Sweepstakes probability with invalid entries How do you take into account if an invalid entry is drawn in a sweepstakes? Basically, I want to know if the math/probability changes in an example like this:
Let's say I have 12 entries in a giveaway on my blog. 2 entries are invalid but I don't delete the entries and still calculate based on 12 entries. Let's also assume that each person only entered once. I'm wanting to compare the probably of winning with the 2 invalid entries included versus if the 2 invalid entries are deleted.
I have this so far:
p(x) = 1 − (y-1 / y)^x
y is the total number of entries
x is the number of times entered
p(x) is the chance of winning
p(x) = 1 - (9 / 10)^1 == p(x) = .1 or 10% [12 original entries but 2 invalid were deleted, giveaway entered once]
p(x) = 1 - (11 / 12)^1  ==  p(x) = .083 or about 8% [12 entries, giveaway entered once, let's pretend 2 entries are invalid but not deleted]
And this is where I'm stuck. I don't know how to write the redraw effect (where I will redraw a new winner from the 12 entries if an invalid entry is chosen.
 A: Let's just check I understand:
You have $N$ entries total (eg 12), including $M$ invalid entries (eg 2).


*

*Method 1: You remove the invalid entries, leaving $N-M$ (=10) remaining. You choose a winner, so my single valid entry has a probability of winning of $1/(N-M)$ (=0.1).

*Method 2: You select from all the entries. If you select an invalid entry, you ignore it, and replace it. Continue until you have a valid winner. Intuitively, the probability of winning must be the same as before, since you are still selecting a random winner from the same number of possible winning entries.
Let's say the probability of picking my single valid entry from all the entries is $p_{my}=1/N$ (=1/12) and the probability of picking a bad entry is $p_{bad}=M/N$ (=2/12).
Then, following the possible tree of events to pick my entry, method 2 is calculating:
mine OR (bad AND THEN (mine OR (bad AND THEN (...))))
$p_{me} + p_{bad}(p_{me} + p_{bad}(p_{me} + p_{bad}(...)))$
Note that this is infinite - you may keep picking bad values. However, it will converge to the $1/(N-M)$ value of method 1.
In general, thinking through the tree of possible events is often a good way to think about probability problems.
