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.
指南以英文編寫，與面試語言一致。
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.
例題 · Two dice
經程式驗證What is the probability that two dice add up to 7?
查看解題過程
- 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).
答案6/36 = 1/6 ≈ 16.7%.
Tool 2: the complement
For "at least one" questions, work out the chance of none and subtract it from 1.
例題 · At least one six
經程式驗證You roll a die four times. What is the probability of at least one six?
查看解題過程
- 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.
答案671/1296 ≈ 51.8%.
Tool 3: conditional probability and Bayes
P(A given B) = P(A and B) ÷ P(B)
例題 · The medical test
經程式驗證1% 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?
查看解題過程
- 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.
答案1/6 ≈ 16.7%.
提示Natural 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.
例題 · Hats
經程式驗證n people throw their hats in a pile and each takes one at random. What is the expected number who get their own hat back?
查看解題過程
- 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.
答案1.
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.
例題 · Two heads in a row
經程式驗證You flip a fair coin until you get two heads in a row. What is the expected number of flips?
查看解題過程
- 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.
答案6 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.
例題 · Keep or re-roll
經程式驗證You 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?
查看解題過程
- 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.
答案$4.25: keep 4 or more.
例題 · Two re-rolls
經程式驗證Now you may re-roll up to twice. What changes?
查看解題過程
- 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.
答案14/3 ≈ $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
常見錯誤
- 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.
Trading and quant