Ace in the Hand

Question

You are dealt five cards from a well-shuffled 52-card deck. What is the probability your hand contains at least one ace?

Accepted Answer

Count the complement, hands built entirely from the 48 non-aces: P(textno ace) = fracbinom485binom525 = frac4852cdotfrac4751cdotfrac4650cdotfrac4549cdotfrac4448 approx 0.6588, so P(textat least one ace) = 1 - 0.6588 approx 0.3412. Roughly one hand in three. The trap, and what it secretly computes The tempting shortcut 4 times frac552 = frac513 approx 0.385 is not the probability, it double-counts hands holding two or more aces. It is, however, exactly the expected number of aces in the hand (linearity of expectation), and therefore an upper bound on the probability (union bound). Knowing precisely what the wrong number means, and that probability le expectation for counts, is the kind of error-awareness interviewers probe. Speed check left(frac4852right)^5 = (0.923)^5 approx 0.67 is the with-replacement approximation; the exact without-replacement answer must be a touch lower for the "no ace" event (each miss depletes non-aces), giving approx 0.66, consistent.

Ace in the Hand

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