there is also a chance, that you will never win anything ;)
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I got ~22.51% odds of you having won anything with 1:1679 odds over 428 giveaways.
1-(n^x) where n = odds of you not winning a giveaway (1678/1679) and x = number of attempts.
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If you take the average. But not every giveaway is 1679 entries. I've entered a couple giveaways where there are sub 10 or 50 entries.
1-(product(g1 g2 g3 * ....)^x) where g is each giveaway chance to lose and x = theoretical times won
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Damn I can't get this to open on OOcalc. Taking forever.
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Damn if you made this, I'm proud of ya. Tons of work put into this.
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ROFL, It's called procrastination before Finals XP
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EDIT:^^^ my point exactly
Jokes about nerds not getting "gf"s (as you call them) are getting weak because of all the nerd-celebrities out there that are either married or close.
Just my opinion. Weak insult.
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Oh relax, he was just trying to be funny. No harm.
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The probability to win nothing is (let's call it NOTHING)
NOTHING := g1 * g2 * g3 * ...
The probability to win at least one item is
1 - NOTHING
The probability to win exactly one item is quite ugly already
ONETHING := sum ((1/g_i - 1) * NOTHING; i = 1, 2, 3, ...)
The probability to win at least two items is
1 - NOTHING - ONETHING
and the probability to win exactly two items is
sum ((1/g_i - 1) * (1/g_j - 1) * NOTHING; i = 1, 2, 3, ...; j = i+1, i+2, i+3, ...)
and so on...
TL;DR: Your formula is not correct
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gi and gj are going to be the same thing, since this is a fixed set of data, and not infinite. What happens when i is giveaway 400 of 400? J has to be giveaway 1, which makes these values the same.
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Based off of the giveaways I've entered, (To give you guys an idea of the kind of giveaways I enter: avg 1679 entries over 428 giveaways [Mostly public, only very recently contributor, a few group])
1-(product(g1 g2 g3 * ....)^x) where g is each giveaway chance to lose and x = theoretical times won
Odds are 56% that I won 1 item, which I haven't :(
31% for 2
18% for 3
9.8% for 4
and 5.5% for 5 games.
Cool story, I know.
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