DIGEST
The Coin-Flip Puzzle: Why the Answer Is 50%
Source: Presh Talwalkar — MindYourDecisions (YouTube)  |  Type: Mathematics / Probability
Explainer  |  Context: Analysis of a banned Google interview question
BOTTOM LINE UP FRONT
The puzzle: In a fictional scenario, every subject flips a fair coin repeatedly until they
get heads, then stops. What is the expected percentage of heads across all flips by all
subjects?
The answer is exactly 50%. Despite the strong intuition that tails should dominate —
since many subjects flip multiple tails before stopping — the expected ratio of heads to
total flips is precisely one-half. The key is that each round of flipping is itself a fair
50/50 split, and a naïve count that ignores the probability weighting of each outcome
produces a biased (and wrong) estimate of 1/3.
The deeper lesson: Raw counting without probability weighting is a fundamental
statistical error. The puzzle also demonstrates that geometric series and round-by-
round decomposition are powerful tools for reasoning about sequential random
processes.
The Puzzle and Its Origin
Years ago, Business Insider published a list of 15 banned Google interview questions. One
stood out as a probability riddle: in a fictional country where families have children until
they have a boy (girls prompt another child, a boy means they stop), what is the expected

ratio of boys to girls? The question was likely banned both for its gendered framing and for
being unrelated to job competency — but it contains a genuinely interesting mathematical
problem beneath the surface.
Rephrased in neutral terms: a king orders every subject to flip a fair coin until it lands heads.
Tails means flip again; heads means stop. What is the expected percentage of heads across
the entire experiment?
The Intuition Trap
The natural first reaction is that tails should vastly outnumber heads. After all, every subject
produces exactly one heads but potentially many tails. It feels obvious that the accumulated
tails across the population will swamp the heads count.
The common mistake: Listing possible outcomes (H, TH, TTH, TTTH, TTTTH, ...),
counting heads and total flips across those outcomes without weighting by probability,
and concluding the ratio is 5/15 = 1/3. This is a biased estimator. The unweighted count
treats a rare event (five flips) as equally likely to a common event (one flip), which
fundamentally distorts the calculation.
Why the Answer Is 50%: The Intuitive Argument
Consider n subjects all flipping simultaneously, round by round.
Round 1: All n subjects flip. Half get heads and stop; half get tails and continue. In this
round, exactly 50% of the flips are heads.
Round 2: The n/2 remaining subjects flip. Again, half get heads and stop, half get tails. Once
more, 50% of this round's flips are heads.

Round 3, 4, 5, ...: The same pattern repeats. Each round, the surviving population halves,
and exactly half the flips in that round are heads.
Core insight: Every single round of the experiment is a collection of independent fair
coin flips. In each round, 50% of the flips are heads. Since the overall result is the
union of all these rounds — each internally 50/50 — the aggregate is also 50%. The
shrinking population changes the volume of each round but not its composition.
Why the Answer Is 50%: The Rigorous Proof
Let n be the number of subjects. The number of heads is trivially n, since each subject flips
until they get exactly one heads.
The total number of flips, F, requires more care. The fraction of subjects who flip exactly k
times is (0.5)
k
, and each contributes k flips. So:
F = 1·(0.5)n + 2·(0.5)
2
n + 3·(0.5)
3
n + 4·(0.5)
4
n + ...
To evaluate this sum, a standard technique is used: compute F − 0.5F by shifting and
subtracting. The shifted version aligns each term with the next, and the subtraction collapses
the coefficients. What remains is a geometric series:
0.5F = n · (0.5 + 0.5
2
+ 0.5
3
+ ...) = n · (0.5 / (1 − 0.5)) = n · 1 = n
Therefore F = 2n. The ratio of heads to total flips is n / 2n = 1/2 = 50%.
The Numerical Sanity Check
With 800 subjects, the expected breakdown illustrates the probability weighting concretely:

400 subjects flip once (400 flips), 200 flip twice (400 flips), 100 flip three times (300
flips), 50 flip four times (200 flips), 25 flip five times (125 flips). Total: 800 heads out of
1,425 flips ≈ 56.1%.
The 56.1% result from truncating at five flips is already close to 50%. As the number of
terms approaches infinity, the ratio converges exactly to 50%. The truncation error comes
from ignoring the (small but nonzero) fraction of subjects who flip six or more times.
Broader Implications
Statistical reasoning: This puzzle is a clean illustration of why unweighted averages
over outcomes with different probabilities produce biased estimates. The "1/3" mistake
is structurally identical to errors that arise in real-world data analysis whenever events
are counted without adjusting for their likelihood.
Interview question design: The puzzle was banned from Google interviews, likely
because clever brainteasers test puzzle familiarity more than engineering ability. It
remains, however, a valuable teaching tool for probability and expectation.
Geometric series as a proof tool: The shift-and-subtract technique for evaluating ∑
k·r
k
is a broadly useful method in probability, combinatorics, and analysis — not just a
one-off trick for this problem.

Further Exploration
Generalize the coin: What if the coin is biased with probability p of heads? The
expected fraction of heads is still p, by the same round-by-round argument. The
geometric series proof generalizes cleanly.
Variance question: The expected ratio is 50%, but what is the variance across
different population sizes? For small n, realized ratios can deviate substantially.
Connection to the negative binomial: Each subject's flip count follows a geometric
distribution. The puzzle is equivalent to asking about the mean of a specific function of
geometric random variables — a standard topic in probability theory.