Three Coins, Partial News

Question

Three fair coins are tossed. You are told that at least one landed heads. What is the probability all three are heads?

Accepted Answer

The 8 equally likely outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. The information "at least one head" eliminates only TTT, leaving 7 equally likely outcomes, of which exactly one is HHH: P(textall heads mid textat least one head) = frac1/87/8 = frac17. The trap Answering frac14 treats the statement as "this particular coin is heads, what about the other two?", a different conditioning event. "At least one" is a property of the whole set and removes only one outcome; "coin 1 is heads" removes four. The two-child problem is the same trap in costume, and interviewers deliberately cycle between the costumes to see whether you understand the mechanism or memorized one answer. General rule For n coins, P(textall heads mid textat least one head) = frac12^n - 1, always strictly smaller than frac12^n-1 (the specific-coin version), and the gap quantifies how much weaker "at least one" is as evidence than identifying an individual.

Three Coins, Partial News

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