IMC Intern Quantitative Trader Interview Experience — Betting Strategy and Election Fund Allocation

IMC·Quantitative Trader·Nov 2025
Technical ScreenInternhard

Q1: Betting game.

The setup was a series of matches with their odds and match times. Each group of matches could be treated as independent. I would receive two events, meaning inside information or statistics, and had to decide from each event whether to change my bet on a particular match. I also needed to calculate my profit and loss. During the final review, I had to explain my betting strategy and reasoning, and say which single bet I would change if I could.

A: The first piece of information was that one evening match would have an unexpected result. The natural idea was to find an evening match and bet on the side with higher odds, since lower odds implied a higher probability of winning.

The second piece of information concerned opponents A and B in one match. We knew A had lost its previous three matches, so my idea was to ignore the odds and bet on B because A was in poor form.

My overall betting strategy was: 1) There were three afternoon matches and three evening matches. Because there was no extra information about the afternoon matches, I bet a fixed amount, about 10 percent of my total funds, on the lower-odds teams and tried to earn a relatively stable return. 2) Because I knew the evening set would contain an unexpected result, I placed a smaller bet, 5 percent of my total funds, on a high-odds outcome to seek a larger payoff while controlling risk. 3) When the odds were 1:1, I chose not to take on the extra risk of betting.

Looking back at the game's overall result, I thought there were several possible problems. 1) In the first round of the game, the size of my bet on the lower-odds side was inconsistent with the size of my lower-odds bet in round five. The interviewer followed up on this, and I did not have a particularly good explanation. Maybe I should have used the same size consistently, or perhaps adjusted it dynamically based on the odds. I was unsure because when I previously played a market-making game, I used the Kelly criterion to decide whether to bet and to calculate the optimal fraction. Here, however, I had no way to know the true probability of winning, so I did not dare try dynamic sizing. I would be interested in whether there is a systematic method. 2) Was 10 percent too conservative? The stated goal was to maximize profit, but my fraction might not produce a large gain even if I won, while the attempt to earn a stable return could still expose me to a large loss. Maybe the reasoning itself was flawed. 3) In this kind of betting game, was it possible to control losses by betting on both sides? Because there was only one set of odds, there was no arbitrage. My understanding was that betting on both sides should be a negative move, but I was not sure whether that was the right way to think about it.

Q2: A brain-teaser-style game.

There were d districts holding an election. We had n1 funds and our opponent had n2 funds, and each side needed to allocate its funds to win the election. We won the election if the number of districts we won was greater than the number our opponent won; individual districts could tie.

Q2a: First came a comprehension check with d = 3 and n1 = n2 = 4. Assuming the opponent allocated funds completely at random, what were my best and worst allocations, and how many possible opponent allocations were there?

A: The number of opponent allocations was a stars-and-bars problem: C(3 + 4 - 1, 3 - 1) = C(6, 2) = 15. My idea for the best allocation was to spread the funds as evenly as possible. Because the opponent was completely random, it could assign zero to a district, and distributing my funds evenly maximized the chance of winning those districts with the least funding. I therefore chose (2, 1, 1) as best. The worst was more obvious: (4, 0, 0), or any rotation of it. That allocation could win at most one district, while the opponent would have at least as many districts as we did, so it could never win the overall election.

Q2b: Was there any difference among (2, 1, 1), (1, 2, 1), and (1, 1, 2)? Which would I choose?

A: I said that because the opponent was completely random, all three should be equivalent and symmetric in probability. Rotation should not matter, so I would randomly choose whichever one I liked. I felt that logic was reasonable, though I was open to being corrected.

Q2c: The warm-up was over. In the new situation, d = 8, n2 = 9, and n1 was unknown. We knew the opponent would distribute its funds as evenly as possible. What was the minimum funding we needed to have a possibility of winning the election? This did not ask for a guarantee, only an allocation under that funding level that could win.

A: My first reaction was that with eight districts, we needed a majority and therefore had to win at least five. Since the opponent spread its funds evenly, the best case was meeting five districts where it had placed one fund. We would need two in each to beat it, giving 2 * 5 = 10.

The interviewer then asked me to think again about whether we really needed to win all five. Based on that hint, I considered using as many ties and deliberate losses as possible and only ensuring that our district count ended one higher. The best plan could be to lose the districts where the opponent placed two, beat two districts where it placed one, and tie all the rest. That gave 2 * 2 + 1 * 5 = 9. The same reasoning also produced two other arrangements, 3 * 2 + 3 * 1 = 9 and 4 * 2 + 1 * 1 = 9. So I concluded that the minimum was nine, not ten.

Q2d: Among those three allocations, which was best?

A: I answered that the arrangement with four 2s and one 1 was best. It left us with three zeros, maximizing the probability that our zero-funded districts met districts where the opponent had put two. If that happened, we could guarantee the win. I thought the other two arrangements should have lower winning probabilities, though I was open to a better answer if this reasoning was wrong.

Q2e: Based on the previous answer, suppose n1 = 9. What was the minimum n2 that would guarantee we could not win, still assuming the opponent spread its funds as evenly as possible?

A: I did not solve this during the interview, so this was my reasoning afterward. Using the same logic, we could deliberately lose districts where the opponent had two and try to win the overall election through a combination of ties and wins elsewhere. With n1 = 9, the most we could do was win four districts by putting two in each and hoping to meet opponent districts with one, then tie one more district. Once the opponent had at least four districts funded with two, it became impossible for us to win with n1 = 9. I therefore concluded that when the opponent had 1 * 4 + 2 * 4 = 12, we no longer had any possibility of winning with nine.

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Interview at a glance

Company
IMC
Role
Quantitative Trader
Level
Intern
Rounds
Technical Screen
Difficulty
hard
Interview date
Nov 2025
Questions from this interview
2 questions

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