Investment Banking puzzles, solved step by step

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.

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.

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.

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.

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.

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.

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?

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 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.