Odds in a competition 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?
 A: 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.
A: Firstly, don't use odds loosely, or chance; let us stick to probability.
Secondly, I take it that where more than one prize is there you mean win at least one prize.
$1\; ok, Pr = \frac1{100}$
$2\; ok, Pr = \frac1{20}$
$3\; ok, Pr =\frac1{20}$, also confirmed using complement: $1- \frac{95}{100}$
$4\; not\;ok\;$ here you must use the complement: $Pr = 1 - \frac{90}{100}\frac{89}{99}\frac{88}{98}\frac{87}{97}.... \frac{81}{91} \approx0.6695$

ADDED
If you so prefer, you could do the last as $Pr = 1 - \frac{\binom{90}{10}}{\binom{100}{10}}$   
A: First case: you are correct, 1/100 chance
Second case: yes, you are still correct, 1/20 chance
Third case: yes you are still correct
Odds of getting first prize: 1/100
Odds of not getting first but getting second: 99/100*1/99=1/100
Odds of not getting first two but getting third: 98/100*1/98=1/100
etc.
5*1/100=1/20
Fourth case: incorrect - the problem here is that you combined the probabilities wrong, but that's okay, I was a little confused at first myself.
What you must do is account for the fact that there is a possibility you'll get multiple prizes.
The difficulty in this problem is that each entry can only get one prize.  As a result, we can just assume that once any entry in or out of your possession wins a prize, that entry just goes home and gets drunk. The easiest way to find the possibility that you'll get at least one prize is to find the probability you get absolutely no prizes and subtract that from 100%.
The first prize is given out, and by a 90/100 chance, you don't get it.  The person who did get it went home and got drunk.  We now only have 99 entries.
The second prize is given out, and by an 89/99 chance, you didn't get that one either.  The person who did get it went home and got drunk.  We now only have 98 entries.
By this pattern, we end up with 90/100*89/99*88/98*87/97*86/96*85/95*84/94*83/93*82/92*81/91= 90!/80!/100!*90!
The odds you don't win any prizes at all are 33.04762%
The odds you do win at least one prize is 66.95238%
There you go.
A: *

*Yes, that is 1:100

*The chance of not winning with any of the five entries is 99/100 * 98/99 * 97/98 * 96/97 * 95/96 = 19/20 (not 99/100 * 99/100 ... because you know the entries have lost, not won). Therefore the chance of winning with one is 1 - 19/20 = 1:20 so that is also correct.

*The chance of not winning is 95/100 (as 95 entries lose) = 19/20 so the chance of winning is also 1:20. Correct.

*The chance of not winning is 90/100 * 89/99 * ... * 81/91 = 2.0759...*10^19/6.2815...*10^19 which is approximately 0.3305 so the chance of winning is about 0.6695. Think about step 2: I get the answer a similar way.

