Investment Banking puzzles, solved step by step

Puzzles: 100
Traced to a firm: 36
Topics: 12
Hard: 29

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018Three cards sit in a hat: one red on both sides, one white on both sides, one red on one side and white on the other. You draw one and see a red face. What is the probability the other side is red?ProbabilityHardBulge bracket IBSales and trading
Try it first
Pick your answer.
Show the worked solution
Two thirds. You are not choosing among cards; you are looking at a face. There are three red faces you could be seeing, all equally likely: two belong to the red-red card and one to the red-white card. In two of the three cases the hidden side is red. The tempting answer of one half treats the two red-capable cards as equally likely and ignores that the red-red card shows red twice as often.
Why is one half so tempting, and where does it go wrong?
Imagine two families at a party: one with two daughters, one with a daughter and a son. If you meet one of the girls at random, she is twice as likely to come from the two-daughter family, because that family brought two girls. The red-red card is twice as likely to be the one showing red, because it has two red faces to show. Treating the two remaining cards as equally likely throws that information away, which is exactly what the question is built to catch.
Of the three red faces you might be looking at, two belong to the card that is red on both sides and one to the red-white card, so the hidden side is red two times in three, not one in two.
How do you prove it with Bayes' rule?
Write it as a conditional probability using Bayes' ruleA formula for updating a probability after seeing evidence: the chance of the evidence given the cause, times the prior chance of the cause, divided by the overall chance of the evidence.. The chance of drawing the red-red card is one third, and if you did, you are certain to see red. The chance of seeing red at all is the share of red faces, three of six, one half. So the chance the card is red-red, given a red face, is one third times one, divided by one half: two thirds.
The relationship
$P(RR \mid \text{red seen}) = \frac{P(\text{red seen} \mid RR)\,P(RR)}{P(\text{red seen})} = \frac{1 \times \tfrac13}{\tfrac12} = \tfrac23$
P(RR) the chance of drawing the red-red card, one third
P(red seen | RR) the chance of seeing red if you hold that card, which is 1
P(red seen) the overall chance of seeing red, 3 red faces out of 6
What it says in wordsWeight each card by how likely it is to produce what you saw, then divide by how likely that sight was overall.
How do you check it without any formula?
Imagine drawing 600 times, each card about 200 times. The red-red card shows red all 200 times, the red-white card shows red about 100 times, and the white card never does. Of the 300 red sightings, 200 have red on the back. Counting outcomes in a large imagined sample is a reliable check whenever a conditional probability feels slippery. The puzzle is a version of Bertrand's box paradox, and the same reasoning sits under the Monty Hall problem.
Where candidates lose it
One half is the answer most candidates give, and they defend it by saying the white card is out, so two cards remain. That is true about cards and irrelevant to faces; the interviewer will ask how many red faces you could be looking at.
The fix is to count the smallest equally likely outcomes. Here that means faces: six of them, three red, and two of those three have a red reverse.
What the interviewer asks next
What if the hat held two red-red cards and one red-white card?
You see a white face. What is the probability the other side is white?
How does this connect to the Monty Hall problem?
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.
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
$P(\text{sum }7 \mid \text{at least one }4) = \frac{2/36}{11/36} = \frac{2}{11}$
2/36 the chance of a 7 that contains a 4: (4,3) or (3,4)
11/36 the 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?
050A screening test flags accounting fraud with 95% accuracy both ways: it flags 95% of companies that commit fraud and clears 95% of companies that do not. In the sample, 2% of companies commit fraud. A company is flagged. What is the probability it is actually committing fraud?ProbabilityHardBulge bracket IBSales and trading
Try it first
Instinct first: how likely is a flagged company to be a fraud?
Show the worked solution
About 28%. Take 1,000 companies. Twenty commit fraud and the test flags 19 of them. The other 980 are clean, and the test wrongly flags 5% of them, which is 49. So 68 companies are flagged and only 19 are frauds: 19 over 68 is 27.9%. The test is accurate, but fraud is rare, so a small error rate on the large clean group produces more false flags than true ones.
Why is the answer not 95%?
Think of a smoke alarm that sounds for 95% of real fires and stays quiet 95% of the time there is no fire. In a building where fires are rare, most alarms you hear are burnt toast, because the quiet days vastly outnumber the fires and 5% of them still ring. A test's accuracy tells you how it behaves when the truth is known; the question asks the reverse, how likely the truth is given the test, and the base rateHow common the thing is before any test is run. Here, 2% of companies commit fraud. sits between the two. With 2% frauds, a 5% error on the 98% of clean companies creates 49 false flags for every 19 true ones.
Of 1,000 companies, the test flags 19 of the 20 frauds and 49 of the 980 clean companies, so 68 are flagged and only 19 of them are frauds, which makes a flag 28% likely to be right rather than 95%.
How do you work it with whole numbers?
Start with 1,000 companies and walk them down the tree. Frauds: 2% of 1,000 is 20, and 95% of those, 19, are flagged. Clean: 980, and 5% of those, 49, are wrongly flagged. 19 true flags and 49 false flags make 68 flagged companies, so the chance that a flagged company is a fraud is 19 over 68, about 28%. Working in counts rather than percentages is faster in the room and harder to get wrong, and it shows the interviewer where every number came from.
The relationship
$P(\text{fraud} \mid \text{flag}) = \frac{0.02 \times 0.95}{0.02 \times 0.95 + 0.98 \times 0.05} = \frac{0.019}{0.068} \approx 0.28$
0.02 x 0.95 the share of all companies that are frauds and get flagged
0.98 x 0.05 the share that are clean and get flagged anyway
What it says in wordsDivide the true flags by all the flags, true and false together.
What moves the number, and what does that mean for a screen?
Two things. The base rate: if 10% of the sample committed fraud, the same test would make a flag 68% reliable. The false-flag rate: cutting it from 5% to 1% lifts the answer to 66% even at a 2% base rate. A screen for a rare event is a filter that earns a closer look, not a verdict, and the second look does the real work: run an independent test of the same accuracy on the 68 flagged companies and a double flag is 88% likely to be a fraud. That is how diligence red flags, transaction monitoring and credit early-warning lists are meant to be read. The working assumes the two tests err independently; if they look at the same ratios, the second adds less than the arithmetic suggests.
Where candidates lose it
Candidates answer 95%, reading the test's accuracy as the answer. The question asks the reverse conditional, and when the event is rare the two numbers can sit far apart.
The second loss is reaching for the formula in symbols and getting lost in it. Say 1,000 companies, 20 frauds, 19 caught, 49 false flags, 19 over 68; the counts carry you through and the interviewer can follow every step.
What the interviewer asks next
The company is also flagged by a second, independent test with the same accuracy. Now what is the probability?
How rare does fraud have to be before a flag is more likely wrong than right?
The test clears a company instead. What is the chance it is actually clean?
065A banker is running five independent live deals, and each has a 30% chance of closing this quarter. What is the probability that at least one closes?ProbabilityWarm upBulge bracket IBSales and trading
Try it first
Pick the probability that at least one deal closes.
Show the worked solution
About 83.2%. Go through the back door. Each deal fails to close with probability 0.7, and because the deals are independent the chance that all five fail is 0.7 to the fifth power, about 16.8%. At least one closing is the only other possibility, so it is 1 minus 16.8%, which is 83.2%. Adding five lots of 30% to get 150% is the trap.
Why can you not add the five chances?
If five friends each have a 30% chance of turning up to dinner, you would not say there is a 150% chance someone comes. Adding counts the evenings when two or three friends arrive several times over. Probabilities of separate events add only when the events cannot both happen, and here two deals can close in the same quarter. Five times 30% is instead the expected number of deals that close, 1.5, which is a useful number but a different question.
Stacking five 30% chances gives 150%, which cannot be a probability, while multiplying five independent 70% chances of failure gives 16.8% for no deal closing, so at least one closes with probability 83.2%.
Why is the complement the fast route?
At least one means one, or two, or three, or four, or five deals closing, and working each case out is slow. The opposite of at least one is none, and none is a single clean case: every deal fails. Independence lets you multiply the five 0.7s, and the number shrinks fast: 70%, 49%, 34.3%, 24.0%, then 16.8%. Subtract from 100% and you have the answer in one line. Whenever a question says at least one, reach for the complement before anything else.
The relationship
$P(\text{at least one}) = 1 - (1 - 0.3)^5 = 1 - 0.7^5 = 1 - 0.168 = 0.832$
0.3 the chance one deal closes
0.7 the chance one deal does not close
5 the number of independent deals
What it says in wordsThe chance that at least one closes is one minus the chance that every deal fails.
What does the independence assumption hide?
The question tells you the deals are independent, and you should say that you are leaning on it. In a real quarter the deals share a market: a rate shock or a credit squeeze that stalls one is likely to stall the others, so the true chance of at least one closing is lower than 83.2%. If you want to show range, add that the chance of exactly one closing is 5 x 0.3 x 0.7 to the fourth, about 36.0%, so most of the 83.2% is a single close rather than a flood.
Where candidates lose it
The fast wrong answer is 150%, and even candidates who know it is impossible sometimes try to rescue it by capping at 100%. The interviewer wants to hear the word complement within the first few seconds.
The second miss is multiplying the 30%s instead of the 70%s, which gives the chance that all five close, about 0.24%. Say which event you are multiplying before you multiply.
What the interviewer asks next
What is the probability that exactly two of the five close?
How many such deals would you need for a 95% chance of at least one closing?
If the five deals are all in one sector, how would you change your answer?
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 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
$P(\text{match}) = 1 - \frac{365}{365} \cdot \frac{364}{365} \cdots \frac{336}{365} \approx 1 - 0.294 = 0.706$
365/365 the first person can have any birthday
364/365 the second avoids the first
336/365 the 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?
098Two associates agree to meet at a coffee shop, each arriving at a random time between 12:00 and 1:00, independently. Each will wait 15 minutes for the other and then leave. What is the probability that they meet?ProbabilityHardBulge bracket IBSales and trading
Try it first
Pick before you draw anything.
Show the worked solution
7/16, about 43.8%. Put one arrival time on each axis of a square from 12:00 to 1:00. They meet when the two times are within 15 minutes, a band along the diagonal. They miss in two corner triangles with legs of 45 minutes, each covering (3/4)^2 / 2 = 9/32 of the square. So the chance of meeting is 1 - 9/16 = 7/16.
Why turn a timing problem into a square?
If you and a friend each pick a random seat in a cinema, every pair of choices is equally likely, so the chance of sitting together is the share of seat pairs that are neighbours. Arrival times work the same way, except they are continuous. When two independent times are uniformly likely, every point in the square of arrival pairs is equally likely, so a probability is simply an area. The question becomes: what fraction of the square has the two times within 15 minutes?
Plotting both arrival times on a square, the associates meet inside the diagonal band where their times are within 15 minutes, which covers 7/16 of the square, while the two grey corners where one arrives more than 15 minutes ahead cover 9/16.
Why is the corner easier to measure than the band?
The band has an awkward shape, but its complement is two right triangles. In the lower right corner, associate 1 arrives more than 15 minutes after associate 2, who has already left. That triangle runs from 12:15 to 1:00 along each side, 45 minutes, or 3/4 of the hour. Each triangle covers half of (3/4) squared, which is 9/32, so the two together cover 9/16 and the band covers the remaining 7/16.
The relationship
$P(\text{meet}) = 1 - (1 - w)^2 = 1 - \Big(\tfrac{3}{4}\Big)^2 = \tfrac{7}{16} \approx 43.75\%$
w the waiting time as a share of the hour, 15/60 = 1/4
(1 - w)^2 the two miss triangles together
What it says in wordsThe chance of meeting is one minus the area of the two corners where one person arrives too late.
The formula gives quick answers to the follow-ups. If each waits 30 minutes, w is a half and they meet with probability 1 - 1/4 = 75%. The limitation is the assumption of uniform arrivals: real people arrive close to the agreed time, which raises the chance of meeting, so the answer is for the model, not for actual associates.
Where candidates lose it
The common wrong answer is 25%, fifteen minutes out of sixty. It forgets that either person can arrive first and that the window shrinks near the ends of the hour.
The second loss is trying to integrate by cases out loud and getting lost. Draw the square, shade the corners, and the arithmetic is two lines.
What the interviewer asks next
How long must each wait for the chance of meeting to be 50%?
What if one associate waits 15 minutes and the other waits 30?
What if they must arrive between 12:00 and 2:00 instead?

P(RR)	the chance of drawing the red-red card, one third
P(red seen \| RR)	the chance of seeing red if you hold that card, which is 1
P(red seen)	the overall chance of seeing red, 3 red faces out of 6

2/36	the chance of a 7 that contains a 4: (4,3) or (3,4)
11/36	the chance of at least one 4

0.02 x 0.95	the share of all companies that are frauds and get flagged
0.98 x 0.05	the share that are clean and get flagged anyway

0.3	the chance one deal closes
0.7	the chance one deal does not close
5	the number of independent deals

Try it first

Why is one half so tempting, and where does it go wrong?

How do you prove it with Bayes' rule?

How do you check it without any formula?

Where candidates lose it

What the interviewer asks next

Try it first

What does being told something actually change?

Why is 1/6 tempting and wrong?

How do you check it quickly?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the answer not 95%?

How do you work it with whole numbers?

What moves the number, and what does that mean for a screen?

Where candidates lose it

What the interviewer asks next

Try it first

Why can you not add the five chances?

Why is the complement the fast route?

What does the independence assumption hide?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the answer feel far too high?

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

Where candidates lose it

What the interviewer asks next

Try it first

Why turn a timing problem into a square?

Why is the corner easier to measure than the band?

Where candidates lose it

What the interviewer asks next