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Anonymous Tue 23 Jun 18:44:57 2020 No.11829164 View Reply Original Report

Quoted By: >>11830917 >>11830935 >>11831816

Well, /sci/?

Anonymous

Anonymous Wed 24 Jun 2020 03:54:35 No.11830917 Report

Quoted By:

>>11829164
10, or I guess anything over 9.3 if fractional balls were a thing

Anonymous

Anonymous Wed 24 Jun 2020 04:03:58 No.11830935 Report

Quoted By:

>>11829164
5

Anonymous

Anonymous Wed 24 Jun 2020 06:43:27 No.11831286 Report

Quoted By: >>11831530

I think its 5, its trying to make you think that pulling 10 white balls is easier if all 10 are white but the odds of all 10 being white are extremely small. So its far more likely there was 5ish balls and you happen to draw the same balls over and over.

But im a brainlet so i could be wrong.

Anonymous

Anonymous Wed 24 Jun 2020 07:06:58 No.11831322 Report

Quoted By: >>11831655 >>11831823 >>11831926

$A$ : You draw 10 white balls
$B_n$ : There are n white balls in the box

$P(B_n|A) = \frac{P(A | B_n) P(B_n)}{ P(A) }$

$P(B_n) = \frac{{10 \choose n}}{1024}$
$P(A | B_n) = \left( \frac{n}{10} \right)^{10}$
$P(A) = \sum_{k=0}^{10} P(A | B_k) P(B_k) = \frac{111304237}{8000000000}$

This gives
$P(B_0) | A) = 0 \\ P(B_1) | A) = \frac{1}{14246942336} \\ P(B_2) | A) = \frac{36}{111304237} \\ P(B_3) | A) = \frac{177147}{3561735584} \\ P(B_4) | A) = \frac{172032}{111304237} \\ P(B_5) | A) = \frac{123046875}{7123471168} \\ P(B_6) | A) = \frac{9920232}{111304237} \\ P(B_7) | A) = \frac{847425747}{3561735584} \\ P(B_8) | A) = \frac{37748736}{111304237} \\ P(B_9) | A) = \frac{3486784401}{14246942336} \\ P(B_{10}) | A) = \frac{7812500}{111304237} $

So $P(B_8 | A)$ is the largest.

Anonymous

Anonymous Wed 24 Jun 2020 09:10:51 No.11831530 Report

Quoted By:

>>11831286
The problem there is that if there are five white balls and five black balls, then you have 5/10 (or 1/2) chance to draw a white ball every time, and if you repeat that ten times, then the odds of getting white ten times are the same as having tossed a coin ten times and gotten the same result everytime.

Anonymous

Anonymous Wed 24 Jun 2020 10:43:48 No.11831655 Report

Quoted By: >>11831659

>>11831322
In English, professor?

Anonymous

Anonymous Wed 24 Jun 2020 10:47:07 No.11831659 Report

Quoted By:

>>11831655
There are most likely 8 white balls in the box.

Anonymous

Anonymous Wed 24 Jun 2020 12:17:37 No.11831816 Report

Quoted By: >>11831823

>>11829164
P(n|10W) = P(n) P(10W|n) / P(10W)

= ((10 choose n)/2^10) (n/10)^10 / P(10W)

So all we need to do is find the maximum of (10 choose n)n^10. This is at n = 8

Anonymous

Anonymous Wed 24 Jun 2020 12:20:08 No.11831823 Report

Quoted By:

>>11831322
Thorough but inefficient. See >>11831816

Anonymous

Anonymous Wed 24 Jun 2020 13:03:09 No.11831926 Report

Quoted By:

>>11831322
Nice. But is there a less tedious way to calculate P(A) ?

Anonymous

Anonymous ## Mod Wed 24 Jun 2020 15:26:34 No.11832290 Report

Quoted By:

This thread was moved to >>>/b/830706645

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