Calculate the odds of losing when there are multiple ways to win I have a game with odds of winning prize A at 1 in 5, prize B at 1 in 10, and prize C at 1 in 20,000.
Do I add these together like this:
1/5 + 1/10 + 1/20000 = 6001/20000 (Decimal 0.30005)
In this case 1 - 6001/20000 = .69995
So there is a approximately 70% chance of not winning at all.
Is this right?
 A: It depends on whether it is possible to win more than one prize, and also on how exact the odds are.
We know there is at most an $80\%$ chance to come up with no prize, because you have at least $20\%$ chance to win prize A.
But if you have to win prize A in order to win either of the other two prizes
(consider a video game where you can't complete level 3 unless you first complete level 1),
then the chance to win nothing is exactly $80\%$.
On the other extreme, if it is not possible to win two prizes in the same game and if  all the odds are exact, you have (as you computed)
exactly a $69.995\%$ chance of no prize.
If these were the posted odds for an actual game, it is quite possible that the conditions for winning prize C are a subset of the conditions for winning prize B which are in turn a subset of the conditions for winning prize A, and you are awarded the least valuable prize for which you qualify. (Compare a lottery where you pick five numbers and get some payoff for matching three numbers, a bigger payoff for matching four, and a much bigger payoff for matching five.)
It might be that $1$ in $10$ is actually the odds for winning either prize B or C,
and the chance of winning B is actually just $9.995\%$ or about $1$ in $10.05.$
But the organizers might round that off to $1$ in $10$ for simplicity (though it's slightly inaccurate that way).
It is also possible for the losing chances to be somewhere between
$69.995\%$ and $80\%$ -- really anywhere you like in that range -- if there are some rules that let you win two or more prizes some of the time with some probability.
(Independent odds of each prize is one way for this to happen, but really any amount of "overlap" of the probabilities is possible, ranging from never happening simultaneously to one event only happening when the more likely event does.)
A: Notice that "winning A" and "winning B" and "winning C", as events, are not disjoint. Therefore, you cannot just add them together. In this case, the proper way is to write
$$
P(\text{no win}) = P(\text{no A}) P(\text{no B}) P(\text{no C}) = 
\left( 1- \frac{1}{5}\right)
\left( 1- \frac{1}{10}\right)
\left( 1- \frac{1}{20~000}\right)
$$
