Fumbling with a ring of keys

Question

You have a ring of 7 keys, exactly one of which opens your door, but you can't tell them apart. In the dark, you try keys one at a time in random order, setting aside each key that fails so you never repeat one.

On average, how many keys do you try until the door opens?

Accepted Answer

Think of the 7 keys as landing in a random order; the correct key is equally likely to be in any of the 7 positions. The number of tries equals the position of the correct key. Since that position is uniform over \{1, 2, \dots, 7\}, its expected value is the average of those positions: E[	ext{tries}] = \frac{1 + 2 + \cdots + 7}{7} = \frac{28}{7} = 4. So you expect to try 4 keys on average. Takeaway. Sampling without replacement makes the winning item equally likely to sit in any position, so the expected number of tries is the midpoint \frac{n+1}{2}. Trying without replacement beats trying w…

Fumbling with a ring of keys

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