Probability against winning a raffle "If there are 10000 raffle tickets, all of which are sold, and you purchased 20 of these tickets, what are the odds against you winning?"
This was a question I got wrong on a recent test which I plan to retake (as an altered version of the original), but I need help understanding how to go about solving this.  I originally thought that all I needed to was take 1 - the probability of winning, but that ended up being incorrect.
Any help?
 A: The subtlety here is one in language. If you were asked to find the probability that you lose the raffle, you'd be correct, just take $1-P(win)$. However, you are being asked for the odds against winning.
Odds represent the ratio of possible outcomes, they are different from probability. In simple examples, this roughly means that probability is $\frac{\text{Number of ways to win}}{\text{Total number of outcomes}}$, whereas odds against are $\frac{\text{Number of ways to lose}}{\text{Number of ways to win}}$.
For example, suppose we have a fair 6-sided die. The odds against you rolling a 2 are 5:1, since there are 5 outcomes in which you lose (fail to roll a two) and only 1 outcome where you win (roll a 2).
In your example, this translates to the following:
9980 of the tickets are other people's tickets; if theirs gets called, you lose. The other 20 are yours. So assuming the lottery is played fairly, the odds against you winning are 9980:20, since in 9980 outcomes, you lose, and in 20 outcomes, you win. This ratio can be reduced to 499:1.
A: The odds for your winning are $20:9980$ or $1:499$ so the odds against you winning are just $499:1$
