulli trials satisfies the following requirements:

• Each repetition of the experiment is independent from the rest.
• There are only two possible outcomes for each repetition of the experiment (call them win and lose).
• The probabilities of winning and losing are the same for each experiment (call the probability of success ).

There are a number of interesting probability distributions associated with Bernoulli trials. These distribution appeared in previous sections. For example, the binomial distribution arises when we are interested in the number of wins among a total of repetitions of the experiment.

The binomial distribution is given by

For example, the binomial distribution appears when computing the probability that a certain outcome of roulette appeared times in repetition of the wheel (recall Section 7.4).

The geometric distribution arises when we want to know how many trials it will take us to get one success,

The geometric distribution is given by

On the other hand, the negative binomial distribution appears when we want to know how many trials it will take us to get successes (therefore, the geometric distribution is a special case of the negative binomial when ).

The negative binomial distribution is given by

The geometric and negative binomial distributions appeared in Section 7.3 when we investigated the number of spins of a roulette wheel that you can make before running out of money if profits are not reinvested. Note that the main difference between the binomial and the negative binomial random variables is what is considered fixed and what is considered random. While the binomial distribution assumes that the number of trials is fixed and the number of wins is random, the negative binomial assumes the opposite.

7.6 Exercises

1. 1. What is the martingale doubling system? How can it fail?

2. 2. What is the Labouchère system? How can it fail?

3. 3. How can minimum and maximum table bets affect the likelihood that you go bust when using a martingale doubling system?

4. 4. Would a martingale tripling system avoid the problems with the martingale doubling system?

5. 5. You decide to play roulette using the martingale doubling system. If your bankroll is $30, your initial bet is $1 and you do not reinvest your winnings, what is the average amount of time you might expect to play?

6. 6. You decide to play roulette using a martingale tripling system. If your bankroll is $90, your initial bet is $1 and you do not reinvest your winnings, what is the average amount of time you might expect to play?

7. 7. How could you use the martingale doubling system in craps?

8. 8. What is the probability of winning 12 even bets in 30 spins of the roulette?

9. 9. What is the probability of winning 12 even bets in 200 spins of the roulette?

10. 10. The probability of getting a 7 in the game of craps is 1/6. The famous craps player known as the dice dominator is said to have avoided a 7 in the point-phase of the game of craps for 30-something consecutive rolls. To keep it simple, just imagine rolling two dice and you are only interested in whether a 7 comes out or not; what is the probability of avoiding 7 in 35 consecutive rolls of a die?

11. 11. Do you think you can be called a dice dominator if you can avoid 7 for 15 consecutive rolls of two dice?

12. 12. Say that you are trying to determine if a given (European) roulette wheel is biased in favor of the number 16. To do that you collect the outcome of 15,000 spins, and find that 400 and 13 of them are 16s. What is the probability of obtaining 413 or more 16s in 15,000 spins of a European wheel if it is not biased? Is there evidence that this particular wheel is biased?

13. 13. In the same setting as the previous question, what is the probability of obtaining 602 or more 16s in 15,000 spins of a European wheel if it is not biased? Is there evidence that this particular wheel is biased?

14. 14. If playing American roulette you bet $1 on red each time, what is the probability that you are ahead by at least $10 after 100 rounds?

15. 15. In the same setting as the previous question, what is the probability that you will be ahead by at least $2 after 500 rounds?

16. 16. [R] Corroborate the value of when is binomial with and provided by the function pbinom(159, 300, 18/38, lower.tail=FALSE) in two ways:

    • By adding up the 141 terms involved in the sum.
    • Using a simulation.

17. 17. [R] Modify the simulation of the Labouchère system to estimate the probability of going bankrupt if your bank is $200 and your list consists of 20 elements, each corresponding to $10 and you are making even bets.

18. 18. [R] Modify the simulation of an even bet in roulette to corroborate the calculation of the expected number of spins before going bankrupt if you do not reinvest your winnings.