Investment Banking puzzles, solved step by step

Puzzles
100
Traced to a firm
36
Topics
12
Hard
29
Topic
All topicsProbability7Mental maths and counting11Growth and compounding9Valuation riddles11Logic and brainteasers11Rates, risk and options9Estimation and market sizing7Accounting riddles9Expected value and games7Enterprise value and dilution6Deal maths6DCF and cost of capital7
Level
AnyWarm upCoreHard
Source
AnyReported at a firmStandard
Showing 1–2 of 2 · filtered from 100Clear filters
  1. 034Two fair dice are rolled and you are told that at least one of them shows a 4. What is the probability that the two dice sum to 7?ProbabilityCoreBulge bracket IBSales and trading

    Try it first

    Pick before you count.

    Show the worked solution

    2/11, about 18.2%. Of the 36 equally likely outcomes, 11 contain at least one 4: six with the first die showing 4, six with the second, less the double 4 counted twice. Of those 11, only 4 and 3 and 3 and 4 sum to 7. The new information shrinks the world to 11 outcomes, and 2 of them qualify.

    What does being told something actually change?

    Imagine a teacher tells you that your friend is one of the 11 students who scored above 90. Your friend's chance of topping the class is now measured against those 11, not the whole class of 36. Conditional probability throws away every outcome the new information rules out and counts only inside what is left. Here the information rules out the 25 outcomes with no 4 anywhere, and 11 remain.

    Told 'at least one 4': the grid shrinks from 36 cells to 11First dieSeconddie11223344556623456734567845678956789106789101178910111236 outcomes, all equally likelyCells with no 4: 5 x 5 = 25, ruled outCells with at least one 4: 36 - 25 = 11Of those, sum to 7: (4,3) and (3,4) = 2P(sum 7 given a 4)2 / 11 = 18.2%Not 1/6 = 16.7%: that answers 'the first die is 4'The double 4 sits in the set once, so there are 11, not 12
    Of the 36 outcomes for two dice, 11 contain at least one 4 and only two of those, 4 and 3 and 3 and 4, sum to 7, so the probability is 2/11, about 18.2%, rather than 1/6.

    Why is 1/6 tempting and wrong?

    1/6 comes from fixing one die at 4 and asking for a 3 on the other, which is the right answer to 'the first die shows a 4'. 'At least one shows a 4' is a weaker statement: it lets the 4 sit on either die, which creates 11 outcomes rather than 6. If you counted the double 4 twice you would have 12 outcomes and get 2 in 12, exactly 1/6. Counted once, as it must be, there are 11, and the answer edges up to 2/11.

    The relationship
    P(sum 7∣at least one 4)=2/3611/36=211P(\text{sum }7 \mid \text{at least one }4) = \frac{2/36}{11/36} = \frac{2}{11}
    2/36the chance of a 7 that contains a 4: (4,3) or (3,4)
    11/36the chance of at least one 4
    What it says in wordsCount the outcomes that satisfy both conditions and divide by the outcomes that satisfy the information.

    How do you check it quickly?

    Use the complement for the denominator. The chance of no 4 on either die is 5/6 times 5/6, which is 25/36, so at least one 4 is 11/36. Two cells sum to 7 and contain a 4, so 2/36 over 11/36 is 2/11. Two routes to the same 11 is what the interviewer wants to hear before you commit to the answer.

    Where candidates lose it

    The trap is answering 1/6 by treating 'at least one die' as 'this die'. The two sound alike in a fast question and give different sets of outcomes.

    The second loss is counting 12 outcomes with a 4 by forgetting that the double 4 sits in both lists. Picture the grid, or say the complement out loud, and the 11 is safe.

    What the interviewer asks next

    • Now you are told the first die shows a 4. What is the probability of a 7?
    • Given at least one 6, what is the probability of a double?
    • Given that the sum is 7, what is the probability that at least one die shows a 4?
  2. 081Your incoming analyst class has 30 people. What is the probability that at least two of them share a birthday? Roughly how large must the class be for the odds to pass 50%?ProbabilityCoreBulge bracket IBSales and trading

    Try it first

    Before working it: what is the chance of a shared birthday in a class of 30?

    Show the worked solution

    About 71% for 30 people, and the odds pass 50% at 23. Work the chance that nobody matches: the second person avoids the first with 364/365, the third avoids both with 363/365, and so on. Multiply 30 of these and you get about 29.4%, so a shared birthday has a 70.6% chance. Repeating the product shows it first passes half at 23 people.

    Why does the answer feel far too high?

    Your instinct pictures yourself in the room: what are the odds someone shares my birthday? That is about 30 in 365, under 10%. But the question allows any two people to match. In a group of 30, every person can be compared with every other, and there are 30 x 29 / 2 = 435 such pairs. The number of pairs, not the number of people, drives the answer, and pairs grow roughly with the square of the group. Four hundred chances to hit a one-in-365 coincidence is a lot of chances.

    The odds climb fast because pairs grow faster than people50%100%0%1102030405060People in the group23 people: 50.7%30 people: 70.6%23 people make 253 pairs30 people make 435 pairseach pair is a chance to match
    The chance of a shared birthday crosses 50% at 23 people and reaches 70.6% at 30, because 23 people already form 253 pairs and 30 form 435, each one a separate chance of a match.

    How do you compute it without a list of 435 cases?

    Count the opposite. Matching can happen in many overlapping ways, but nobody matching happens in exactly one: each new person must miss every birthday already taken. When at least one is messy, compute none and subtract from one.

    The relationship
    P(match)=1−365365⋅364365⋯336365≈1−0.294=0.706P(\text{match}) = 1 - \frac{365}{365} \cdot \frac{364}{365} \cdots \frac{336}{365} \approx 1 - 0.294 = 0.706
    365/365the first person can have any birthday
    364/365the second avoids the first
    336/365the thirtieth avoids the 29 taken
    What it says in wordsThe chance of at least one match is one minus the chance every person lands on a fresh day.

    For a quick estimate in your head, use pairs: the chance of no match is close to e raised to minus pairs over 365. With 253 pairs that is e to the minus 0.69, almost exactly a half, which is why 23 is the crossing point. The method assumes birthdays are spread evenly across the year; real birthdays bunch a little, which only nudges the odds higher. By 60 people the chance is 99.4%.

    Where candidates lose it

    The fast wrong answer is 30 over 365, under 10%. It quietly fixes one person and asks whether anyone matches them, which is a much smaller event than any two people matching.

    The second loss is trying to add up the ways a match can happen, which double counts groups with two or three matches. Say complement first, then multiply.

    What the interviewer asks next

    • What is the chance that someone in the class shares your birthday specifically?
    • How large must the class be for a 99% chance of a shared birthday?
    • Where does the same pairs logic show up in hashing or in risk of correlated failures?