Addition rule for mutually exclusive events Why does the addition rule not work here?
$$P(\text{win this week}) = 0.5$$
$$P(\text{win next week}) = 0.5$$
They are mutually exclusive, and you can only win once (although it is possible that neither occurs). What is the probability that either occurs?
$P(A \cup B) = P(A) + P(B)$ clearly does not give the correct answer, unless you are guaranteed to win (but that is not the case here).
 A: The question makes sense in a scenario as under.
A school (say) gives a prize to some student every fortnight, but if someone has got the prize in the first fortnight, she is not considered for the prize in the second fortnight. Then, for any particular student,
P(doesn't get any prize that month) = $0.5\times0.5 = 0.25$
P(doesn't get a prize the $1^{st}$ fortnight, but does in the $2^{nd}) = 0.5\times 0.5 = 0.25$
P(gets a prize in the first fortnight) = $0.5$
As it must, the probabilities add up to $1$

Generalization added
Let probabilities of getting prize in first and second fortnight be $a$ and $b$ respectively, then
P(don't get any prize) = $(1-a)(1-b) = 1-a-b+ab$
P(get prize only first fortnight) = $a$
P(don't get $1^{st}$ fortnight, get in $2^{nd} = (1-a)b = b-ab$
Adding together, we get $1$, as we must
A: I think you confuse Independency & Exclusiveness. These are seems to be Mutually independent, not exclusive. Or there is some other information about game or lottery as @miracle173 noticed.
By informations of comments:
So, Yes, the probability of wins in whole $2$ weeks is $1$. But if you have problem to be winner surely, you can suppose some scenario in which you don't:
Suppose you randomly pick a number in $[0,1]$ and if it's a irrational number in $(0,1/2)$ you become the winner of the first week, and if it's a irrational number in $(1/2,1)$ you would be the winner of the next week. So you win by probability of $1$, but not surely, you lose if you be a infinitely bad luck rational picker!
