Last night on Australia’s “The Footy Show”, a supporter from each team in the Australian Football League (AFL), 16 in all, were rounded up. Sitting in front of them was a barrel with 16 keys, and a new car. They each took turns on walking up to the barrel and picking out a key, before returning to their seat. There were 16 keys in the barrel, one of which, opened the new car. Whoever picked that key from the barrel, won the car.

It got me wondering, which pick would have given someone the best chance of winning the car? Would it make any difference which pick they had?

They all have an equal chance, at the start of the game. Granted, if the first guy fails the remaining players chances are improved, but that is balanced by the probability that the first guy won’t fail.

I’m sure somebody else will explain it more coherently shortly

Yeah, they all have the same chance of 1/16 of winning.

A quick example of how to show this: Let’s figure the probability of, say, the fourth picker getting the car.

In order for that to happen, the first, second, and third all had to get the wrong key. The first guy has a 15/16 chance of being wrong, for the second guy it’s 14/15 (since one key is now gone, given that the first guy missed), for the third it’s 13/14 (given that the first two missed). So the probability they are all wrong is

(15/16)(14/15)(13/14) = 13/16.

The fourth guy must now pick one correct key out of the remaining 13; obviously a 1/13 chance at this point.

So the probability of the first three being wrong, the fourth getting it right, is (13/16)*(1/13) = 1/16.

[hijack]Doc there’s a chance we might have a downunder Dopefest, with Australian groundrules (basically same as in the US, just less hugging). Are you interested? Possibly Sydney early May[/hijack]

If they keep picking until one has found the key as stated in the OP, and if the order of picks on each round stays the same, then it’s an advantage to go first. That’s because if, after four rounds no one has found the key, you win by default.