Expected value round, often followed by 'and the last ace?'.
Cards until the first ace
The question
You deal a shuffled 52-card deck one card at a time. What is the expected number of cards dealt before the first ace appears?
Have a go before you open the solution. If you are stuck after a few minutes, the first paragraph below is usually enough of a nudge.
Show the worked solution
The answer
9.6 cards before it, so the first ace is the 10.6th card on average.
The four aces split the other 48 cards into five gaps, before the first ace, between consecutive aces, and after the last.
By symmetry every gap has the same expected size, so each holds 48/5 = 9.6 cards.
The expected number of non-aces before the first ace is therefore 9.6, and the first ace arrives at position 10.6.
The symmetry argument is the point. Anyone can write the sum; the interviewer is watching for whether you avoid needing to.
What it is testing
Symmetry as a substitute for computation.
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