Where the first spade appears in a shuffled deck

Question

A standard 52-card deck is shuffled uniformly. You deal cards one at a time until the first spade appears. There are 13 spades. On average, what is the position of the first spade (including that card)?

Accepted Answer

Sampling without replacement again. The 13 spades split the 39 non-spades into 14 gaps, and by symmetry each non-spade is equally likely to fall in any gap. So on average \frac{39}{14} non-spades precede the first spade, and the first spade is the next card: \frac{39}{14} + 1 = \frac{39 + 14}{14} = \frac{53}{14} \approx 3.79 	ext{ cards.} This is the general formula \frac{N+1}{m+1} = \frac{53}{14} with N = 52, m = 13. Takeaway. With 13 spades the first appears at expected position \frac{53}{14} \approx 3.79; more special items means a shorter wait.

Where the first spade appears in a shuffled deck

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