# 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:

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.

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I understand what you're saying. With Method 2, the result is basically infinite, so how do you think the probability of winning compares between the two methods? Do you have the same probability of winning no matter if the invalid entries are deleted or not? Thank you for your reply, by the way! – Jenny May 5 '11 at 4:58
You're right: the probability of a good entry winning is the same both ways because essentially you're doing the same thing - randomly choosing one from all the good entries. Working through the infinite sum above (possible but hard!) would give the same probability. – Aaron Lockey May 8 '11 at 17:05
Thank you Aaron! – Jenny May 10 '11 at 3:56