Trading and quant
Probability questions: six tools that solve most of them
Counting, complements, Bayes, linearity of expectation, states and working backwards: the tools behind almost every trading-firm probability question, with exact answers.
Die Leitfäden sind auf Englisch, der Sprache der Interviews.
What interviewers want
Trading firms ask probability on the phone and in person. They want three things: the method out loud, an exact answer (a fraction, then a decimal), and a quick check that the answer makes sense.
Tool 1: count equally likely outcomes
P(event) = favorable outcomes ÷ total outcomes
Only when every outcome is equally likely.
Rechenbeispiel · Two dice
Per Code geprüftWhat is the probability that two dice add up to 7?
Lösungsweg anzeigen
- 1.There are 6 × 6 = 36 equally likely outcomes.
- 2.Six of them sum to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
Antwort6/36 = 1/6 ≈ 16.7%.
Tool 2: the complement
For "at least one" questions, work out the chance of none and subtract it from 1.
Rechenbeispiel · At least one six
Per Code geprüftYou roll a die four times. What is the probability of at least one six?
Lösungsweg anzeigen
- 1.P(no six in one roll) = 5/6, so P(no six in four rolls) = (5/6)4 = 625/1296.
- 2.P(at least one six) = 1 − 625/1296 = 671/1296.
Antwort671/1296 ≈ 51.8%.
Tool 3: conditional probability and Bayes
P(A given B) = P(A and B) ÷ P(B)
Rechenbeispiel · The medical test
Per Code geprüft1% of people have a condition. A test catches 99% of cases, but also flags 5% of healthy people. Someone tests positive. What is the chance they have the condition?
Lösungsweg anzeigen
- 1.Imagine 10,000 people: 100 have the condition and 99 of them test positive.
- 2.Of the 9,900 healthy people, 5% (495) also test positive.
- 3.So 99 of the 594 positives have the condition: 99 ÷ 594 = 1/6.
Antwort1/6 ≈ 16.7%.
TippNatural frequencies (imagine 10,000 people) make Bayes questions much easier to do in your head, and easier for the interviewer to follow.
Tool 4: linearity of expectation
E[X + Y] = E[X] + E[Y]
True even when X and Y depend on each other, which is why it cracks so many "expected number of…" questions.
Rechenbeispiel · Hats
Per Code geprüftn people throw their hats in a pile and each takes one at random. What is the expected number who get their own hat back?
Lösungsweg anzeigen
- 1.Each person has a 1/n chance of getting their own hat.
- 2.Add the n expectations: n × 1/n = 1, whatever n is.
Antwort1.
Tool 5: states and first-step analysis
When a process can be in a few situations ("states"), write one equation per state for what happens after the next step, then solve.
Rechenbeispiel · Two heads in a row
Per Code geprüftYou flip a fair coin until you get two heads in a row. What is the expected number of flips?
Lösungsweg anzeigen
- 1.Let E be the expected flips from the start and F the expected flips after one head.
- 2.E = 1 + ½F + ½E (a head moves you to F, a tail sends you back).
- 3.F = 1 + ½ × 0 + ½E (a head finishes, a tail sends you back).
- 4.Substituting: E = 6 and F = 4.
Antwort6 flips.
Tool 6: work backwards
For games where you choose when to stop, start from the last decision. Its value tells you what to do one step earlier.
Rechenbeispiel · Keep or re-roll
Per Code geprüftYou roll a die and win its face value in dollars. You may re-roll once, but then you must keep the second roll. What is the game worth, and what's your strategy?
Lösungsweg anzeigen
- 1.A re-roll is worth 3.5 on average, so keep 4, 5 or 6 and re-roll 1, 2 or 3.
- 2.Value = (4 + 5 + 6)/6 + (3/6) × 3.5 = 2.5 + 1.75 = 4.25.
Antwort$4.25: keep 4 or more.
Rechenbeispiel · Two re-rolls
Per Code geprüftNow you may re-roll up to twice. What changes?
Lösungsweg anzeigen
- 1.With one re-roll left the game is worth 4.25 (above), so on the first roll keep only 5 or 6.
- 2.Value = (5 + 6)/6 + (4/6) × 4.25 = 11/6 + 17/6 = 28/6 ≈ 4.67.
Antwort14/3 ≈ $4.67: keep 5 or 6 on the first roll, then 4 or more.
How to answer out loud
- 1
Restate
Repeat the question in your own words and state assumptions: fair dice, independent flips, cards without replacement.
- 2
Name the tool
"This is an at-least-one question, so I'll use the complement."
- 3
Compute
Keep fractions exact until the end.
- 4
Check
Is it between 0 and 1? Does it behave at the extremes (n = 1, a certain event)? Is it symmetric where it should be?
Common mistakes
Häufige Fehler
- Adding probabilities of events that can happen together ("at least one six in four rolls is 4/6").
- Treating draws without replacement as independent.
- Confusing P(A given B) with P(B given A).
- Giving only a decimal when an exact fraction is easy.
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Trading and quant