Probability Interview Questions: Solved Riddles Step by Step
These riddles test reasoning about conditional probabilities, expected value and misleading intuitions. Here are 4 classic riddles with their solutions, including the famous Monty Hall problem.
Monty Hall Problem: Should You Switch Doors?
Probability question from finance job interviews
Three doors are closed. Behind one of them is a car, and behind the other two, a goat. You choose a door without opening it. Then the host, who knows where the car is, opens one of the two remaining doors, revealing a goat. He now offers you the chance to switch doors: is it in your interest to switch?
Need a hint?
Compare the two cases: at the start you have a one-in-three chance of picking the car, and the host always acts the same way.
Show the answer and the solution
Answer: Yes: switching gives a 2 in 3 chance of winning the car, versus a 1 in 3 chance if you keep your original door.
The reasoning
At the start, your door has a one-in-three chance of hiding the car, so the other two doors together hide it with two-thirds probability. The host, who knows where the car is, always opens a door with a goat among the two you did not choose. If you chose the car at the start, which has a one-in-three chance, switching makes you lose. In the other two cases, the car is behind the other remaining door, and the host is forced to reveal the only remaining goat: switching then makes you win. Switching therefore gives two chances in three, against one chance in three for keeping your door.
Birthday Paradox: How Many People Are Needed for 50%?
Classic probability problem
How many people must be in a room for the probability that two of them share the same birthday to exceed 50%? Ignore people born on February 29, and assume that the 365 days of the year are equally likely.
Need a hint?
It is easier to calculate the probability that no pair of people shares a birthday, then take the complement.
Show the answer and the solution
Answer: 23 people, with a probability of about 50.7% that at least two of them share the same birthday.
The reasoning
It is easier to calculate the complementary probability, that no pair of people has the same birthday. The first person can be born on any day. The second must avoid that day, which gives 364 choices out of 365; the third must avoid two days, which gives 363 choices out of 365, and so on. For 23 people, this probability equals 365/365 × 364/365 × ... × 343/365, which is about 0.493. The probability that at least two people share a birthday is therefore 1 minus 0.493, or about 0.507, that is about 50.7%. The 23 people form 253 possible pairs, which explains why the probability rises quickly.
Dice Expected Value with One Reroll: The Optimal Strategy
Trading floor interview style (expected value)
You roll a fair six-sided die and win, in euros, the value of the face you get. You may roll the die once more, if you wish; the second result is then final. What is the expected winnings with the best strategy?
Need a hint?
First calculate the average of one roll, then compare it with the faces you would keep after a reroll.
Show the answer and the solution
Answer: €4.25: reroll if the first result is 1, 2 or 3, and keep the first result if it is 4, 5 or 6.
The reasoning
Without a reroll, the expected value is the average of the six faces, that is (1 + 2 + 3 + 4 + 5 + 6) / 6 = €3.50. With one reroll, the best strategy is to reroll if the first result is 1, 2 or 3, since €3.50 is higher than each of those faces. If the first result is 4, 5 or 6, keep it. When you keep it, the gain averages €5 with probability 1/2; when you reroll, it averages €3.50 with probability 1/2. The expected value is therefore 1/2 × 5 + 1/2 × 3.50 = 2.50 + 1.75 = €4.25.
Two Children Paradox: What Is the Chance of Two Boys?
Classic probability paradox
A family has two children. We know that at least one of the two is a boy. We assume that the birth of a boy and that of a girl are equally likely and independent. What is the probability that both children are boys?
Need a hint?
List all the possible two-child families, taking the birth order into account, then eliminate those with no boy.
Show the answer and the solution
Answer: 1/3: of the three possible families (boy-boy, boy-girl, girl-boy), only one has two boys.
The reasoning
There are four possible birth orders: boy-boy, boy-girl, girl-boy and girl-girl. The statement “at least one of the two is a boy” rules out girl-girl, so three equally likely cases remain. Only the first, boy-boy, has two boys. The probability is therefore 1/3 and not 1/2, because the information given does not specify which child is the boy. If we knew that the older child is a boy, the answer would be 1/2, since the remaining cases would be boy-boy and boy-girl.
Frequently asked questions
Which probability notions do you need to master for these riddles?
These riddles require counting the possible outcomes, reasoning with a conditional probability and working out an expected gain. The birthday problem is solved by finding the complementary probability, that no pair shares the same day.
Do you need special formulas for these riddles?
No: they rely on a few basic tools, counting the possible cases, using the complementary probability and computing an expected value, with no hard formula. The most important step is to define the event you are studying clearly before counting.
How can you avoid intuition traps in these riddles?
Write out the full list of outcomes before concluding, then check your reasoning on a smaller case. Several riddles on this page go against intuition, such as the birthday one, where the number of people needed is smaller than you might think.
What next?
Reasoning improves like memory: by practising often. Next, test yourself on our general knowledge quizzes.