Probability of number appearing on a board of randomly generated numbers? Let's say I have a board that generates 23 number randomly (0000 - 9999).
1st prize has 1 number
2nd prize has 1 number
3rd prize has 1 number
4th prize has 10 number
5th prize has 10 number
Let's say I pick number $1234$. The probability that my $1234$ will appear can be calculated in different way
1st
1 - $P(1234 \text{ not appearing}) = 1 - (9999/10000)^{23} = 0.002297471770114836...$
2nd
$P(\text{appear in }1st) + P(\text{appear in }2nd) ... + P(\text{appear in }5th) = 
(1/10000)*3 + 2(1 - (9999/10000)^{10}) = 0.002299100239958...$
Probability that 1234 appear in 1st, 2nd or 3rd are 1/10000 each. So 1/10000*3
Probability that 1234 appear in 4th are 1 - P(Not appear in 4th). Using binomial it is 1 - 0.9999^10
Probability that 1234 appear in 5th are 1 - P(Not appear in 5th). Using binomial it is 1 - 0.9999^10
Adding them together should yield probability of 1234 appearing on the board
3rd
I can simply say there are 23 numbers on the board so my number has 23/10000 chance = 0.0023 
Why are these three methods giving slightly different value? Which method is the most accurate?
 A: Your third calculation assumes that the numbers are drawn without replacement, that no number can be selected twice.  That is the usual way lotteries are done.  
Your first approach assumes the numbers are drawn with replacement, that the first number is drawn and we check whether it is $1234$, then another number is drawn from all the numbers and we check that one, and so on.  The probability is a little less than $\frac {23}{10000}$.  The expected number of draws of $1234$ is $\frac {23}{10000}$ but we can draw it more than once.  Therefore the chance we draw it at least once must be less than $\frac {23}{10000}$.  
You don't explain the second calculation.  It appears the numbers are drawn with replacement.  You can't just add the probabilities because the events are not disjoint.  If you draw with replacement you could win both first and second.  It sounds like you are computing the chance of winning at least one prize, in which case you must use inclusion/exclusion for this route.
