Brain Teasers & Mental Math Interview Questions

Sample Answer

The answer is $\frac{142}{143}$. You compute the common denominator as $13 \times 11 = 143$, then the numerator becomes $7 \times 11 + 5 \times 13 = 77 + 65 = 142$. Since $142 = 2 \times 71$ and $143 = 11 \times 13$, they share no common factors, so the fraction is already reduced. Notice how close this is to 1, which serves as a good sanity check.

Sample Answer

You could try direct long division of 999 by 1024, or you could rewrite it as $1 - \frac{25}{1024}$. The second approach wins here because $\frac{25}{1024}$ is much easier to estimate. You know $\frac{25}{1000} = 0.025$, and dividing by 1024 instead of 1000 makes it slightly smaller: $\frac{25}{1024} \approx 0.02441$. So the answer is approximately $1 - 0.02441 = 0.976$ to three decimal places. This "subtract from 1" reframing is a technique you should reach for whenever the numerator is close to the denominator.

Sample Answer

You could compute this by enumerating all ordered draws and filtering, or you could condition directly on the first draw being red and compute the second draw probability. The second approach wins here because the conditioning is already given to you. Given the first ball is red, there are 4 balls left: 2 red and 2 blue. So the probability the second is also red is $\frac{2}{4} = \frac{1}{2}$.

Sample Answer

OK so the first person can pick anything, that is $\frac{10}{10}$. The second person needs to avoid the first person's number, so that is $\frac{9}{10}$. The third person needs to avoid both previous numbers, giving $\frac{8}{10}$. Multiplying these together you get $\frac{10 \times 9 \times 8}{10^3} = \frac{720}{1000} = \frac{18}{25}$. This sequential conditioning approach is the cleanest way to handle "all distinct" problems.

Sample Answer

This question is checking whether you can set up recursive expected value equations from scratch rather than pattern matching to a memorized answer. Let $E_0$ be the expected flips from the start, $E_1$ be the expected flips given you just flipped one head. From $E_0$: flip a coin, if tails (prob $\frac{1}{2}$) you spent 1 flip and return to $E_0$, if heads (prob $\frac{1}{2}$) you spent 1 flip and move to $E_1$. From $E_1$: if heads you are done (1 more flip), if tails you spent 1 flip and reset to $E_0$. This gives $E_1 = \frac{1}{2}(1) + \frac{1}{2}(1 + E_0)$ and $E_0 = \frac{1}{2}(1 + E_0) + \frac{1}{2}(1 + E_1)$. Solving yields $E_0 = 6$.

Sample Answer

Let's reason through this step by step. If you place your first quarter exactly at the center of the table, you establish a symmetry advantage. Now, for every subsequent move your opponent makes, you can mirror their placement through the center point of the table. Since the table is round and your first quarter sits at the center of symmetry, any valid position your opponent finds will always have a corresponding valid mirror position available to you. This means your opponent will always be the first player unable to place a quarter, so going first and playing the center is a guaranteed win.

Sample Answer

This question is checking whether you can identify modular structure in sequential games and apply backward induction cleanly. The losing positions are multiples of 3, because if you face a multiple of 3, whatever you add (1 or 2), your opponent can add the complement to advance by exactly 3. Since $100 = 33 \times 3 + 1$, Alice should add 1 on her first move to put the total at 1, then maintain the invariant that after her turn the total is always $\equiv 1 \pmod{3}$. This forces Bob onto multiples of 3 every round, and Alice reaches 100 exactly.

Sample Answer

This question is checking whether you can anchor on a few physical quantities and chain them together without getting lost. A Boeing 777 burns roughly 5 to 6 liters of fuel per kilometer, and the distance from JFK to Heathrow is about 5,500 km, so total consumption is approximately $5.5 \times 5{,}500 \approx 30{,}000$ liters. You can sanity check by noting the plane's max fuel capacity is around 170,000 liters and a transatlantic leg should use a modest fraction of that, so 30,000 liters (about 18%) checks out nicely. Stating the answer as roughly 25,000 to 35,000 liters shows you have the right order of magnitude.

Sample Answer

The standard move is to estimate contract volume and multiply by notional per contract: the E-mini trades roughly 1.5 to 2 million contracts per day, and each contract is worth $50 times the S&P 500 index level, so at an index level around 5,000 the notional per contract is $250,000. But here, the micro E-mini matters because it now absorbs significant volume at one-tenth the notional, so you should note that roughly 1.5 million E-minis at $250k each gives about $375 billion per day in notional, and adding micro contracts might push the combined figure toward $400 to $500 billion. Sanity check this against US equity market daily volume of roughly $500 to $600 billion in cash equities: futures notional being comparable in magnitude is consistent with how institutional hedging and speculation work.

Sample Answer

The standard move is to compute the expected value: $E = \frac{1+2+3+4+5+6}{6} = 3.50$, so you should pay at most $3.50 for a single play. But here, the number of repetitions matters because with 10,000 plays, the law of large numbers guarantees your average payout converges tightly to $3.50, so you could comfortably pay slightly below $3.50 per game and lock in near-certain profit. For a single play, risk aversion might make you demand a bigger discount, say paying only $3.00 or so, depending on your utility function.

Sample Answer

Get this wrong in production and you blow up your bankroll despite having a massive edge. Betting $50 on a $100 bankroll means a single loss cuts you in half, and two consecutive losses (probability $0.4^2 = 0.16$) effectively wipe you out. The right call is to use Kelly sizing: $f^* = 2p - 1 = 2(0.6) - 1 = 0.20$, so you bet 20% of your current bankroll each flip, which is $20 initially. This maximizes the expected logarithmic growth of your wealth and protects you from ruin while still aggressively compounding your edge over 50 flips.

Sample Answer

Get this wrong in production and you get run over by informed flow, selling cheap contracts all day while your P&L bleeds out. The right call is to immediately shift your market upward because the aggressive buyer likely has information suggesting the true value is above your offer of 2.9. You should move your entire market up, not just widen it. A reasonable update might be to re-center around 3.1 or higher, quoting something like 2.9 at 3.3, with the magnitude of your adjustment depending on how much you trust the signal (size of the order, identity of the counterparty). The key insight interviewers want is that you recognize adverse selection: when someone eagerly trades at your price, it is evidence your price is wrong.

Sample Answer

You might think $50 since you saw one heads and there are "two coins that could show heads." But that breaks under Bayesian updating because the double-headed coin is twice as likely to have produced the heads you observed. The prior probability of picking each coin is $\frac{1}{3}$, but the double-tails coin is eliminated (probability of heads is 0). The posterior for the double-headed coin is $\frac{1/3 \cdot 1}{1/3 \cdot 1 + 1/3 \cdot 1/2} = \frac{2}{3}$, and the posterior for the fair coin is $\frac{1}{3}$. That leaves you with an expected value of $\frac{2}{3} \cdot 100 + \frac{1}{3} \cdot 50 = \$83.33$ as the fair price. A naive $50 answer ignores the likelihood weighting, which is exactly the mistake the interviewer is testing for.