/sci/ - Science & Math » Thread #13985744

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Anonymous Mon 20 Dec 04:18:12 2021 No.13985744 View Reply Original Report

Quoted By: >>13985755 >>13985851 >>13985862 >>13985877 >>13985886 >>13985902 >>13985908 >>13985929 >>13986292 >>13986294 >>13986360 >>13986392 >>13986838 >>13986843 >>13987122 >>13987530 >>13989393 >>13989409 >>13989562

Let's say I have a 100-sided die.
If I roll that die 100 times, what are the odds that I will roll a 100 at least once? I don't care how many times, it just has to be at least once.

I wrote a python script to estimate and it seems to converge to around 57.7%

57.7% sounds awfully arbitrary. Is there anything more to be said about this answer? I just find it really interesting.

Anonymous

Anonymous Mon 20 Dec 2021 04:20:36 No.13985755 Report

Quoted By: >>13985815

>>13985744
You might be encountering an artifact of how the Python math library generates random numbers. Have you tried any other languages?

Anonymous

Anonymous Mon 20 Dec 2021 04:29:50 No.13985793 Report

Quoted By: >>13985815

Try again using something more reliable amd acfually random like how bustabit gets its numbers

Anonymous

Anonymous Mon 20 Dec 2021 04:34:27 No.13985803 Report

Quoted By: >>13985815

The odds you roll a 100 at least once is 1 minus the odds you don't roll it at all, so 1 - (99/100)^100 = 63.3%. If you fix your random number gen like other anons say, it'll get you closer to that number.

Anonymous

Anonymous Mon 20 Dec 2021 04:35:03 No.13985805 Report

Quoted By: >>13985815

1-(99/100)^100 = 0.63396765872...

I think this is the correct answer, correct me if I'm wrong

Anonymous

Anonymous Mon 20 Dec 2021 04:37:22 No.13985815 Report

Quoted By: >>13985847 >>13985860 >>13986303

>>13985755
>>13985793
>>13985803
Didn't consider that. Thought I could trust something as simple as rolling 1-100, keeping count of rolls and wins and dividing rolls/wins.

>>13985803
>>13985805
That's a lot simpler, thanks. Still, why is it ~63.3%?

Anonymous

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Anonymous Mon 20 Dec 2021 04:45:29 No.13985831 Report

Quoted By:

i don't know, so let's reason it out. suppose you have an N sided die and want to know the likelihood of getting an N with M rolls.

let's count the number of possible outcomes and the number of outcomes with an N.

M = 1 roll: N outcomes, N - 1 outcomes fail
M = 2 rolls: N^2 (assume order matters), [N - 1]^2 outcomes fail
M = 3 rolls, N^3 outcomes, [N - 1]^3 outcomes fail
M rolls, N^M outcomes, [N - 1]^M outcomes fail

subtracting the total from the failed outcomes gives the passing outcomes, so given M rolls, N^M - [N - 1]^M outcomes pass.

takes the ratio with the total to compute the prob

P(N in M rolls) = [N^M - [N - 1]^M] / [N^M] = 1 - [1 - 1/N]^M

plugging in your numbers
P(100 in 100 rolls) = 1 - [1 - 1/100]^100 = 0.63397ish

seems somewhat close to your value. does my logic make sense. btw, i think this is markov chain shit

Anonymous

Anonymous Mon 20 Dec 2021 04:47:58 No.13985836 Report

Quoted By: >>13985882

Tried it in C# using the built in random number generator.
For exactly one roll of 100 per 100 tries, I got 36.9679% over a million trials (100 rolls per trial).
For at least one roll of 100 per 100 tries I got 63.0687% over a milion trials (100 rolls per trial).

Anonymous

Anonymous Mon 20 Dec 2021 04:50:42 No.13985847 Report

Quoted By: >>13985895 >>13985904 >>13985915

>>13985815
It's about 1 - 1/e. Comes up a lot in the asymptotic behavior of discrete distributions

Anonymous

Anonymous Mon 20 Dec 2021 04:51:11 No.13985851 Report

Quoted By:

>>13985744
How many times did you run the script? You might want to check it out something.
The probably of getting 100 at least once is
1-P(never 100)
= 1 - P(not 100 on 1st roll)*P(not 100 on 2nd roll)*...*P(not 100 on 100th roll)
= 1 - (1-1/100)^100
About 0.634

Anonymous

Anonymous Mon 20 Dec 2021 04:54:34 No.13985860 Report

Quoted By:

>>13985815
Be sure you don't have a off by one error. When you call the random number generator, be sure that it can return both 1 and 100 but not zero. Sometimes a call like Random(100) returns values 0-99 or 1-99. You have to check the library documentation to be sure.

Anonymous

Anonymous Mon 20 Dec 2021 04:55:21 No.13985862 Report

Quoted By: >>13985891

>>13985744
What's your python code?

Anonymous

Anonymous Mon 20 Dec 2021 04:59:51 No.13985877 Report

Quoted By:

>>13985744
$ bc -l <<< '1 - 0.99 ^ 100'
.63396765872677049507

Anonymous

Anonymous Mon 20 Dec 2021 05:01:28 No.13985882 Report

Quoted By:

>>13985836
After posting about checking for off-by-one errors, I found one in my own C# code. Now it comes out with 63.406597% over a million trials of 100 rolls each. (36.9153% if looking for exactly one roll of 100 per trial instead of one or more.)

Anonymous

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Anonymous Mon 20 Dec 2021 05:01:41 No.13985886 Report

Quoted By:

>>13985744

Anonymous

Anonymous Mon 20 Dec 2021 05:02:33 No.13985891 Report

Quoted By: >>13985962 >>13985977 >>13986840

>>13985862
Not sure if code tags work here but here you go:

[code:lit]
import random

i = 0
m = 100
ctr = 0
rep = 10000
wins = 0
losses = 0
bigwins = 0
biglosses = 0
random.seed()

while True:
ctr += 1
i = random.randrange(1,101)
if i == m:
wins += 1
else:
losses += 1

if ctr % 100 == 0:
if wins > 0:
bigwins += 1
else:
biglosses += 1
wins = 0
losses = 0

if ctr >= rep:
ctr = 0
print(str(float(biglosses)/float(bigwins)))
[/code:lit]

Anonymous

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Anonymous Mon 20 Dec 2021 05:03:38 No.13985895 Report

Quoted By: >>13985904 >>13985915

>>13985847
I was just going to post this. The answer approaches 1 - 1/e when the number of the sides on the dice and times rolled approach infinity.

Anonymous

Anonymous Mon 20 Dec 2021 05:06:29 No.13985902 Report

Quoted By: >>13986326

>>13985744
>around 57.7%
I can get the same result if I use a 118 sided die instead of a d100. Probably not helpful to you but interesting to me.

Anonymous

Anonymous Mon 20 Dec 2021 05:06:50 No.13985904 Report

Quoted By: >>13986374

>>13985847
>>13985895
I had a feeling it would be a relatively simple answer if e was involved. I suck at math so that constant is still fascinating to me. Or at least, how that constant manages to be irrational despite how easy it is to express.

Anonymous

Anonymous Mon 20 Dec 2021 05:07:31 No.13985908 Report

Quoted By:

>>13985744
>57.7%
wrong

1-P("no 100's") ? 0.634

Anonymous

Anonymous Mon 20 Dec 2021 05:09:47 No.13985915 Report

Quoted By:

>>13985895
>>13985847
Nice

Anonymous

Anonymous Mon 20 Dec 2021 05:15:41 No.13985929 Report

Quoted By: >>13985942

>>13985744

Seems to generally hit at least once with 100 attempts in ruby.

irb(main):053:0> 100.times.map { |o| [o, (1..100).to_a.sample] }.select { |r, n| n == 100 }.map { |r, n| "hit 100 on attempt #{r}" }
=> ["hit 100 on attempt 34"]

irb(main):054:0> 100.times.map { |o| [o, (1..100).to_a.sample] }.select { |r, n| n == 100 }.map { |r, n| "hit 100 on attempt #{r}" }
=> ["hit 100 on attempt 74"]

irb(main):055:0> 100.times.map { |o| [o, (1..100).to_a.sample] }.select { |r, n| n == 100 }.map { |r, n| "hit 100 on attempt #{r}" }
=> ["hit 100 on attempt 91"]

Anonymous

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Anonymous Mon 20 Dec 2021 05:19:10 No.13985941 Report

Quoted By:

This comes up with results near 63.4%.

Anonymous

Anonymous Mon 20 Dec 2021 05:19:21 No.13985942 Report

Quoted By:

>>13985929
P(0) = 0.366
P(1) = 0.367
P(2) = 0.185
P(3) = 0.061

Anonymous

Anonymous Mon 20 Dec 2021 05:27:27 No.13985962 Report

Quoted By: >>13985977

>>13985891
>print(str(float(biglosses)/float(bigwins)))
Your error is here.

the god form..

the god form.. Mon 20 Dec 2021 05:27:39 No.13985963 Report

Quoted By: >>13985977 >>13985978

everyone here is a retard. If you have a bubble inside the dice it will lean towards a specific number or couple of numbers depending where the bubble is and how deep it is.

Anonymous

Anonymous Mon 20 Dec 2021 05:30:48 No.13985977 Report

Quoted By: >>13986183 >>13986592

>>13985963
While your statement is correct, since OP is simulating the die roll in a computer, an ideal die is assumed.
>>13985891
>>13985962
Try
>print(str(float(bigwins)/float(bigwins+biglosses)))

Anonymous

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Anonymous Mon 20 Dec 2021 05:30:48 No.13985978 Report

Quoted By:

>>13985963

Anonymous

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Anonymous Mon 20 Dec 2021 06:32:55 No.13986183 Report

Quoted By:

>>13985977
>While your statement is correct, since OP is simulating the die roll in a computer, an ideal die is assumed.
Ugh, why did I never THINK of that?

Anonymous

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Anonymous Mon 20 Dec 2021 06:48:21 No.13986239 Report

Quoted By: >>13987795

Could you kill someone with a hundred sided dice?

Anonymous

Anonymous Mon 20 Dec 2021 07:04:46 No.13986292 Report

Quoted By:

>>13985744
First do the real math when you come up with exact solution then compare with your program's results.
Goddamn, zoomers these days.

Anonymous

Anonymous Mon 20 Dec 2021 07:05:34 No.13986294 Report

Quoted By:

>>13985744
If you roll the dice, there is 1% chance of 100.

Anonymous

Anonymous Mon 20 Dec 2021 07:07:50 No.13986303 Report

Quoted By:

>>13985815
>That's a lot simpler, thanks. Still, why is it ~63.3%?

Do you have ANY math understanding or are just a zoommie script kiddie?

Anonymous

Anonymous Mon 20 Dec 2021 07:14:06 No.13986326 Report

Quoted By: >>13986337

>>13985902
is a proper 118 sided die even possible?

Anonymous

Anonymous Mon 20 Dec 2021 07:17:25 No.13986337 Report

Quoted By:

>>13986326
Not sure but I was using a computer program to simulate it. Turns out OP's problem was that he calculated the ratio of failures to successes instead of successes to number of attempts.

Anonymous

Anonymous Mon 20 Dec 2021 07:26:41 No.13986360 Report

Quoted By:

>>13985744
1-0.99^100

Anonymous

Anonymous Mon 20 Dec 2021 07:30:57 No.13986374 Report

Quoted By: >>13986377

>>13985904
>I suck at math so that constant is still fascinating to me
I'm good at maths and I find it more fascinating each year.

Anonymous

Anonymous Mon 20 Dec 2021 07:33:06 No.13986377 Report

Quoted By: >>13986388 >>13986656

>>13986374
I am great at maths and I find myself fascinating

Anonymous

Anonymous Mon 20 Dec 2021 07:37:33 No.13986388 Report

Quoted By:

>>13986377
e

Anonymous

Anonymous Mon 20 Dec 2021 07:40:29 No.13986392 Report

Quoted By: >>13986592 >>13989587

>>13985744
I can't believe everyone is talking about how random numbers are generated. There are so many other problems with using python for probabilities. One of the biggest is that it uses floats instead of doubles by default. Fix that and your answer will get much closer than if you use a cryptographically secure random

Anonymous

Anonymous Mon 20 Dec 2021 09:04:23 No.13986592 Report

Quoted By: >>13986603 >>13986655

>>13986392
OP's problem was fixed in >>13985977
He performed the wrong calculation.

the god form..

the god form.. Mon 20 Dec 2021 09:09:07 No.13986603 Report

Quoted By:

>>13986592
i know im awesome

Anonymous

Anonymous Mon 20 Dec 2021 09:46:28 No.13986655 Report

Quoted By:

>>13986592
that is hilarious

El Arcón

El Arcón Mon 20 Dec 2021 09:47:24 No.13986656 Report

Quoted By:

>>13986377
Nice.

Anonymous

Anonymous Mon 20 Dec 2021 11:20:13 No.13986838 Report

Quoted By:

>>13985744
1 - (99/100)^100

Anonymous

Anonymous Mon 20 Dec 2021 11:21:14 No.13986840 Report

Quoted By:

>>13985891
Do mathlets really?

Anonymous

Anonymous Mon 20 Dec 2021 11:25:31 No.13986843 Report

Quoted By:

>>13985744
>what are the odds that I will roll a 100 at least once?
What are the odds that you roll anything BUT 100, 100 times in a row?

Anonymous

Anonymous Mon 20 Dec 2021 13:17:15 No.13987122 Report

Quoted By: >>13987506 >>13989355

>>13985744
How are you rolling it? When does it land each time and how much force do you apply?

Anonymous

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Anonymous Mon 20 Dec 2021 15:28:31 No.13987506 Report

Quoted By: >>13987519

>>13987122
I roll it down a slope of 40 degrees with a small ramp upwards at 23 degrees from a height of 1 meter. I apply no force, just let it go by its own weight of 40 grams. It usually lands after 0.43s on average over a trial of 20 000 rolls.

Anonymous

Anonymous Mon 20 Dec 2021 15:31:41 No.13987519 Report

Quoted By: >>13987548

>>13987506
Do you position the die the exact same way each time? Like with specific numbers in specific locations?

Anonymous

Anonymous Mon 20 Dec 2021 15:33:33 No.13987530 Report

Quoted By:

>>13985744
1 - 0.99^100

Anonymous

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Anonymous Mon 20 Dec 2021 15:37:35 No.13987548 Report

Quoted By: >>13987620

>>13987519
yes, sorry I forgot to mention it. I place it with 19 on top. Rolling resistance is 0.02 and everything happens on Earth with normalized air density and pressure.

Anonymous

Anonymous Mon 20 Dec 2021 15:53:22 No.13987620 Report

Quoted By:

>>13987548
Let's see, I suppose it comes down to how many arrangements you can have with a 19 on top. I'm gonna estimate that since 19 isn't in the OP pic, its on the bottom bracket.

Anonymous

Anonymous Mon 20 Dec 2021 16:46:20 No.13987795 Report

Quoted By: >>13987806

>>13986239
most likely if you shoved it in their throat

Anonymous

Anonymous Mon 20 Dec 2021 16:50:14 No.13987806 Report

Quoted By: >>13987810

>>13987795
or arse

Anonymous

Anonymous Mon 20 Dec 2021 16:51:44 No.13987810 Report

Quoted By: >>13987835

>>13987806
kek reminds me of the pear of anguish

Anonymous

Anonymous Mon 20 Dec 2021 16:57:50 No.13987835 Report

Quoted By: >>13987878

>>13987810
the sex toy varian or the medieval torture?

Anonymous

Anonymous Mon 20 Dec 2021 17:10:50 No.13987878 Report

Quoted By:

>>13987835
mostly torture because we're talking about offing someone with a giant die, but i guess if said victim was into really kinky shit it could be seen as a sex toy lol

Anonymous

Anonymous Mon 20 Dec 2021 17:12:38 No.13987884 Report

Quoted By:

finish high school then post here

Anonymous

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Anonymous Mon 20 Dec 2021 23:14:26 No.13989355 Report

Quoted By:

>>13987122
I just chunk them in a dice tower but you'd need an extra large one for a d100. Easier to use a computer to simulate.

Anonymous

Anonymous Mon 20 Dec 2021 23:28:07 No.13989393 Report

Quoted By: >>13989832

>>13985744
yoo, this got me thinking. If I got big enough dice (in the sense of sides N), is there a number that would break pythons rng? In the sense that, the ratio of rolls less then N would not converge to 1-((N-1)/N)^N as the number of test rolls goes to infinity?

Anonymous

Anonymous Mon 20 Dec 2021 23:33:01 No.13989409 Report

Quoted By:

>>13985744
>Let's say I have a 100-sided die.
>If I roll that die 100 times, what are the odds that I will roll a 100 at least once?
1-0.99^100 ~ 63.4% ~ 1-1/e

lim as n-> inf of 1-(1-1/n)^n = 1-1/e

Anonymous

Anonymous Tue 21 Dec 2021 00:16:30 No.13989562 Report

Quoted By:

>>13985744
What is the probability that you don’t roll a 100 even once? It’s (1-1/100)^100.
Notice that (1+n/k)^k is roughly e^n for large k. So the probability of not rolling a 100 is roughly e^-1.

Anonymous

Anonymous Tue 21 Dec 2021 00:25:19 No.13989587 Report

Quoted By:

>>13986392
>One of the biggest is that it uses floats instead of doubles by default.
It uses double precision floats, you retard.

Anonymous

Anonymous Tue 21 Dec 2021 01:46:29 No.13989832 Report

Quoted By:

>>13989393
n=9,007,199,254,740,992 is the max for a 53 bit number, which is what Python uses for the random function. After that, it starts skipping numbers.

Anonymous

Anonymous Tue 21 Dec 2021 01:48:52 No.13989839 Report

Quoted By:

>Almost all module functions depend on the basic function random(), which generates a random float uniformly in the semi-open range [0.0, 1.0). Python uses the Mersenne Twister as the core generator. It produces 53-bit precision floats and has a period of 2**19937-1. The underlying implementation in C is both fast and threadsafe. The Mersenne Twister is one of the most extensively tested random number generators in existence. However, being completely deterministic, it is not suitable for all purposes, and is completely unsuitable for cryptographic purposes.
For our purposes, it works just fine. Also funny to note that it is implemented in C under the hood.

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