# Winning A Car in a series of Drawings

True Story (Happened a few days ago):

A car was given away by a major corporation at an event.

First Drawing: All 230 attendees were given a key to unlock a lock. Only 10 of the keys would work. Those winners of the first drawing became 10 finalists.

Second Drawing (10 Finalists): The 10 finalists were given a chance to pick a key fob from a table, one at a time. The finalists would go in order according to the same order that they had unlocked the lock in the first drawing.

The table started with 10 key fobs, only one of which would start the car. The first person picked a key fob, tried it, and it did not work. The key fob was not returned to the table. The 2nd person now had 9 key fobs to choose from. He picked a key fob and it did not start the car. His key fob was not returned to the table. The 3rd person now had 8 key fobs to choose from. The same pattern repeated itself until there were only 2 people left, and 2 key fobs on the table. The 9th person picked one of the 2 key fobs. It did not work. The 10th person won the car with the remaining key fob.

What are the odds that an attendee would become a finalist, and then become the 10th person to try the key fob and win the car, after all other finalists had tried before hand?

• It depends on how your question is read. The odds that some attendee would become a finalist and then win the car on the tenth try is 1/10. Arguably, that is a better assessment of the improbability of what happened than an alternative reading, which asks about the odds that a particular pre-specified attendee would do this. It's like the difference between putting a billiard ball in some pocket, and pointing to the pocket beforehand. I would bet that nobody pointed publicly to a particular individual and said "this person will be a finalist and they will open the lock at the very last." – whuber Sep 2 '16 at 21:51