Draws from a hat until the first winning ticket

Question

A hat holds 20 raffle tickets, of which 4 are winners. Tickets are drawn one at a time without replacement, in a uniformly random order. On average, how many draws until the first winning ticket appears (including that ticket)?

Accepted Answer

This is sampling without replacement, so use the gap argument. The 4 winners split the 16 losers into 5 gaps, and by symmetry each loser is equally likely to land in any gap. So on average \frac{16}{5} = 3.2 losers come before the first winner, and the first winner is the next draw: 3.2 + 1 = 4.2 	ext{ draws.} Equivalently, the expected position of the first of m special items among N is \frac{N+1}{m+1} = \frac{21}{5} = 4.2. Takeaway. First of m winners among N tickets sits at expected position \frac{N+1}{m+1} = \frac{21}{5} = 4.2.

Draws from a hat until the first winning ticket

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