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Example Probability & Statistics Questions

Coin, dice, and card probability questions involving expected value, Bayes, Markov chains, and more — all at quant-trading-firm difficulty.

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Probability and statistics are central to most quant-trading screens, and the same themes come up across firms: Bayes' theorem, expected value, variance of sums, coupon collectors, random walks, card and dice problems, and more.

The questions below are real examples from our full bank, each with a complete solution. Give a problem a proper attempt before you expand it — and if you catch yourself peeking early, add it to your study plan and come back to it cold.

Sample Questions & Solutions

Each question is a real interview problem. Try it yourself first, the full solution is revealed below.

Q1

Bankrupt

Easy
Two players, A and B, play a game in which the winner receives 1 dollar from the other player. Player A has 1 dollar and player B has 2 dollars. Player A is better in this game and wins 2/3 of the games. They play until one of them is bankrupt.

What is the probability that player A wins?
Show solution
This is a pretty straightforward Markov chain problem. There are 4 states. The transition graph is given in Figure 1.

Figure 1 - Transition graph for this problem.


The problem starts at state 1. As has been explained in the lessons of this course, we use the following equation:
\begin{equation}
s_1 = \sum_{i=0}^{3}p_{1,i}s_i
\end{equation} \begin{equation}
s_2 = \sum_{i=0}^{3}p_{2,i}s_i
\end{equation} Furthermore, $s_0=0$ and $s_3=1$. Then we have
\begin{equation}
s_1 = \frac{1}{3}*0 + \frac{2}{3} * s_2
\end{equation} \begin{equation}
s_2 = \frac{1}{3}* s_1 + \frac{2}{3} * 1
\end{equation} Solving these equations gives us $s_1=4/7$ and $s_2=6/7$. So, starting with 1 dollar, player A has a 4/7 chance of winning.

Proof
If we substitute Equation 4 in Equation 3, we have

\begin{equation}
s_1 = \frac{1}{3}*0 + \frac{2}{3} * (\frac{1}{3}* s_1 + \frac{2}{3} * 1)
\end{equation} \begin{equation}
s_1 = \frac{2}{9} * s_1 + \frac{4}{9}
\end{equation}  \begin{equation}
\frac{7}{9} * s_1 =  \frac{4}{9}
\end{equation}  \begin{equation}
s_1 =  \frac{4}{9} / \frac{7}{9} = \frac{4}{7}
\end{equation}
Asked at: IMC Da Vinci All Options Akuna Capital SIG Jane Street Maven Citadel Securities Virtu DRW
Category: Markov Chain Probability View full question page →
Q2

All Faces

Easy
What’s the expected number of throws to see all of the faces of a die?
Show solution
This is called the coupon collector problem. For a fair n-sided die, the expected number of attempts needed to get all n values is \begin{equation}
n \Sigma_{k=1}^n \frac{1}{k}
\end{equation} which, for large n is approximately n log n.

The time until the first result appears is 1. After that, the random time until a second (different) result appears is geometrically distributed with parameter of success 5/6, hence with mean 6/5 (recall that the mean of a geometrically distributed random variable is the inverse of its parameter). After that, the random time until a third (different) result appears is geometrically distributed with parameter of success 4/6, hence with mean 6/4. And so on, until the random time of appearance of the last and sixth result, which is geometrically distributed with parameter of success 1/6, hence with mean 6/1. This shows that the mean total time to get all six results is \begin{equation}
6 \Sigma_{k=1}^6 \frac{1}{k} = \frac{147}{10} = 14.7
\end{equation}
Asked at: IMC Da Vinci All Options Akuna Capital SIG Jane Street Maven Citadel Securities Virtu DRW
Category: General View full question page →
Q3

Multiply 3 Dice

Easy
You roll 3 dice and you multiply all face values. What is the probability that the outcome is even?
Show solution
For the outcome to be even is the same as the outcome not being odd. For the outcome to be odd, all three dice need to be odd, respectively:
  • 1*1*1 = 1 (odd)
  • 3*3*3 = 27 (odd)
  • 5*5*5 = 125 (odd)
and also all combinations of the odd numbers result in odd outcomes, like:
  • 1*1*3 = 3 (odd)
  • 3*3*5 = 45 (odd)
  • etc.
Basically, multiplying odd numbers with each other results in an odd outcome. If ANY even number comes in play, then we have "a number" multiplying "an even number", which will always result in an even number. So, "odd * even * odd" will also result in even.

So, in any other scenario than multiplying three odd numbers, the product will be even. Therefore, the probability that the product is even is, given that half of the sides of the dice are odd ({1,3,5} vs {2,4,6}) equal to \begin{equation} P(even \; product) = 1 - P(only \; odd \; product) \end{equation} \begin{equation} P(even \; product) = 1 - (\frac{1}{2})^3 = \frac{7}{8} \end{equation}
Asked at: IMC All Options SIG Jane Street Maven Citadel Securities Virtu DRW
Category: General View full question page →

Why practise these

Trading rewards a particular way of thinking — weighing probabilities, reasoning about expected value, updating as new information comes in. These puzzles aren't the job itself, but they run on that same logic, and practising it makes the reasoning feel natural rather than something you're working out for the first time in the interview.

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