Probability of being selected in a raffle There's a raffle with 1,000 names in a bucket. 600 of those names are in there once, and 200 are in there twice. So, just to reiterate, there are 800 unique names in the raffle, and 1000 names total. 500/1000 names will be selected from the raffle.
Q1: What is the probability of being selected if your name is in the bucket 
once? 
Q2: What is the probability of being selected if your name is in the bucket twice?
The answers are:
Q1: 500/100 = 50% chance of being drawn if your name is in the bucket once
Q2: 500/1000 + (500/1000)*(499/999) = 75% chance of being drawn if your name is in the bucket twice.
How do these answers make sense? Is there another way of arriving at these answers? I can't seem to understand these conceptually. Thank you!
 A: 
Q1: What is the probability of being selected if your name is in the bucket once? 
The answers is: 500/1000 = 50% chance of being drawn if your name is in the bucket once

If your name is in the bucket once, it is either among the 500 drawn, or it is among the 500 not drawn.
$P(\text{Drawn} \mid \text{In Once}) = \frac{500}{1000} = \frac 12$

Q2: What is the probability of being selected if your name is in the bucket twice?
The answer is: 500/1000 + (500/1000)*(499/999) = 75% chance of being drawn if your name is in the bucket twice.

That's the probability of drawing one of the names, plus the probability of not drawing that name times the probability of drawing the other name given that.  (Though there's 500 places left, not 499.)
$P(\text{Drawn} \mid \text{In Twice}) = \frac {500}{1000} + \frac {500}{1000}\cdot\frac{500}{999} \approx \frac34$
You have two names in the bucket.  Either the first is among the 500 of 1000 drawn, or it is among the undrawn.  If it is drawn we don't need to know where the other is.  If it is not drawn, there are 999 remaining tickets, and still 500 of which may be drawn.  Your second name may be one of the 500 drawn or one of the 499 not drawn.

*$P(D_1)=\frac{500}{1000}$
*$P(\overline{D_1} \cap D_2) = P(\overline{D_1})\times P(D_2\mid \overline{D_1})=\frac{500}{1000}\times\frac{500}{999}$Where $P(D_2 \mid \overline{D_1})$ is the conditional probability that the second name is in the draw, given that the first name is not in the draw.
A: Q1 is pretty easy. You have one ticket in there, they draw half of them out, so you have a 50% chance.
Q2 is more interesting. You can see it as $1-$ the probability of you not getting picked after 500 times, which is $1-\frac{998}{1000}*\frac{997}{999}*...*\frac{498}{500}=1-\prod_{k=1}^{500}\frac{998-k}{1000-k}=1-.249249249...\approx.75$.
A: The number of games is $\binom{1000}{500}$.
The number of games where a single ticket (say ticket no. $1$) wins is $\binom{999}{500}$, and $\frac{\binom{999}{500}}{\binom{1000}{500}}=\frac12$.
Wikipedia
The number of games where two tickets (say tickets no. $1$ and $2$) win is $\binom{998}{500}$, and $\frac{\binom{998}{500}}{\binom{1000}{500}}\to\frac14$ as the number of tickets increases.
