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pi question

Anonymous Sun 04 Aug 20:34:30 2019 No.10863606 View Reply Original Report

Quoted By: >>10863629 >>10863632 >>10864333 >>10864868 >>10865004

What's the probability that there's a Graham's number of zeroes in a row at some point in the decimals of pi? 0%? 100%? 50%?

Anonymous

Anonymous Sun 04 Aug 2019 20:37:22 No.10863615 Report

Quoted By:

100%

Anonymous

Anonymous Sun 04 Aug 2019 20:39:20 No.10863624 Report

Quoted By: >>10863636

50%, either it's there or it isn't

Anonymous

Anonymous Sun 04 Aug 2019 20:41:04 No.10863629 Report

Quoted By:

>>10863606
100%

Anonymous

Anonymous Sun 04 Aug 2019 20:42:10 No.10863632 Report

Quoted By: >>10864338 >>10865001 >>10865035

>>10863606
100%. Every combination of digits exists within Pi and they are all normal numbers.

Anonymous

Anonymous Sun 04 Aug 2019 20:43:12 No.10863636 Report

Quoted By: >>10863654

>>10863624
>either it's there or it isn't
That translates to "100% or 0%", not 50%

Anonymous

Anonymous Sun 04 Aug 2019 20:44:54 No.10863643 Report

Quoted By: >>10864333

There's the sentence "if you've calculated this far, you're a giant faggot" in binary at digit 4.2*10^10^59

Anonymous

Anonymous Sun 04 Aug 2019 20:47:28 No.10863654 Report

Quoted By: >>10863727

>>10863636
(100+0)/2

Anonymous

Anonymous Sun 04 Aug 2019 21:17:38 No.10863727 Report

Quoted By: >>10864333

>>10863654
You can't average mutually exclusive truths. If something eternally is or eternally never was, it's 100% or 0%. You're talking about whether a certain piece of truth exists, not the result of a coin flip. The OP's question is stupid in that he asks about the 'probability' of something that either is or isn't, which may have confused you.

Anonymous

Anonymous Mon 05 Aug 2019 01:20:52 No.10864333 Report

Quoted By:

>>10863606
47.368%
>>10863643
prove it
>>10863727
BASED

Pi !!DjPgmuTe1Oe

Pi !!DjPgmuTe1Oe Mon 05 Aug 2019 01:23:26 No.10864338 Report

Quoted By: >>10864365 >>10864979

>>10863632
>Every combination of digits exists within Pi and they are all normal numbers.
This is false. Because for a number n digits long, it would have to end at least n digits into pi. Yet there are 10^n numbers that are n digits long. And only one of those 10^n numbers can exist in the first n digits of pi. So that means that only 1 out of 10^n of all possible numbers exist in pi, with n approaching infinity.

Anonymous

Anonymous Mon 05 Aug 2019 01:37:25 No.10864365 Report

Quoted By:

>>10864338
Right from the pi hole.

Anonymous

Anonymous Mon 05 Aug 2019 04:26:39 No.10864868 Report

Quoted By: >>10864876

>>10863606
Zero and 100. In theory the idea of pi is a square /_\ that’s been rotated already once. Since it has a flat surface. That means that in 3.14 the number one represent a reflection or a rotation to the same spot. So Pi already measures a circle via a triangle. So it must keep summarizing itself the longer it goes on the more 0s in a row you’ll see. It means it’s recapping its discernment. There’s a specific logic Pi uses and it keeps going back to 1 to graduate itself again and again and again a graham amount of times.

Anonymous

Anonymous Mon 05 Aug 2019 04:29:30 No.10864876 Report

Quoted By:

>>10864868
Pi is greater than Graham’s equivalent. Because there needs to be way way more than a graham amount of digits to even have Graham’s amount of 0s. Ultimately the graham equivalent is you. When you stop calculating.

Anonymous

Anonymous Mon 05 Aug 2019 05:22:54 No.10864979 Report

Quoted By:

>>10864338
that's actually a cool way to look at it. Thanks anon.

Anonymous

Anonymous Mon 05 Aug 2019 05:32:31 No.10865001 Report

Quoted By:

>>10863632
Only true if ? is a normal number, which is speculated yet unknown.

Anonymous

Anonymous Mon 05 Aug 2019 05:34:52 No.10865004 Report

Quoted By:

>>10863606
100%

Anonymous

Anonymous Mon 05 Aug 2019 05:50:48 No.10865035 Report

Quoted By:

>>10863632
>Every combination of digits exists within Pi and they are all normal numbers.
Prove it, faggot.

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