Radio contest... (Math and probablility experts wanted!)

Arrrgh! What the ^$&#% happened to my OP?!?

Anyway, here’s what I was TRYING to write:

A few years ago, I passed through a mall where the culmination of a summer radio promotion was taking place. On a board were 104 keys (the same number as in the station’s frequency; I don’t remember what it was, exactly, but that sounds about right). In the crowd were 104 qualifiers who’d called in at the right time for their opportunity.

These keys all fit in a sound box attached to a brand new, bright red Harley-Davidson Fat Boy. The contestants would be called up in random order and asked to pick a key and try it in the lock. Only one (I assume; let’s assume here that it is only one) would cause the box to produce the siren noise that indicated that that person had won the motorcycle.

Here’s something I’ve always wondered: which of the 104 positions has the best chance of winning? Obviously, it’s not the first, since that person has 104 possible keys to choose from; not very good odds. Neither is it the last person; (s)he may have a 100% chance of picking the right key, but the odds that the winning key would be the last one picked are just as laughable as picking the right one out of 104.

So who DID have the odds on his/her side?

Intuitively speaking, there are two effects that decrease your chances of success, both of which you note. If you’re early on, then you have more keys to pick from. If you’re later, then there’s a chance someone else has already gotten it. As it happens, these two effects exactly cancel each other out, so that everyone has the exact same chance of success - 1/104. This is not terribly hard to show mathematically, and I or someone else can if you like, but that’s the basic answer.

They all have a 1/104 chance of winning at the start of the contest.

Each person has 1/(# of people remaining to choose) once they choose their key. The last person would have a 100% chance of choosing the right key, but a 1/104 chance of actually getting the oppurtunity.

They all had equal odds of winning. The first person has a 1/104 chance of choosing the right key. When the second person comes up, there’s a 103/104 chance that the right key is still in the box, and if it is, he/she has a 1/103 chance of choosing the right one. So the chance of him/her ending up with the correct key is (103/104)*(1/103)=1/104. The same result holds for all the following contestants.

Wow, so quick! Thanks, guys!

See, this is why I became an English major, so I wouldn’t HAVE to bother with math… I guess this time my ignorance is showing. :slight_smile:

Thanks again!