Probability Puzzles
Live probability tasks — Bayes, dice, binomial, Poisson, and coupon-collector reasoning.
12 questions
MiddleTheoryVery commonFraud base rate 1%, model recall 90%, false-positive rate 5% — chance a flagged user is truly fraud?
Fraud base rate 1%, model recall 90%, false-positive rate 5% — chance a flagged user is truly fraud?
Weight each branch by the base rate — true positives 0.01 × 0.90 = 0.009, false positives 0.99 × 0.05 = 0.0495. The posterior is 0.009 / (0.009 + 0.0495) ≈ 0.154, about 15%. A 90% recall still yields mostly false flags because fraud is rare.
Common mistakes
- ✗Reporting recall (
0.90) as the posterior, ignoring how rare fraud is - ✗Skipping the base rate, so the huge legitimate population's false positives are never weighted in
- ✗Confusing
P(flag | fraud)with the asked-forP(fraud | flag)— reversing the conditional
Follow-up questions
- →How high must recall go to make a flag more likely fraud than not, holding the other rates fixed?
- →How does raising the fraud base rate to 10% change the posterior?
MiddleTheoryVery commonThree users each convert independently, p = 0.2 — chance at least one converts, and why not 3 × 0.2?
Three users each convert independently, p = 0.2 — chance at least one converts, and why not 3 × 0.2?
Use the complement — P(at least one) = 1 − P(none) = 1 − 0.8³ = 1 − 0.512 = 0.488, about 48.8%. 3 × 0.2 = 0.6 is wrong because it sums overlapping events and double-counts the cases where two or all three convert.
Common mistakes
- ✗Summing
0.2three times, which can exceed 1 for largernand clearly cannot be a probability - ✗Multiplying the three rates, computing
all convertinstead ofat least one - ✗Forgetting that
at least oneis cleanest as the complement ofnone convert
Follow-up questions
- →For what value of
ndoes1 − 0.8^nfirst exceed 90%? - →When does summing per-event probabilities give a valid answer instead of the complement?
JuniorTheoryCommonFive visitors each convert independently with p = 0.2 — chance that exactly two convert?
Five visitors each convert independently with p = 0.2 — chance that exactly two convert?
Counting successes in fixed independent trials uses the binomial formula C(n,k) · p^k · (1−p)^(n−k). Here C(5,2) · 0.2² · 0.8³ = 10 · 0.04 · 0.512 ≈ 0.205, about 20.5%. The C(5,2) = 10 counts which two of five convert.
Common mistakes
- ✗Dropping the
C(n,k)combinatorial factor and forgetting the different ways two of five can be the successes - ✗Forgetting the
(1−p)^(n−k)factor for the visitors who do not convert - ✗Reading
exactly twoas the rate2/5rather than a binomial probability
Follow-up questions
- →What is the probability that at least one of the five visitors converts?
- →How does the formula simplify the expected number of conversions to
n · p?
JuniorTheoryCommonA coin lands heads with probability 0.3 — on average, how many flips until the first head?
A coin lands heads with probability 0.3 — on average, how many flips until the first head?
Flips until the first success follow a geometric distribution, whose mean is 1/p. With p = 0.3 that is 1/0.3 ≈ 3.33 flips on average. Rarer successes take longer — a p = 0.1 coin averages 10 flips.
Common mistakes
- ✗Reporting
pitself instead of its reciprocal1/pas the expected wait - ✗Using
1/(1 − p)(the failure rate) rather than1/p(the success rate) - ✗Ignoring the bias and defaulting to 2 flips as if the coin were fair
Follow-up questions
- →What is the probability the first head arrives on exactly the third flip?
- →Why does the geometric distribution have no memory of past failed flips?
JuniorTheoryCommonA page averages 2 signups per hour — under the count model Poisson, chance of zero signups in an hour?
A page averages 2 signups per hour — under the count model Poisson, chance of zero signups in an hour?
Counts of independent events over a fixed interval follow a Poisson distribution, P(k) = λ^k e^(−λ) / k!, with λ the mean. With λ = 2, zero signups is P(0) = e^(−2) ≈ 0.135, about 13.5%. Mean and variance both equal λ.
Common mistakes
- ✗Confusing the mean
λ = 2with a probability and computing1 − λ - ✗Modeling counts with a coin flip per event instead of the Poisson formula
- ✗Forgetting that
P(0) = e^(−λ), thek = 0case of the Poisson mass function
Follow-up questions
- →What is the probability of seeing exactly three signups in that hour?
- →When does a binomial count converge to a Poisson approximation?
JuniorTheoryCommonYou roll two fair dice — what is the probability their sum is at least 10?
You roll two fair dice — what is the probability their sum is at least 10?
There are 6 × 6 = 36 equally likely ordered outcomes. Sums of 10, 11, 12 come from (4,6)(5,5)(6,4), (5,6)(6,5), (6,6) — six of them. So the probability is 6/36 = 1/6 ≈ 16.7%. Count ordered pairs.
Common mistakes
- ✗Treating the 11 sum-values (2–12) as equally likely instead of the 36 face-pairs
- ✗Counting
(4,6)and(6,4)as one outcome, undercounting the ordered sample space - ✗Assuming every qualifying outcome needs both dice ≥ 5, which drops
(4,6)and(6,4)
Follow-up questions
- →What is the probability the sum is exactly 7, the single most likely total?
- →How does the answer change if you ask for a sum strictly greater than 10?
JuniorTheoryCommonFrom a 36-card deck you draw two cards — probability that both are hearts?
From a 36-card deck you draw two cards — probability that both are hearts?
A 36-card deck holds 9 hearts. You draw without replacement, so multiply the two conditional chances — 9/36 × 8/35 = 2/35 ≈ 5.7%. Equivalently, favorable pairs over all pairs, C(9,2)/C(36,2) = 36/630 = 2/35.
Common mistakes
- ✗Treating the draws as independent and squaring
9/36instead of using8/35for the second - ✗Assuming a 36-card deck has 13 hearts (it has 9 — six ranks per suit is wrong; it is nine)
- ✗Adding the two draw probabilities rather than multiplying them
Follow-up questions
- →How would the answer change if the first card were returned to the deck before the second draw?
- →What is the probability that the two cards share the same suit, whatever suit it is?
MiddleTheoryCommonOf 100 coins one is two-headed. You pick one and flip 10 heads in a row — chance it's the rigged coin?
Of 100 coins one is two-headed. You pick one and flip 10 heads in a row — chance it's the rigged coin?
The rigged prior is 1/100 and its likelihood is 1; a fair coin gives ten heads only with (1/2)^10 = 1/1024. The posterior is (1/100) / (1/100 + (99/100)(1/1024)) = 1024/1123 ≈ 0.912, about 91%. The long streak swamps the very low prior.
Common mistakes
- ✗Treating the prior
1/100as the answer and ignoring the likelihood ratio the streak provides - ✗Equating the fair coin's
1/1024with the posterior for the rigged coin - ✗Splitting the answer 50/50 once the evidence leans one way, ignoring the prior
Follow-up questions
- →How many heads in a row would you need before the posterior first passes 50%?
- →How does the answer shift if 5 of the 100 coins are two-headed instead of one?
MiddleTheoryCommonIn a room of 23 people, what's the probability at least two share a birthday, and why so high?
In a room of 23 people, what's the probability at least two share a birthday, and why so high?
About 50.7%. Use the complement — P(all distinct) = 365/365 × 364/365 × … × 343/365 ≈ 0.493, so a match is ≈ 0.507. It feels high because you count pairs, not people — 23 people already make C(23,2) = 253 pairs.
Common mistakes
- ✗Comparing 23 to 365 directly instead of counting the
C(23,2) = 253pairs - ✗Computing the chance of matching one fixed date rather than any pair matching
- ✗Believing you need ~183 people before the probability reaches 50%
Follow-up questions
- →Roughly how many people push the shared-birthday probability past 99%?
- →How does the complement trick simplify the calculation here?
MiddleTheoryCommon70% of churned users used feature X — does that prove X drives churn? Which probability do you need?
70% of churned users used feature X — does that prove X drives churn? Which probability do you need?
No. That figure is P(used X | churned), but you need the reverse — P(churn | used X) versus P(churn | didn't use X). If 70% of all users use X anyway, the churned group mirrors the base rate — correlation, not causation.
Common mistakes
- ✗Reading
P(used X | churned)as if it wereP(churn | used X)— reversing the conditional - ✗Never comparing against
P(churn | didn't use X)or the overall X-usage base rate - ✗Jumping from an observed association straight to a causal claim without an experiment
Follow-up questions
- →What comparison group and base rate would isolate whether X actually lifts churn?
- →What experiment would move this from correlation toward a causal answer?
MiddleTheoryCommonIn the game-show puzzle Monty Hall you pick 1 of 3 doors; the host opens a losing door — switch, and why?
In the game-show puzzle Monty Hall you pick 1 of 3 doors; the host opens a losing door — switch, and why?
Switch — it wins with probability 2/3 versus 1/3 for staying. Your first pick is right 1/3 of the time, so the prize is behind the other two doors 2/3 of the time. The host always opens a losing door, moving that 2/3 onto the last door.
Common mistakes
- ✗Assuming two closed doors mean a symmetric
50/50, ignoring the host's informed choice - ✗Forgetting the host never opens the prize door, which is what shifts the odds
- ✗Splitting the freed
1/3evenly to1/2each instead of concentrating it on the switch door
Follow-up questions
- →How do the switch odds change with 100 doors where the host opens 98 losing ones?
- →Why does the host knowing where the prize is matter to the result?
SeniorTheoryRareA loot box gives 1 of 8 distinct items uniformly at random — on average how many boxes to collect all 8?
A loot box gives 1 of 8 distinct items uniformly at random — on average how many boxes to collect all 8?
About 21.7 boxes. With k items already held, each box gives a new one with probability (8−k)/8, so the wait is 8/(8−k). Summing gives 8 · (1 + 1/2 + … + 1/8) = 8 · H₈ ≈ 21.7. The last item alone averages 8 boxes.
Common mistakes
- ✗Multiplying
8 × 8 = 64, treating each item as an independent 8-box wait - ✗Answering just
8, ignoring the growing chance of duplicate draws - ✗Missing that the wait for the final item alone averages 8 boxes
Follow-up questions
- →How does the expectation scale as the number of distinct items
ngrows large? - →Which stage of the collection contributes the most boxes to the total?
