Current Events > Can anyone tell me the probability of this happening?

(+) Yeah! (+)

Theres a ~30-40% chance of me getting something. After 33 tries, I only got said thing once :<

I want to know how bad my luck is because the online calculator thing doesnt go past 1/10,000 <_<

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Yeah, you're going to have to be more specific. Depending on the action that needs to be done, and your involve in it.

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If I'm thinking this right, in the best (or worst for you) case scenario:

(0.4)((0.6)^32)*33

which is an astronomically low number (0.0001050543266512924916739735552%)

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Let;s call it 35%

35/100
...
7/20 ^ 33 = .0000000000000009 to 1

BUT, this presumes you got the item on the 33rd try.

In any case, this highlights the problems with RNG in most video games.

Robot2600 posted...

Let;s call it 35%

35/100
...
7/20 ^ 33 = .0000000000000009 to 1

BUT, this presumes you got the item on the 33rd try.

In any case, this highlights the problems with RNG in most video games.

That would be the probability of getting the item 33 times

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RNGesus hates me :<

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YugiNoob posted...

Theres a ~30-40% chance of me getting something. After 33 tries, I only got said thing once :<

I want to know how bad my luck is because the online calculator thing doesnt go past 1/10,000 <_<

Always go for Wolfram. (https://is.gd/8khDa3)

Anyhow, we want to focus on a binomial distribution with probability p = 0.35 (to keep it nice and round). Assuming you didn't just stop after getting the first, we consider the number of ways in which you could get one success, which would be given by 33-choose-1, and multiply by the relevant probabilities.

The end result is C(33,1) * (0.35)^1 * (0.65)^30, or about 0.0028%.

But this is a flawed calculation for a number of reasons, but it gives a rough estimate I suppose.

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Eevee-Trainer posted...

(0.65)^30

This should be (0.65)^32

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Eevee-Trainer posted...

Always go for Wolfram. (https://is.gd/8khDa3)

Anyhow, we want to focus on a binomial distribution with probability p = 0.35 (to keep it nice and round). Assuming you didn't just stop after getting the first, we consider the number of ways in which you could get one success, which would be given by 33-choose-1, and multiply by the relevant probabilities.

The end result is C(33,1) * (0.35)^1 * (0.65)^30, or about 0.0028%.

But this is a flawed calculation for a number of reasons, but it gives a rough estimate I suppose.

Yeah, I wouldve kept going but I ran out of skip tickets at 33 and like fuck Im gonna farm those stages manually >:(

I have terrible luck in games sometimes. Like with this draw in Arknights; literally half was the same character, and four times were in a ROW



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LordFarquad1312 posted...

This should be (0.65)^32

Yeah my bad

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