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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 ?2011-05-03
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    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.2011-05-05

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