Probability Theory in Quantitative Finance Interviews: Part 1

Different from traditional interviews for finance jobs, quant companies typically test more advanced mathematics topics during their interview processes. Chances are that you will face at least a couple of probability problems in most quantitative interviews. Probability theory is the foundation of every aspect of quantitative finance. As a result, it has become a popular topic in quantitative interviews.

Although good intuition and logic can help you solve many of the probability problems, having a thorough understanding of basic probability theory will provide you with clear and concise solutions to most of the problems you are likely to encounter. Furthermore, probability theory is extremely valuable in explaining some of the seemingly- counterintuitive results. Armed with a little knowledge, you will find that many of the interview problems are no more than disguised textbook problems (Zhou, 2008).

To being with some basic probability notations that will be useful in solving the question below:

Outcome (𝔀): the outcome of an experiment or trial.

P(𝔀): Probability of an outcome. (∑P(𝔀) = 1)

P(A): Probability of an event A

Now let’s look at an introductory level probability question.

Two gamblers are playing a coin toss game. Gambler A has (n +1) fair coins; B has n

fair coins. What is the probability that A will have more heads than B if both flip all their coins?

Many might have the instinct that the answer to this question is 0.5, which is true. However, let’s look at the detailed probability approach for a more rigorous solution.

First, notice that the difference between gambler A and B is that A has (n+1) coins while B has n coins. Let’s assume that we remove the one extra coin from A, now A and B have equal amounts of coins. There are now three possible outcomes:

1. E1: A’s n coins have more heads than B’s n coins;

2. E2: A’s n coins have equal number of heads as B’’s n coins;

3. E3: A’s n coins have fewer heads than B’s n coins.

By symmetry, the probability that A has more heads is equal to the probability that B has more heads. So we can write P(E1) = P(E3). For easier presentation, let’s set P(E1) = P(E3) = x, and

P(E2) = y. We know that ∑P(𝔀) = 1(the probability of all events is 1), so we have 2x+y = 1.

Now, let’s discuss the three different scenarios:

For event E1, A will always have more heads than B no matter what A’s (n+ 1)th coin’s side is; For event E3, A will have no more heads than B no matter what A’s (n +1)th coin’s side is.

E2 is the special case here. If A’s (n+ 1)th coin is a head, which has a 0.5 probability, it will make A have more heads than B. So the (n + 1)th coin increases the probability that A has more heads than B by 0.5y and the total probability that A has more heads is x+0.5y. We know that 2x+y = 1, so we can substitute y with (1–2x). So x+0.5y = x+0.5(1–2x) = 0.5 when A has (n +1) coins. Therefore the answer is 0.5.

Thanks for reading!

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