Martingala

On this page, and as an example of what we will find in the book "Mathbet, win or not win", the chapter dedicated to the famous betting method known as Martingale is reproduced. All the formulas and mathematical developments described here are studied in depth in previous chapters of the book, but if something is not understood, do not hesitate to leave your question (or your criticism) in the comments section.

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Annex C: Case study of the Martingale method

In this third annex we will study the famous betting method known as Martingale, which consists in placing higher and higher bets backing the same result until it is won. In the following sections, we will first describe the method mathematically and after we will prove why it is not a method with a positive expected value and, therefore, why we should never apply it.

Annex C1: Description of the Martingale method

In order to consider a method of the Martingale type, the stakes are selected so that each one is enough to both recover the money lost in previous bets and to obtain a certain profit.

 An example of this method, applied to a balanced coin toss (50% probability (p) for each option) could be defined by the following rules:

• We start from an initial bankroll (b0) of 100 units.
• We place the 0.5% of our bankroll (x) backing heads at odds of 2.00 (c).
• In case of failure, we back heads again placing a stake twice the previous one (f) at odds of 2.00, until the maximum number of attempts (nl) is reached.
• If we succeed before that, the total gain is added to our bankroll.

Each time heads are obtained, an iteration (i) (not to be confused with the attempts) will be completed and the process will start again.

In the following sections, and based on the specific method defined above, we will deduce the equation that describes the common ratio (fm) of a generic Martingale method. That is, the formula for a Martingale of any factor (f), initial bankroll (b0), fraction of it (x) and odds (c).

Annex C2: Stake wagered in each attempt

First of all, we analyse how much will be placed on each successive attempt to guess the coin toss.

The first stake of each iteration (a1) is simply the fraction (x) of the initial bankroll of said iteration (b0). That is:

The second stake is the first one multiplied by the factor (f) of the Martingale; in our case 2.

The third stake is, in turn, the second one multiplied again by the factor (f):

That is, being (n) the number of the corresponding attempt, (f) the factor of the Martingale and (b0) the initial bankroll of each iteration, the stake to place on each attempt (an) can be defined as follows:

Annex C3: Relation between the odds (c) and the factor (f)

In a pure Martingale method, the net benefit that we will obtain when winning our bet is independent of the attempt (n) on which we succeed. That is, the stake (a) on each attempt has not only to recover what was lost in the previous ones, but also to win a certain amount (rn). In order to mathematically define this characteristic, we start raising the following equations:

The equation (rn1) defines the net result of the method when winning on the first attempt, while the equation (rn2) does so when winning on the second one. As described, the net result (rnn) has to be the same regardless of the attempt (n) on which we succeed. So:

Finally, solving the odds (c) as a function of factor (f) and vice versa we have:

In our case, for odds of 2.00 (c), the factor (f) of the Martingale should be:

Annex C4: Gain ratio

In the first place, as seen in the previous section, the gross gain when guessing the coin toss at any attempt (ran) will be the stake (an) multiplied by the corresponding odds (c) minus 1:

And every time we fail, the loss (rfn) of each attempt will simply be the stake:

Next, we deduce the formula that defines the net result (rnn) obtained when guessing the coin toss regardless of the number of attempts (n). If we succeed on the first attempt, the net result (rn1) will simply be the stake multiplied by the odds minus 1:

If doing so on the second one, the net result (rn2) will be the won amount on the second bet (ra2), minus the lost amount on the first (rf1):

The net result after succeeding on the third one (rn3), will be the won amount with the third bet (ra3) minus the lost amounts on the first (rf1) and on the second (rf2) ones:

If we continue to calculate for the cases of winning on the fourth attempt, on the fifth, etc. we will see that the net result after guessing at any attempt (rna) follows this expression:

As a reminder, any number raised to the power of zero (which would be the first iteration of the summation) is equal to one:

This summation of the powers of factor (f) can also be expressed as follows:

And, therefore, the net result (rna) when winning after any attempt can be expressed as:

If we finally substitute the odds (c) by the equation that relates them to the factor (f), we have that the net result (rna) when winning on the nth attempt (n) would be:

In our particular example, we will have a net gain of 0.5 units regardless of the attempt (n) on which we succeed.

Finally, the gain ratio (fg) for each iteration is:

where (bi0) is the initial bankroll of iteration (i).

In our example, the gain ratio (fg) is:

Annex C5: Loss ratio

In order to calculate the loss ratio (fp), we must first know the number of bets that can be placed before the stake (an) becomes higher than the available bankroll (bn). The following table shows this limit for an initial bankroll (b0) of 100 units, an initial stake (a1) of 0.5 and a factor (f) of 2:

Table 34. Initial bankroll (bn-1), stake (an), final bankroll (bn) and net result (rn) for each attempt of a Martingale of factor (f) 2

From the previous table, we obtain that the maximum number of available attempts (nl) in our example is 7. That is, if we get tails 7 times in a row, we would not have enough funds to place the eighth bet (a8) of 64 units, since our bankroll (b7) would have been reduced to 36.5 units. That is, we would have lost the 63.5% of it.

This can be mathematically expressed by calculating to which value our bankroll would be reduced when all attempts are failed (rnf). In this case, for 7 attempts, we have:

Again, the summation of the powers of (f) can be expressed as follows:

And therefore, the net result (rnf) in case of failing all attempts would be:

Finally, the loss ratio (fp) would be defined by the following expression:

In our particular case (with an initial bet of 0.5% of our bankroll and a factor of 2) we have a maximum number of attempts (nl) of 7 (Table 34). With these values, the loss ratio (fp) of our example is:

Annex C6: Probability of success

In the example described at the beginning, the probability of succeeding (pa) is given by the probability of getting heads before the eighth attempt. To calculate this probability, we add the chances of getting heads on the first, the second, the third, the fourth, the fifth, the sixth or the seventh attempt; since any of those results would work for us.

Being the probability of getting heads equal to (p),

the probability of succeeding on the first attempt will simply be the probability of getting heads (50%) in the first coin toss.

The probability of succeeding on the second attempt will be the probability of getting tails in the first coin toss multiplied by the probability of getting heads in the second one.

The probability of succeeding on the third attempt will be, in turn, the probability of getting tails in the first coin toss, multiplied by the probability of getting tails again on the second one and multiplied again by the probability of getting heads on the third.

And so on. Therefore, we have that the probability of success is:

We have again a summation of powers, with the difference that instead of adding powers of factor (f), we now add powers of (1-p). Applying the same solution as for the gain ratio we have:

In our specific example, the chances of succeeding before the eighth attempt are:

can check this result by calculating its complementary probability, that is, the probability of getting tails 7 times in a row:

That is, in 99.21875% of iterations we will succeed in the method, while in the other 0.78125% of them we will fail it.

Annex C7: Maximum number of attempts

Both the loss ratio (fp) and the probability of success (pa) depend on the maximum number of attempts (nl); and it is easy to see that these depend on the initial bankroll (b0), the fraction placed of it (x) and the factor (f) of the Martingale. That is, for the same factor, if we start from a bankroll of 200 units initially betting 0.5, we will have more attempts than if starting with a stake of 1 unit of a bankroll of 50.

The amount of money that we can lose by failing all our attempts (rnf), an amount already calculated above, may be at most our initial bankroll (b0). Thus, we equal this amount to the initial bankroll with negative sign (representing a lost amount):

Solving for the maximum number of attempts (nl) we have that:

Using logarithms, we have that the maximum number of attempts (nl) is:

However, since the number of attempts has to be an integer, we have to round the solution obtained to its immediately lower integer. In the given example, the maximum number of attempts would be:

Annex C8: Common ratio of the Martingale

Finally, as done with other geometric progression methods (as this is the case), we can define the common ratio of the method (fm) as follows:

In our particular case, the common ratio (fm) is:

As the common ratio (fm) is less than 1, the expected value of this method is negative and, therefore, it will make us lose money in the long-term.

Let’s suppose that with an initial bankroll (b0) of 100 units we carry out 200 iterations of this method; out of which we succeeded in 199 and failed in 1 (99.5% of wins, better than the theoretical 99.22% calculated before). We would end with the following units:

If we failed two iterations (a rate of success of 99%; closer to the theoretical 99.22%) the loss would be much greater:

The result expected after 200 iterations of this method would be the initial bankroll (b0) multiplied by the common ratio (fm) raised to the power of the total iterations (i):

That is, a loss halfway between failing 1 or 2 iterations; in the same way that the percentage of theoretical success (99.22%) is halfway between winning 199 (99.5%) or 198 (99%) out of 200.

Annex C9: Kelly Criterion

We have already checked that the Martingale method, under the conditions described at the beginning, does not result in a method with a positive expected value. However, the fraction of the bankroll (x) to wager has been determined randomly and, therefore, we could wonder if there is any other value that makes the common ratio of the method (fm) greater than one.

To apply the Kelly Criterion to the Martingale of our example, we need to define the equation that describes its common ratio (fm) as a function of the bankroll’s fraction (x), as done with other geometric progression methods:

Since finding the derivative of the previous equation as a function of (x) and solving it is impossible, we will graph the values taken by the common ratio (fm) for different values of (x). When calculating the common ratio, the rounding of the maximum number of attempts to the immediately lower integer has been taken into account.

Graph 20. Common ratio (fm) of a Martingale as a function of the bankroll’s fraction (x)

In graph 20, we see that no value of the common ratio (fm) is greater than 1. Therefore, there are no values of (x) for which the Martingale method has a positive expected value. That is, in the long-term, regardless of the fraction of our bankroll chosen to wager, we will always lose money.

It will get worse if we try to apply this method to any sports bet or casino game, since, as seen in chapter 2, these do not offer fair games to the users. This means that, either an event with 50% success rate is not paid at odds of 2.00 (like a bet on the final odd or even points of a basketball match) or that something offered at odds of 2.00 has a real probability of happening below 50% (like betting on colours in a roulette, where the probability is less than 50% due to the 0 and the double 0 boxes).

That is, if the Martingale method does not work in the fair game studied here, even less if applied to any unfair real game.

Example 49

We want to apply the Martingale method betting on whether the total number of points in successive basketball matches will be even or odd. Both results are offered at odds of 1.90 (c), but we know that their real probabilities (pr) are not the inverse of their odds, but 50%. This can be calculated using equations 3.2 and 3.1.

On the other hand, the factor (f) of our Martingale so that the net benefit obtained would be the same regardless of the attempt (n) on which we succeed is:

With both results we can already calculate the common ratio (fm) of our method:

For the factor (f) and the real probabilities (pr) calculated above, graph 21 shows the common ratio (fm) as a function of (x):

Graph 21. Common ratio (fm) of a Martingale method as a function of the bankroll’s fraction (x)

We can see how the common ratio (fm) in graph 21 (red curve) is less than 1 for the entire domain of (x); thus, the method will never give us long-term benefits. In turn, since it is a less fair game than the one described in the beginning of Annex C, the resulting common ratio is even worse than the one obtained before (blue curve).