Hey, I recently had to prepare for a probability and problem solving assessment for a global investment and trading firm. In order to do that, I reviewed and collected different relevant formulas and useful tips. I thought I’d share them here in case they’re useful for someone else.

Note that in the different following sections there are standardized formulas that solve each problem. Even though I do provide them, my goal is is to take one step further and explain the logic behind the formulas. With this, instead of pure memorization I encourage you to actually understand the approach (since from personal experience it is then much easier to remember stuff). Hope you enjoy the read :).

Understanding over mere memorization

Tactical Game Theory

Before the math starts, you need a strategy. Most trading assessments (like Maven, Jane Street, or Optiver) use negative marking (e.g. +1 for correct, -1 for incorrect), which turns the test into a risk management exercise.

The Confidence Threshold

When negative marking is present, calculate beforehand how confident you need to feel in order to provide an answer.

For example, in a 4-option Multiple Choice Question with -1/+1, you should only answer if you can eliminate at least two of the options. Otherwise the Expected Value of the grade of your answer is negative, and it’s better for you to skip the question.

The Three-Pass Rule

Going through the exercises in the given order might lead in losing easy points that are in the last few exercises or in general distributing your limited time in an inefficient way. Instead use the three-pass rule:

1. First Pass (Easy): Instant solves. Get points on the board.
2. Second Pass (Medium): You know the method, but it needs scratchpad work.
3. Third Pass (Hard): High-complexity problems. Only touch these if you have time and high confidence.

Combinatorics

Here we go over the formulas you can use to calculate the number of ways you can pick k elements out of n. I quite proud of the explanations I provide here :D, so hopefully they do help you in understanding the logic behind the formulas :).

Permutations (order matters)

• In the upper part $(n!)$ the term represents the permutations of all elements.
• With the term $(n-k)!$ in the divisor, we remove the internal ordering of the $(n-k)$ elements we didn’t select, as their ordering does not affect our result.

Combinations (order does not matter)

• Here we keep the permutations of all $(n)$ elements in the upper part (as before)
• And in the divisor:
  • apart from removing the permutations of the $(n-k)$ not selected elements (as before)
  • we also remove the permutations of the $k$ selected elements (since their ordering does not affect our solution)

Arrangement with repetitions

• Here we refer to the number of different arrangements of $n$ different elements, where some of the elements are repeated.
• Divisor includes the number of repetitions $(n_{i}!)$ the different elements have in the given sequence.
• Take for example the word “SUCCESS“:
  • n = 7
  • |’S’| = 3, |’C’| = 2 (and |’U’| = |’E’| = 1, so the presence of these terms in the divisor isn’t necessary)
  • Solution: $\frac{7!}{3!2!}=\frac{5040}{12}=420$

Expected Value

Average outcome if you play the game infinitely many times.

where:

• $x_i$ is the value of the outcome and $P(x_i)$ its probability
• $EV > 0:$good, beneficial game to play
• $EV = 0:$fair game
• $EV < 0:$bad game to play (i.e. the casino way :D)

Conditional Probability

What is the probability of “A” happening, given that “B” has happened.

One way:

• In the context where “B” is happening, how often does “A” also happen?

Bayes’ Theorem:

Base 100/1000 way (The Bayes’ shortcut):

Avoid the complex Bayes’ formula and the many fractions that usually come with it, and instead imagine a population of 100 or 1000 people. I will go with 1000 here.

Example question: A disease affects 1%. A test is 90% accurate (both ways). You test positive, what is the probability you are actually sick?

• 1000 people:
  • 10 are sick (1%)
  • 990 are healthy (99%)
• Out of the 10 sick:
  • 9 test positive (90% accuracy)
  • 1 tests negative
• Out of the 990 healthy:
  • 891 test negative (90%)
  • 99 test positive

So from the 1000 people we have (getting a bit repetitive here, for clarity’s sake):

• 9 sick and positive
• 1 sick and negative
• 99 healthy and positive
• 891 healthy and negative

Now we have the data and focus on what the question asks:

• It says positive test… therefore we focus on the
  • 9 sick and positive and
  • 99 healthy and positive
• What is the probability I am actually sick?
  • Well out of the 108 people that tested positive, 9 are actually sick which means…
  • $\frac{9}{9+99} = \frac{9}{108}=0.083$
  • and there is your answer
• With this we avoid dealing with too many fractions, simplify things a bit and avoid making unnecessary mistakes.

Geometric & Binomial Distribution

Here $p$ represents the probability of success in a random trial.

Binomial (exactly k successes in n trials):

where:

• $\binom{n}{k}$ represents the number of distinct ways you can pick k success out of n trials
• and the latter part captures the probability of getting exactly k successes

Geometric (first success on the n_th trial):

• simply $n-1$ failures, and a success afterwards (no other way around it)

Expected number of trials until success:

• e.g. expected number of throws of a dice until you get a six: $\frac{1}{\frac{1}{6}}= 6$
• Tip: Asked for the probability of having at least one success? It’s much faster to calculate $1-P(zero\_successes)$
• A fun, more complex example: Q: How many throws do you have to do to get every number of the dice shown (at least) once?

The Stars and Bars method

The question is: In how many different ways can we distribute $n$ identical items into $k$ distinct bins?

       Note that what we are discussing here is for when we deal with identical items

The Stars and Bars method is essentially a useful, and brilliant in my eyes :D, reformulation of this problem. Instead of trying to work out complex combinations or picking combinations, we reformulate the problem as follows:

• We have n items and k distinct bins
• In order to separate these n items into the k bins, it’s enough to place k-1 “group” separators between the n items.
• Then, we can say we have $n + (k-1)$ (in total) items to be placed in any order, out of which the n and the k-1 are identical/repeated.
• And that’s the arrangements with repetitions formula
• So we can do $\frac{(n+(k-1))!}{n!(k-1)!}$ and find our solution

Note that there might additional requirements to this, like for example: “while making sure every bin gets at least one item”:

• There is more than one way to approach this but the most straightforward for me is to simply take care of the requirements first, and then proceed with the logic above
• i.e. pre-allocate $k$ (out of $n$) items, one for each bin, therefore making sure the requirement is satisfied
  • note we are able to do that without any permutation/choice concerns here, because the items are identical
• and then proceeding with allocating the $n-k$ items to the $k$ bins
• i.e.: $n-k$ items and $k-1$ separators => $\frac{((n-k)+(k-1))!}{(n-k)!(k-1)!}$

Invitation for another perspective

How can you use the above to calculate the number of different value combinations that form

• $x_1 + x_2 + x_3 + x_4 = 12$ , where $x_i \ge 0$
• $x_1 + x_2 + x_3 = 10$, where $x_i \ge 1$
• $x_1 + x_2 + x_3 \le 15$ where $x_1 \ge 2, x_2 \ge 1, x_3 \ge 0$

Have a go, and let me know in the comments what you think.

Extra tips:

Sum of 2 dice pyramid

For two dice, the probability of sum $X$ increases linearly up to 7, then decreases. Number of ways to get $X$ formula:

• $X \le 7 \rightarrow X -1$
• $X \ge 8 \Rightarrow 13 – X$

Heads-up: you don’t have to memorize the formulas. Instead, notice that this forms an 1-point-per-step pyramid starting from 1 way (for 2 and 12) and scaling up to 6 (for 7). So if you remember the two ends (e.g. for me it’s easy to remember that for 7 there are 6 ways) you call quickly find the number you are looking for within a few steps.

Square difference

Use this to quickly calculate the multiplication of numbers equidistant (fancy word isn’t it) from a base number.

Example: $47 \times 53$

• You notice that this is equal to $(50 – 3)(50 + 3)$
• Which leads to $50^2 – 3^2= 2500 – 9 = 2491$ which is a much easier calculation

Main challenge in real tests: Reformulation

• knowing the formulas is the necessary and easier part
• in a real test the real challenge will be to find out how you can apply the things you know to the given problem.
• My advice: master the basics. Remember everything that you know. And when a challenging exercise comes think: Okay, I know combinatorics, I know how to distribute identical items into bins, I know binomial, geometric distributions (etc. etc.)… Now… how can I turn this problem into a one of these things? How can I rephrase the problem in a way that the question asked is something I can answer with what I know?
• One extra advice: Of course practice and learn the formulas, but also have fun while doing it. I had quite some fun revisiting these concepts after a while, and solving fun, short problems again. So maybe you can do the same, you know, if you want.

Closing remarks

All in all, I hope this is useful to somebody, and that this was an enjoyable read :).

Thank you very much for stopping by and best of luck with your endeavors.

Let me know which concept or explanation you needed or liked the most, or if you want an additional clarification on anything. I would be happy to answer.

As a last “gift”, I have added a practice quiz for you. So if you wanna put what you learned to the test, have a go! 🙂