Chance of winning a raffle with a special rule A person is hosting a raffle event. There are 1000 participants in the raffle.
The raffle draw will produce one winner.
The Special Rule
The host is also 1 of 1000 participants, but he announces he will not claim the prize, so, if the host wins the raffle, a re-draw will happen and he will be removed from the second draw which makes the total number of participant to 999.
Question
What is the probability of me winning the raffle?
Note
This might be a really silly question, but i cant seem to come up with an answer :).
My attempts:
Answer 1. Chance is $1/1000 + 1/1000 * 1/999$ = (chance of me winning first round + chance of host winning * chance of me winning second round)
Answer 2. Chance is $1/999$ because logically speaking, there are 999 people who can win, chance is just 1/999.
Edit: just did a calculation to actually calculate above 2 answers, they have the same result :)
 A: Well, as you found out:
$${1 \over 1000} + {1\over 1000}{1\over 999} = {1\over 1000}(1 + {1\over 999}) = {1\over 1000}{1000\over 999} = {1 \over 999}$$
so the two answers agree, and both answers are correct.
In a sense, answer 1 is more mechanical and is completely non-controversial.  Answer 2, however, is actually a symmetry argument - it is in fact arguing that (1) someone among the $999$ must win, and (2) each of them has the same chance.  
In particular, answer 2 is not concerned about certain details of the drawing, as long as symmetry is preserved.  E.g. imagine a modified process: if the host draws his own number, he puts it back into the hat and redraws, and if he redraws his own number, he puts it back into the hat and redraws, ... and he only throws away his own number (in exasperation) after $17$ consecutive draws of it.  Then you can still write a stupidly long expression for answer 1, and it will evaluate to $1/999$, or you can argue, again by symmetry, that it is $1/999$.
In fact, if the host never throws away his own number, you can still write an infinite expression for argument 1, and it will (in the limit) evaluate to $1/999$.
IMHO the symmetry argument, i.e. answer 2, is obvious and preferred.  However, the problem with "proof by obviousness" is that not everyone agrees it is obvious.  :)  So if you don't like answer 2, then please consider answer 1 as a way to justify answer 2.
BTW, not "seeing" an "obvious" symmetry happens somewhat often when dealing with probability questions.  Here and here (esp. Ned's answer) are more examples where there is a very simple symmetry argument, but if you don't see it or disagree with it, then you need more complicated algebraic manipulations.
