I don't know, maybe i am paranoid or i am salty as ****, what are the chances of overdosing 4 times in 9 days? What would be the math behind all of this?
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Might be paranoia
Started by ExKioz [2122347] on in General Discussion.
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I don't know, maybe i am paranoid or i am salty as ****, what are the chances of overdosing 4 times in 9 days? What would be the math behind all of this?
[image: i.imgur.com]
![[image: i.imgur.com]](https://i.imgur.com/JXh3N4q.png){kind=link}
Just like that guy in dumb and dumber movie - 1 in a million - so you’re saying there’s a chance!!!
dont know the math, but no matter what math says, it happened regardless
i've seen worse, but this is definately bad luck..
It happens, I OD'd twice in a row during the start of elimination the year before last.. Just bad luck.. [:sick:]
Considering the data available, I'd say the chance that this can happen is 100%.
The first two Ecstasy I took resulted in overdose, and the first two Xanax resulted in overdose too. Shit happens and you are allowed to get salty.. you just aren't allowed to stay salty. (JustJanet's rule)
Doing several OD in a row happens quite often.
I'm tempted to think that the probability of ODing increases after an OD.
But overall it's a certified solid 3% (without perks).
I've OD'd twice in 24hrs before (once on ecstasy after a 1000 stack and 5 E DVDs and then again on xanax). It's the luck of the draw but being in a faction with good toleration is better.
I OD'd the other day after 1000 stack and a bunch of porns.
You could say I cried.
I've overdose 4 times in a week before. Wasn't fun. It happens though, oh well.
I'd say over half of my ODs have been this exact situation...I've shed many a tear and my window now has bars on it to catch my laptop when I chuck it out of it
Ugh, nothing worse than that. Let's cry together.
Quick math for this…
It looks like you took about 30 drugs during this time.
And lets assume each drug has a 2.5% chance of an overdose (pretty conservative).
The probability that one would have seen 4 or more overdose is then…
= 1 – P(0 OD) – P(1 OD) – P(2 OD) – P(3 OD)
= 1 – (0.975^30) – 30(0.975^29)(0.025) – 435(0.975^28)(0.025^2) – 4060(0.975^27)(0.025^3)
= 1 – 0.4679 – 0.3599 – 0.1338 – 0.0320
= 0.0064; 0.64%
Or about 1 in 157
So not too outrageous.
Doing a shitload of candies will increase your chance of OD