KindiakMath

The Gambler’s Ruin

Using our basic ideas of probability, let’s discuss a simple yet deep problem in probability theory, which laid the foundational ideas for quantitative finance; the gambler’s ruin.

Disclaimer. This page does not promote or encourage online gambling in any form. The information presented here is for educational purposes only.

The gambler’s ruin can be formulated simply. Let L < M < W be pre-determined integers and X_t denote your wealth, in dollars, at time t:

  • You start with X_0 = M.
  • On each turn t, you toss a fair coin.
    • If the coin lands ‘Head’, you win \$1, so that X_t = X_{t-1} + 1.
    • If the coin lands ‘Tail’, you lose \$1, so that X_t = X_{t-1} - 1.
  • The game ends at the first time \tau when X_{\tau} \in \{L, W\}.

Suppose the simplest case L = 0, M = 1, and W = 2.

Example 1. What is the probability that X_1 = 2? What about the result X_1 = 0?

Solution. By definition, X_0 = M = 1. Since the coin is fair,

\begin{aligned} \mathbb P(X_1 = 2) &= \mathbb P(\text{Head}) = 1/2, \\ \mathbb P(X_1 = 0) &= \mathbb P(\text{Tail}) = 1/2. \end{aligned}

In Example 1, getting a ‘Head’ yields X_1 = 2 = W, and getting a ‘Tail’ yields X_0 = 0 = L. No matter the outcome, we have X_1 \in \{L, W\}. Furthermore, X_0 = M \notin \{L, W\}. Therefore, rather trivially, the game ends at time \tau = 1.

Remark 1. The fun really begins when we vary our setup. In fancy quantitative financial language, we call \tau a stopping time for the time series process. The quantity L models our stop loss, M models our capital, and W models our take profit. The coin being fair is a simplistic starting point for discussion—in the real world the price movements are far less predictable than we would hope for.

Now let’s set L = 0, M = 2, and W = 4.

Example 2. What is the probability that X_2 = 4? How about X_2 = 2?

Solution. Since we are dealing with multiple coin tosses, it gets exponentially difficult to intuit the solution. Since this process involves consecutive time steps, we can visualise its process using a probability tree diagram. As its name suggests, it is a diagram that resembles a tree that is described using probabilities.

The number 1/2 on the lines denote the probability that the wealth increases or decreases by $1 respectively. There are four possible paths that the wealth can take, and they all occur with equal probability. Denote the sample space by the wealth evolutions:

\Omega = \{ ( 2,3,4 ), ( 2,3,2 ), ( 2,1,2 ), ( 2,1,0 ) \}

We note that X_2 = 4 corresponds to wealth evolutions of the form (*, *, 4). Since the only outcome when this result takes place is (2,3,4), we write

\{X_2 = 4\} = \{ (2, 3, 4) \}

Therefore,

\displaystyle \mathbb P(X_2 = 4) = \frac{|\{ (2,3,4) \}|}{|\Omega|} = \frac 14.

Similarly, X_2 = 2 corresponds to wealth evolutions of the form (*, *, 2). Since there now are two such wealth evolutions, so that

\{X_2 = 2\} = \{ (2,3,2), (2,1,2) \},

the required probability is

\displaystyle \mathbb P(X_2 = 2) = \frac{|\{ (2,3,2), (2,1,2) \}|}{|\Omega|} = \frac 24 = \frac 12.

Now suppose L = 0, M = 2, W = 5.

Example 3. Evaluate \mathbb P(X_3 = 3).

Solution. We extend our probability tree diagram as follows.

This time, the “total number of possibilities” argument fails. Why? Because the game ended for wealth trajectory (2,1,0).

Nevertheless, we can still solve the problem. Each coin toss doesn’t impact the next one, so that the sample space

\{ (\mathrm H, \mathrm H), (\mathrm H, \mathrm T),(\mathrm T, \mathrm H), (\mathrm T, \mathrm T) \}

contains outcomes that all share the same probability. For instance,

\displaystyle \mathbb P( (\mathrm H, \mathrm T) ) = \frac 14 = \frac 12 \cdot \frac 12 = \mathbb P(\mathrm H) \cdot \mathbb P(\mathrm T).

In fancier language, we say that the coin tosses are independent of each other. This logic holds no matter the coin toss; letting \xi_1,\xi_2,\xi_3 \in \{\mathrm H, \mathrm T\} denote individual coin tosses, we have

\mathbb P( (\xi_1,\xi_2,\xi_3) ) = \mathbb P(\xi_1) \cdot \mathbb P(\xi_2) \cdot \mathbb P(\xi_3).

In particular, each path with length 3 will automatically have a probability of 1/8 of occurring. Notice that there are three paths that get us to X_3 = 3:

\{X_3 = 3\} = \{ (2,3,4,3), (2,3,2,3), (2,1,2,3) \}.

Therefore, \mathbb P(X_3 = 3) = 3 \times 1/8 = 3/8.

Remark 2. When presenting our work, you do not need to be so long-winded as per this writeup. As long as you communicate your thought process through your calculation, you can obtain full credit.

Example 4. How would the answer in Example 3 change if we are working with a biased coin with probability p? That is, \mathbb P(\mathrm H) = p and \mathbb P(\mathrm H) = 1-p. Give your answer in terms of p.

Solution. We modify our probability tree diagram as follows.

The visual is basically the same, and we would follow the same trajectories

\{X_3 = 3\} = \{ (2,3,4,3), (2,3,2,3), (2,1,2,3) \}.

However, each step has as slightly different probability to compute:

\begin{aligned} \mathbb P((2,3,4,3)) &= p \cdot p \cdot (1-p) = p^2(1-p), \\ \mathbb P((2,3,2,3)) &= p \cdot (1-p) \cdot p = p^2(1-p), \\ \mathbb P((2,1,2,3)) &= (1-p) \cdot p \cdot p = p^2(1-p). \end{aligned}

Since all paths are distinct, and turn out to have the same probabilities, we can sum the probabilities up as follows:

\begin{aligned} \mathbb P(X_3 = 3) &= \mathbb P(\{ (2,3,4,3), (2,3,2,3), (2,1,2,3) \}) \\ &= \mathbb P( (2,3,4,3) ) + \mathbb P( (2,3,2,3) ) + \mathbb P( (2,1,2,3) ) \\ &= p^2(1-p) + p^2(1-p) + p^2(1-p) \\ &= 3p^2(1-p). \end{aligned}

Remark 3. This process can be generalised to what is known in probability and statistics as the binomial distribution. The multiplication procedure is known as the multiplication principle, used to calculate probabilities of in-sequence events. The addition procedure is known as the addition principle, used to calculate probabilities of disjoint events.

Example 5. Toss a biased coin with probability p three times. What is the probability that you get two ‘Heads’? What do you notice?

Solution. Draw the probability tree diagram as follows.

By following the chosen trajectories, the required probability is

\begin{aligned} \mathbb P(\#(\mathrm H) = 2) &= p \cdot p \cdot (1-p) + p \cdot (1-p) \cdot p + (1-p) \cdot p \cdot p \\ &= p^2 (1-p) + p^2 (1-p) + p^2 (1-p) \\ &= 3p^2 (1-p). \end{aligned}

The answer in Example 5 matches the answer in Example 4. This observation should not be a surprise—the wealth trajectories in Example 4 are directly determined by the coin tosses in Example 5 and vice versa.

Remark 3. These example suggest that studying probability is less inherently about the underlying random process, but more so about the distributions of the possible outcomes.

Example 6. You have 4 black socks and 6 white socks in your drawer. You choose two socks at random, without replacement (obviously). What is the probability that you get two socks of the same color?

Proof. We draw the following probability tree diagram.

Notice that if the first sock chosen is black, then the probability of the second sock chosen would change (since there is one less black sock present). Following the two routes of matching-coloured socks, the required probability is given by

\displaystyle \mathbb P(\text{matching colors}) = \frac 4{10} \cdot \frac 39 + \frac 6{10} \cdot \frac 59 = \frac{7}{15}.

Of course, the coin toss is one of the simpler examples to begin our discussions on probability theory. Another common toy that we use to discuss probabilities would be that of dice. A fair six-sided die has 6 faces: 1, 2, 3, 4, 5, 6, each occurring with equal probability 1/6.

Example 7. Roll two fair die simultaneously. Assume that the outcomes of the dice are independent. What is the probability that the numbers sum to 8? Which integer n is the most probability sum?

Solution. Let X_1 denote the outcome of the first die and X_2 denote the outcome of the second die. We want to evaluate \mathbb P(X_1 + X_2 = 8). We can illustrate the outcome of sums using the possibility diagam below.

For example, if X_1 = 3 and X_2 = 5, then X_1 + X_2 = 8. Since the dice are independent, each cell has a probability of 1/6 \times 1/6 = 1/36, which also agrees intuitively with the possibility diagram. Since there are 5 cells whose sum yields 8, the required probability is

\mathbb P(X_1 + X_2 = 8) = 5 \times 1/36 = 5/36.

Among all possible sums, 7 has the largest number of cells, namely 6. Therefore, the required integer is n = 7, and in fact,

\mathbb P(X_1 + X_2 = 7) = 6 \times 1/36 = 1/6.

Remark 4. If the dice are unfair, we can still manually compute \mathbb P(X_1 + X_2 = n) by adding up the separate cases

\mathbb P(X_1 + X_2 = n) = \mathbb P(X_1 = k) \cdot \mathbb P(X_2 = n-k)

one after another for k = 1,\dots, 6. Of course if n-k < 1 or n-k > 6, then \mathbb P(X_2 = n-k) = 0. This process is known as taking the discrete convolution between two probability mass functions.

If there is one topic that I insist on discussing applications, it would most certainly be probability. I do think it is a good idea to illustrate probability in the real world. To do that, I’ll need to discuss matrices, which also happens to be our final topic in O-Level mathematics. Of course, we can extend these ideas at great length into the study of stochastic processes and Markov chains, but let’s just touch base with some simple examples to augment our understanding.

—Joel Kindiak, 18 Mar 26, 1539H

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