The unusual question Goldman Sachs is asking in 2024 job interviews

19 November 2024

3 minute read

A goldman sachs banker getting on a plane while drunk

Goldman Sachs' quant strats team has been known to ask some rather unusual questions during job interviews. Questions can be as simple as dice game probabilities, or as obtuse as this question involving lions and steak. The firm might have a new favorite question in 2024, however.

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According to interviews logged on Glassdoor, a repeated question among quantitative analysts is the following:

"A line of 100 passengers is waiting to board a plane. Each holds a ticket to a specific seat. The first passenger to board is drunk, and sits in a random seat. The rest of the passengers are sober, and will sit in their proper seat unless it is occupied; in that instance, they will randomly choose a free seat. What is the likelihood that the final passenger to board sits in their allocated seat?"

The question isn't new, and you may have come across it before in previous interviews. Think you know how to answer it?

Leave it in the comments below, and don't cheat. 🤨

...

The answer is a 50% chance.

The solution lies in breaking the problem down to its simplest form and working up... same as most of these kinds of questions.

At 2 passengers, there's a 50% chance the drunk passenger picks the correct seat, and so a 50% chance the last passenger will sit in the correct seat.

At 3 passengers, there's an equal probability of him sitting in his own seat, or the last passenger's seat, so they cancel each other out. If he sits in the second passenger's seat, there's a 50% chance that passenger will go on to sit in the drunkard's seat or the last passenger's.

This logic continues the longer the line is, meaning there will always be a 50% chance the final passenger is correctly seated.

Let us know your method of working, or your different answer, if you got one.

If you're looking for other questions to cut your teeth on, here are other questions quants have said they were asked in Goldman Sachs job interviews:

"127 pirates stand in a circle. They start shooting alternately in a cycle such that the 1st pirate shoots the 2nd, the 3rd shoots the 4th and so on. The pirates who got shot are eliminated from the game. They continue in circles, shooting the next standing pirate, till only one pirate is left. Which position should someone stand to survive?"
"You have a bag with 100 identical golf balls in it. Each ball is labelled by a number between 1 and 100 (no duplicates). By paying $1 per draw, you can pick one ball from the bag. Then you can either stop, and you will be paid an amount in $ equivalent to the number on the ball, or you can reinsert the ball in the bag and extract again. What is the optimal strategy to maximize the gain?"

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AUTHORAlex McMurray Reporter

IH
IHTFP92
19 November 2024
100 passengers (one of whom is drunk): this depends on the total number of seats on the plane. This must be >=100 (or the final passenger can't sit down, and the answer is 0%). If the number of seats is >100 then the probability of one of the remaining 99 passengers having been assigned the seat number of the drunk declines as the number of seats rises (i.e. the answer changes depending on the size of the plane). For the question to have a single answer the number of seats on the plane must exactly match the number of passengers. This was not stated as a part of the problem, but it must be assumed for a definitive answer to be calculated.
So, this devolves into 2 cases:
1) the drunk sits in their assigned seat (probability = 1%) and all the other passengers also sit in their assigned seats. The chance of the final passenger being in the correct seat = 100%.
2) the drunk sits in the wrong seat (probability = 99%). Among the remaining 99 passengers there must be one who has been assigned the seat of the drunk passenger. As the seats fill up eventually this passenger comes on board, sees their seat is taken, and then sits in one of the remaining empty seats. This changes nothing - there is still one passenger among those who have not yet boarded who can't sit in their assigned seat. Eventually you reach the final 2 passengers:
a) the penultimate passenger is able to sit in their assigned seat (probability = 50%), and the final passenger must sit in the wrong seat.
b) the penultimate passenger is unable to sit in their assigned seat and randomly chooses among the final 2 seats:
b.1) they choose the seat of the final passenger (probability = 25%), and the final passenger must sit in the wrong seat.
b.2) they don't choose the seat of the final passenger (probability = 25%), and the final passenger is able to sit in the correct seat.
The combined probabilities are 0.01 * 100% + 0.99 * 25% = 25.75%
(at least I think that's the case, my brain has gone numb thinking about traumatic aircraft boarding strategies)
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