Probability & Statistics

The First Ace

Flip a shuffled deck one card at a time. On average, how many cards turn over before the first ace appears?

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You turn over the cards of a well-shuffled 52-card deck one at a time. On average, how many cards do you turn over to reach the first ace?

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#Count the non-aces ahead of it

The position of the first ace is one more than the number of non-ace cards lying ahead of it. Label the 48 non-ace cards and let Ii=1I_i = 1 when card ii falls before the first ace, so

X=1+∑i=148Ii.(1)X = 1 + \sum_{i=1}^{48} I_i. \tag{1}

#Five regions, each equally likely

The four aces split the deck into five regions, the stretch before the first ace, the three gaps between consecutive aces, and the tail after the last. By symmetry a given non-ace card is equally likely to land in any of the five, and it beats the first ace only from the leading region, so

E[Ii]=15.(2)\E[I_i] = \frac{1}{5}. \tag{2}
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first acebefore it
The four aces split the deck into five regions, and any one of the 48 other cards is equally likely to land in each. It beats the first ace only from region one, with probability 1/5, so the first ace sits on average at 1 + 48/5 = 10.6.

#Add them up

E[X]=1+∑i=14815=1+485=535=10.6.(3)\E[X] = 1 + \sum_{i=1}^{48} \frac{1}{5} = 1 + \frac{48}{5} = \frac{53}{5} = 10.6. \tag{3}

The same count works for any deck. With mm ordinary cards and nn special ones, the first special card sits on average at 1+mn+1=m+n+1n+11 + \tfrac{m}{n+1} = \tfrac{m+n+1}{n+1}, which is 535\tfrac{53}{5} here.