Human random numbers

[evilbeanfiend]

lets do some science and see how good a human based RNG this fora is

(And no fair looking at the results before choosing a number.) - gm

[Technical Ben]

Could we take this a step further and see if we can make other processes out of forumites? Then we can build a central processor, then some sort of program all run on forum posts... http://xkcd.com/505/

[Blatm]

I like these sorts of polls. I can't wait to see how it will turn out (if we ever get enough people to participate).

[Velifer]

My hypothesis (pre-vote)

Spoiler:

bell-curve. Skewed toward a mean around 7.

My hypothesis (post-vote)

Spoiler:

I suck at guessing.

[Ptolom]

Spoiler:

I predict a profusion of 17, 18, 19's people tend to pick a bigger number, not a "round" number like 1 ,5 ,10 or twenty, not an even number and possibly a prime. In an attempt to be more random, they do the very opposite.

[Sana]

The limited choices affect the results, and therefore the test is flawed.

[itaibn]

After voting and looking at the results, I decided the majority of xkcd is too clever to fall for any of the usual tricks.

[gmalivuk]

The limited choices affect the results, and therefore the test is flawed.

It's not flawed. The test is to see if the general population of people taking this poll are capable of picking numbers randomly. If there is a strong preference for some numbers over others, it's strong evidence against that.

Personally, I think

Spoiler:

numbers like 7 and 17 that first seem "more random" to people will actually be picked less than average in this case, because people are consciously avoiding them.

This gives me an idea for a forum game...

[Charlie!]

Spoiler:

13 is unlucky, apparently.

[spdqbr]

I chose 14 as it's the closest to (pi %1), which is fun.

[Zeno's Pair of Ducks]

I am very disappointed in all of you.

[yy2bggggs]

I picked 14 because in the word "STAN" using modulo 20 arithmetic, T is the additive identity and S and A are inverses, leaving N which is 14. "STAN" came from my makeshift entropy selection strategy.

[evilbeanfiend]

After voting and looking at the results, I decided the majority of xkcd is too clever to fall for any of the usual tricks.

i don't know, there seems to be a slight bias towards numbers > 10. i was going to assume a model for error in each bin of the sqrt of the number in the bin, so the distribution is slightly flatter then it looks but still not flat within errors (feel free to suggest better error models btw)

[Knightshire]

I think there are too many choices with not enough people to do some real statistical analysis.
It would be more meaningful if there were only 5 choices.

[Technical Ben]

Spoiler:

Hmm. I chose 14. I was watching snooker, so that may have influenced me (scores on the tv and all that). Also 4 is my favourite number, but I thought that was too obvious a choice. It's great to see "my" number is winning currently!

[evilbeanfiend]

I think there are too many choices with not enough people to do some real statistical analysis.
It would be more meaningful if there were only 5 choices.

well the results already look statistically significant to me, if it were flat we would expect with 100 votes every bin to contain 5 +/- sqrt(5), some bins are clearly way off this.

[d0nk3y_k0n9]

I'll vote as soon as I can find my d20... or a decent dicegen.

Lol it was sitting buried on my desk. That didn't take long.

[quintopia]

2

[Ieatsoap6]

I find it interesting that 6 and 14 are winning since they're symmetric about 10.

I chose 16 because it was the right choice.

[gmalivuk]

well the results already look statistically significant to me, if it were flat we would expect with 100 votes every bin to contain 5 +/- sqrt(5), some bins are clearly way off this.

Yeah, but isn't that only one standard deviation? Being outside of one sigma is not exactly significant, I don't think.

[evilbeanfiend]

yes its not amazing, and since then it already looks flatter.

[BlackSails]

I actually chose mine randomly, by closing my eyes and scrolling up and down. My mouse landed on 12, so thats what I chose

[scikidus]

1. Because I can.

[Yakk]

If you rolled a d20 100 times, the 95% conf. interval would be about:
2 * sqrt(.95*.05*n) = 4.35889894
with an expected value of 5.

So for a given random digit, you'd expect somewhere between 0.85 and 9.35 votes on it.

However, the problem is we are looking at all 20, and searching for one that is outside that range. It is perfectly reasonable for 1 or more attempts, out of 20, to be outside the 95% confidence interval, even if the results where uniformly selected. That is what the 95% confidence interval means.

The fact we have only 2 outside of that interval is a sign that we aren't being that crappy of a RNG, using this naive test.

[megabnx]

Interesting. As of the time that I voted (randomly by scrolling mouse without looking) I voted for 14, giving 14 a lead with 14 votes. Now if it were the mean and median too I would be really spooked...Its not...

[fwlbg]

Get about 20480 poll answers for this one, then you can analyse it quite well.

I answered, but I'm not saying what I put.

[Bassoon]

3. Prime number FTW.

[yy2bggggs]

This is simple, so I did it.

Each column but the first represents a tabulation of 127 "votes" for the number in the first column, using random.org to generate the numbers.

Code: Select all

1 | 3| 4| 3| 5| 4|
2 | 9| 8| 4| 5| 4|
3 | 9| 7| 5| 9| 3|
4 | 7| 6| 7|10| 5|
5 | 4| 8| 3| 8|11|
6 | 5| 8| 5| 1| 8|
7 |10|10| 7| 6| 9|
8 | 6| 6| 8| 7| 7|
9 | 4| 6| 8| 3| 4|
10| 5| 6| 9| 1| 1|
11| 3| 8| 9| 7| 8|
12| 5| 6| 8| 6| 8|
13| 4| 1| 8| 6| 3|
14| 8| 7| 8| 7| 7|
15|10| 1| 9| 7| 3|
16|17|11| 8| 6|13|
17| 4| 4| 4|10| 3|
18| 4| 4| 3| 4| 9|
19| 6| 7| 5| 9| 8|
20| 4| 9| 6|10| 9|


Enjoy!

[Iv]

import random

random.choice(range(20))+1

It told me 16.

[gmalivuk]

If you rolled a d20 100 times, the 95% conf. interval would be about:
2 * sqrt(.95*.05*n) = 4.35889894
with an expected value of 5.

So for a given random digit, you'd expect somewhere between 0.85 and 9.35 votes on it.

Why the 2, if this is one-sided as implied by your last sentence?

[Yakk]

The 95% confidence interval is about 1.96 or so (or was it 1.97? I forget), IIRC. I find using 2 (as in within 2 SD) to be easier, and (in any case) less bound to the arbitrary 95% number. And only off by 2%.

[gmalivuk]

Ah, right. I just saw twice the standard deviation and was thinking, well, one on each side of the mean. (Forgot temporarily what a confidence interval even is, it seems.)

[Yakk]

So there are 2^20 subsets of the above. Minus 1, because we don't care about the empty subset. That is about 1 million subsets.

So we want on the order of an all but 1 in a million confidence interval for an arbitrary subset.

To be quick about it, this the SD is maximal at 1/2 of the entries (or 10 sides), at which the SD is about 1/2 * sqrt(1/N) in percentage terms.

Being strict about it, we have Chebyshev's inequality, which states at last (1 - 1/k^2) fraction of observations are within k standard deviations of the mean. Which places it at ~1000 standard deviations. (I kid!)

Assuming normally distributed data (which at the scales involved, we'll have hit), 5 sigma is enough for 1 in a million or so (actually 1 in 2 million, but who is counting).

So that is 5/2 * sqrt(1/N) is an (overly generous) standard. If you can find many subsets that are more than 5/2 * sqrt(1/N) standard deviations away from the expected value of that subset (given a random population), you have some strong evidence that the results aren't random.

With 1 million subsets and a 1 in 2 million chance that a given subset would randomly lie outside the 5 sigma range, call it a 1 in 3 chance that a given set of random numbers has an element outside of the 5 sigma range. Sadly, these correlate (sigh).

So... how about a single subset more than 6 sigma away. That's 3 * sqrt(1/N).

On 200 elements, that means +/- 21%. Ie, find a subset of N die sides that is at least N*5%+21%, or less than N*5%-21%.

[theta4]

It's a random number generator. If everybody picked 7, it would still be random.




And, lv:

import random

random.choice(range(20))+1

It told me 16.

afaik, that doesn't qualify as a human random number

[Tass]

I chose the wrong forum for my experiment I see. I should have tried to sell it as science, then I migth have gotten better statistics.

Interestingly it seems that people browsing the science forum is much more random than those in the silly games forum.

[PM 2Ring]

Allegedly, if you ask people to pick a random number between 10 & 20, you get a lot of 17s. But I don't think this works so well on a bunch of scientists, mathematicians & programmers. The idea is that odd numbers are considered more random than even ones, eleven is special because it has repeated digits, 13 is special because it's unlucky, 15 is too round, and so the first "non-special" number in the range is 17.

[Velifer]

And people like "M" better than "Q". (The Pepsi Challenge, for those of you too young to remember the 8-track.)

There are some fun mentalism tricks that exploit these quirks. You're thinking of a carrot, aren't you?

[Simius]

Maybe this should be made into a Global Announcement for a little while... for SCIENCE!

[MarshyMarsh]

I didn't choose my number randomly, I chose 9 because it is 9.

You're experiment is now void as not every number has been chosen by the being choosing them randomly due to me! I picked 9 because it is 9!

This also gives me an idea for a forum game, rather than team post racing. Number racing!

[quintopia]

i think your non-randomness still ends up being pretty random in the scheme of things. on the other hand, the person who used a software RNG IS interfering with the test.