# 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.

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if I understood correctly don't you just minus the total entries by one? Am I missing something ? – Mark May 3 '11 at 23:12
Yes, sort of. I'm not throwing the invalid entry out. I am still redrawing from the original 12. So essentially I have the chance to redraw the bad entry. I am basically wanting to know if I should be spending the time to check all of my comments to see if they are valid or not. I want to be fair to my readers. – Jenny May 5 '11 at 5:01

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:

$p_{me} + p_{bad}(p_{me} + p_{bad}(p_{me} + p_{bad}(...)))$
Note that this is infinite - you may keep picking bad values. However, it will converge to the $1/(N-M)$ value of method 1.