Solution
Peeking at a Fair Coin
Show the problem again
A fair coin is flipped 10 times. After each flip the experimenter checks the running difference D = (heads) - (tails), and declares the coin "biased towards heads" the first moment D reaches +4 (and then stops). What is the exact probability, for a truly fair coin, that the experimenter ever declares it biased within the 10 flips? Give a reduced fraction.
Worked solution
The event is that the simple random walk reaches +4 within 10 steps. By the reflection principle P(max ≥ 4) = P(S10 ≥ 4) + P(S10 > 4) = P(S10 ≥ 4) + P(S10 ≥ 6). S10 = 2H - 10, so S10 ≥ 4 means H ≥ 7 (probability 176/1024) and S10 ≥ 6 means H ≥ 8 (probability 56/1024). The sum is 232/1024 = 29/128 ≈ 0.227. A single look at the end with the threshold D ≥ 4 would declare "biased" with probability only 176/1024 ≈ 0.172; the extra looks inflate the false-positive rate by about a third.
Source: Discrete optional-stopping illustration via the reflection principle. Statement written for AxiomIQ.