I'm trying to work out odds in a variety of competition scenarios.
$1.$ Let's start with the easy one: I've entered a competition. I know that there are 99 other entrants, all with an equal chance of winning a single prize. The odds of winning this must be 1:100. I hope I'm right on that one at least.
$2.$ Now, consider a similar competition. Once again there are a hundred entries and one prize, but this time I have five of those entries instead of one. By my reckoning, buying five entries should mean that I now have a 1:20 chance of winning. Am I still right?
$3.$ Third case: This time we still have a hundred entries. I've only got one of them, but there are five identical prizes, which each will go to different winners. My understanding is that this should also give me a 1:20 chance of winning. This is where I'm not quite so certain of myself. Am I right?
$4.$ Now the reason I'm not so confident about the third case here is because it doesn't really work so well if you combine the second and third cases. So the fourth case is a competition with 100 entries, of which I have 10, and there are 10 prizes. By my reckoning earlier, either of those alone should give me a 1:10 chance to win, but applying them both would mean I've got a 1:1 chance, meaning I'm guaranteed to win. This clearly isn't the case. The odds would be good, but not that good.
So clearly I'm wrong somewhere. But what is my mistake?
Best Answer
Lets take the third case. The winners are drawn, one after another.
For winning the first price, you have a $\frac{1}{100}$ chance of winning it. The lucky entrant of this price is out of the game, after all he has already won (you didn't point out that in the rules very clearly, but I assume nobody is allowed to win twice).
For the next price (In case you didn't win the first), there are only 99 entrants left, one of them is you: $\frac{1}{99}$ chance to win, $\frac{99}{100}$ to loose.
Same goes for the next prices. The chance to win the third price is $\frac{1}{98}$, for the forth $\frac{1}{97}$ and for the fith you even have a chance of $\frac{1}{96}$.
So what is your chance of winning a price? You actually calculate that by asking "What is the chance that I don't win a price?". This chance is $\frac{99}{100} \cdot \frac{98}{99} \cdot \frac{97}{98} \cdot \frac{96}{97} \cdot \frac{95}{96} = 0.95$ (The odds of loosing every single draw).
Winning a price is the opposite of loosing every single draw, so that would be $$1 - \frac{99}{100} \cdot \frac{98}{99} \cdot \frac{97}{98} \cdot \frac{96}{97} \cdot \frac{95}{96} = 1 - 0.95 = 0.05 = 1:20$$ Indeed, you are right.
If someone who already won a price is allowed to win a second one, your chances get worse: You have more opponents in rounds 2-4.
$$1 - \frac{99}{100} \cdot \frac{99}{100} \cdot \frac{99}{100} \cdot \frac{90}{100} \cdot \frac{99}{100} = 1 - (0.99)^5 = 0.0490099501 = 4.90099501\% < 1:20$$
Not much of a deal, though.