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What is the probability of at least 1 ace in a 5 card hand?

This is a hypergeometric probability question because cards are drawn without replacement from a fixed deck, so each draw changes what remains in the deck.

Probability result

34.1158%

Exact fraction: 18,472/54,145

Decimal: 0.341158

The probability of at least 1 success is 34.1158%.

You want every 5-card hand that contains 1, 2, 3, or 4 aces. The probability is the number of those favourable hands divided by all possible 5-card hands.

Worked steps

Show your work

  1. Inputs: population = 52, success states = 4, draws = 5, target successes = 1, event = at least.
  2. Formula: P(X in range) = sum of [C(K, k) C(N-K, n-k)] / C(N, n) terms.
  3. Substitute: use C(4, k), C(48, 5-k), and C(52, 5).
  4. Steps: count the favorable hands for the requested success range, then divide by all 5-draw samples.
  5. Result: The probability of at least 1 success is 34.1158%.

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You want every 5-card hand that contains 1, 2, 3, or 4 aces. The probability is the number of those favourable hands divided by all possible 5-card hands.

Formula

Probability model

P(E) = favorable outcomes / total outcomes

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Probability

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Calculation work