Akis Androulakis

After dedicating two articles to the underlying theory of limits and to relatively abstract examples, it’s time to shift our focus to some real-life applications. Specifically, we’ll see the maths law all casinos and betting companies exploit in order to make profit. 

As a first step, we need a mathematical model of a possible game or bet which gives probabilities for every outcome. For example, consider the following game:

You roll a fair six-sided die. If the outcome is any number from 1 to 5 then you earn a pound, otherwise you lose 6. Essentially there is a 5/6 chance you win the pound (as 5 out of 6 rolls win and the die is fair) and thus only a 1/6 chance of losing the 6. If we denote by X your earnings after one game, then the above information is summarised below. 

Value of X	1	−6
Probability	5/6	1/6

Such variables like X, whose possible values are assigned respective probabilities are called random variables. A measure of whether you should play the game is the “expected value” of X – denoted as E(X) – which is the “mean” earnings per game played. This is calculated by the “weighted average” of all the possible values of X, where the coefficient of each value is the respective probability. This makes intuitive sense as the values of X more likely to occur should play a greater role in determining the average.  In our case we have:

E(X) = 5/6 * 1 + 1/6 * (-6) = -1/6

Which we can interpret to mean that on average you lose 1/6 pounds per game you play. To better understand this, assume you play 6 times. Given the probability of winning, you are “expected” to win 5 games and lose 1, which would overall result in you losing a pound . Therefore on average, you will have lost 1/6 pounds per game.

Let’s consider a second example. Imagine you toss a coin and if it lands tails then you win 10 pounds but if it lands heads you lose 8. Since each outcome has a probability of 1/2 then:

E(X) = 1/2 * 10 + 1/2 * (-8) = 5 – 4 = 1

Meaning that you are “expected” to win one pound on average per game. Although the method of calculating E(X) is reasonable, the concept of the expected value may seem a bit theoretical as in our example(s) you can’t physically lose 1/6th of a pound or win exactly one, by playing each game once. The main reason for this is that by playing once or even just a few times, your “win rate” may be very different from what probability suggests.

Specifically in the example with the die, if you play twice, lose one and win one, then you will have lost 5 pounds instead of 2/6, which is expected because you will have won only 50% of the games whereas the probability suggests that you should be winning approximately (5/6) * 100% = 83.33% of them. We can instead think of it as the more times you play the game, then the closer the ratio of wins to losses will become to the one predicted by the probability. For instance, by only playing twice, your win percentage may well be 50%, whereas if you play 100 times, it is much more likely to be closer to the “theoretical” 83,333…%.

Equivalently, the average earnings after two games won’t be close to the expected value of -1/6, but after 100 games, we would expect it to be. This idea is the core of a fundamental theorem in probability, namely the law of large numbers. As we did with limits, we will explore the law mathematically, beyond intuition.

In mathematical language the average earnings (total earnings divided by number of games) of n games is:

A = (X₁ + X₂ + X₃ + … + X_n) / n

Where X₁ denotes the earnings of game 1, X₂ denotes the earnings of game 2 and so on. It is important to note that X₁, X₂… are independent as winning or losing game 1 doesn’t affect your chances of winning game 2 etc. We also say that X₁, X₂, … are “identical” in the sense that you have the exact same chance of winning game 1 as you do for game 2 etc. In such scenarios, where we have independent and identical random variables, the law of large numbers states that the actual average A converges to the expected value E(X), as the number of the variables n tends to infinity. In mathematical language:

For every positive quantity d, the probability that  “|A – E(X)| < d”, converges to 1 as n increases.

Essentially, no matter how small d > 0 we choose; as n increases (in our example that means as we play more games), then it becomes almost certain (i.e. with probability that converges to 1) that the difference between the actual and theoretical average becomes less than the very small value d. 

Applying this to gambling, it means that the more times you play the same game, your average earnings per game will get closer to their expected value. Thus, all casinos and betting companies have to do is create games/bets such that a player’s average expected earnings *as defined by probability* are negative (like the game with the die). In the short term a player may actually win money but the probability of making profit in total, gets smaller the more you play. 

Taking things further, let Y = n * A = X₁ + X₂ + X₃ + … + X_n, which gives your total earnings after playing a game n times. Then your “expected” total earnings, as calculated by probability, are n * E(X), (number of games expected earnings per game). You are considered “lucky” if your total earnings Y are more than the expected ones n * E(X), that is if their difference [Y – n * E(X)] is positive. However, in order to make profit you only need Y > 0. If we start from this condition for profit and subtract n * E(X) from both sides, we have:

Y – n * E(X) > – n * E(X),

And dividing by n:

Y / n – E(X) > − E(X)   ⇔  A – E(X) > −E(X)  (since Y = n * A as defined above)

Now, in all gambling games your expected average earnings E(X) will be negative as casinos and betting companies make the rules. Hence, the right-hand side will always be positive. On the other hand, the left-hand side converges to 0 by the law of large numbers. Clearly, as the left-hand side gets closer to 0 the more games you play, it gets less likely that the inequality is satisfied and that you make profit. Moreover, as A converges to E(X), then

A / E(X) converges to 1 (E(X) < 0, so division is valid), then equivalently

n * A / [n * E(X)] converges to 1, so substituting in the numerator, Y = n * A we obtain:

Y / [n * E(X)] converges to 1.

Thus, even if you are lucky and your earnings are more than the expected amount, ie. Y > n * E(X) which means you have earned Y – n * E(X) more than expected, that amount in the long term will get insignificant* compared to the amount that you will have lost. 

*See again the conclusion from the answer to question 3 of article 2. 

Now that we’ve looked at the law of large numbers, let’s turn our attention to another problem – that of the Gambler’s Fallacy. The Gambler’s Fallacy is the incorrect belief that if a particular event occurs more frequently than normal during the past, it is less likely to happen in the future (or vice versa). For example, in the game with the die, playing twice and losing both times (something highly unlikely) doesn’t mean that the chance of winning the next game will be any higher in order to “balance the results”. As we said from the beginning, the outcome of each game is independent of the outcomes of the other ones. The law doesn’t state that such “anomalies”, like losing four times in a row, will immediately cancel out with others like winning 3 in a row. Instead, as these anomalies are rare, they will have little effect in the long run after many games.

To conclude, it is worth mentioning that probability and the law of large numbers are used very frequently in all sorts of practical scenarios: in statistics for example when deriving conclusions from polls or experimental data. Likewise, limits have numerous applications – alongside the law of large numbers – across (pure and applied) mathematics, physics, economics, engineering… the fascinating world of mathematics is infinite and is waiting for you to explore it further!

As a final remark I would like to pose the following philosophical question about the interpretation of probability:

Think about what does it truly mean to say the probability of rolling a 4 on a fair die is ⅙ ? 

A popular interpretation is that the probability of a random event denotes the relative frequency of occurrence of an experiment’s outcome when the experiment is repeated indefinitely. This essentially means that if kₙ is the number of 4’s we have after n dice rolls: 

Then kₙ / n converges to ⅙.

If Y is a random variable taking the value 1 if we roll a 4 and taking the value 0 otherwise, is there any connection to what we obtain by the law of large numbers applied on Y and the interpretation of probability?

Have fun – and keep exploring mathematics!