Current Events > Do any mathemagicians on CE want to explain some probability bullshit to me?

(+) Yeah! (+)

My friend is trying to calculate some shit for a video game he plays competitively I think. Basically though, his problem comes down to:

Given a 1/3 probability of success, what are the odds that over 60 attempts he'll succeed exactly 6 times?

No, I'm not asking you to do my homework for me. I did some googling for him and found out about Bernoulli Trials to calculate this for me, and found that the answer was roughly ~0.002%. Incredibly unlikely.

However, we changed a couple of the parameters to see. What if we reduce the number of successes and attempts from 6 and 60 respectively, to 1 and 10, does that affect things?

Turns out it affects them a FUCKTON. It boosted the probability of getting that many successes from ~0.002% to ~8.6%. Literally 4300 times more likely to succeed.

I even ran some tests of both to see if were correct. Sure enough...

https://cdn.discordapp.com/attachments/206112433809522689/717425849711722496/unknown.png
https://cdn.discordapp.com/attachments/206112433809522689/717429517454540820/unknown.png

(ignore the comments in the second one lol)

This is just wild to me. It's the same ratio of successes to total attempts. But somehow it being a little bit smaller of sample size, even if it repeats over a million times, results in a 4300 times better success rate. There's gotta be some of that math magics involved here somewhere, right?

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If you're trying 60 times, you're bound to hit the 1/3 probability more than 6 times, so it makes complete sense to me

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Assume that it's not random, and instead every third attempt you succeed.

So if you try 60 times, you'd be succeeding 20 times.

Meanwhile, if you try 10 times, you'd succeed only 3 times.

3 is a lot closer to 6 than 20.

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parabola_master posted...

If you're trying 60 times, you're bound to hit the 1/3 probability more than 6 times, so it makes complete sense to me

That's not the question I'm asking though. Of course that's the case.

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My head hurts ;_:

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Make sense to me. With 60 trials there are way possible combinations. Also, why does it have to be exactly 6? Why not 6 or more?

Anyway, this seems like a combinatorics problem.

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Jabodie posted...

Make sense to me. With 60 trials there are way possible combinations. Also, why does it have to be exactly 6? Why not 6 or more?

I dunno, this is just the question he asked our discord server, and it's basically the exact kinda problem that bernoulli trials solve.

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Nanaimo-Bar posted...

, and it's basically the exact kinda problem that bernoulli trials solve.

bernoulli trials are if each interaction has a success or fail. You've created two success outcomes. 1.) that thing x happens 2.) that thing x happens exactly y times in z attempts.

if you were only doing 1 it would be relevant, since you aren't, it isn't and it isn't any probability bullshit, it's as others explained to you, changing Z is determining the chance of exactly y happening.

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But yeah, with more trials, the pdf is going to concentrate more tightly around the true probability of success. It's like when you do Monte Carlos with more complex scenarios. If you take this out to infinitely many trials, the probably for less or more than 1/3 of them being successes goes to zero.

Edit: but yeah, if you've ever run Monte Carlos it's basically just saturating to the true probability.

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I won't try to get into any specifics, but think it this way:

For 60 attempts you should succeed ~20 times and for 10 you should succeed ~3.

So, 6 out of 60 is not very likely, and much the same 1 out of 10 isn't either.

But 6 out of 60 is like doing the 1 out of 10 6 times in a row. So you take something unlikely and then ask what the odds are of doing that unlikely thing even more times in a row and you can see why 6 out of 60 is much more unlikely.

I haven't done this in a long time, but I'll try.
Babymode: how many different "number of successes" can you hit? With 10 attempts, 11 (0-10), with 60 attempts, 61. If each number of successes was equally likely, you'd see 9.09% chance at any one out of 11 or 1.64% for any one out of 61.
This is just to demonstrate that there should be a difference.

Going back to the question, each number of successes is not equally likely, you're going to have a bell curve. To demonstrate why your numbers are what they are, I would make a table of other values and the chance at each.

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Been a long time since high school, but here goes:

For 6 successes out of 60 attempts:
(60! / (6! * 54!)) * (1/3)^6 * (2/3)^54 = 0.00212%

For 1 success out of 10 attempts:
(10! / (1! * 9!)) * (1/3)^1 * (2/3)^9 = 8.67076%

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Weird that this just came up in an unrelated video I watched: https://youtu.be/0-ux3G3Tq8A?t=139

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