David's Epic Estimation Challenge

beingforitself

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Dec 14, 2008
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Yo,

Take every 6-color 3x3x3 ever produced. Of the approximately 4.3252 x 10^19 possible unique states a cube can be in, what percentage of these has ever been actually physically attained by any cube for any duration of time? This includes scrambling, solving, messing around, etc. by anyone anywhere at any time.

I am posting my calculations in white below; I'd like to compare with what/how other people estimate.


Some (not really) random numbers that may or may not be useful in this calculation:

According to this and this article "more than 350 million" cubes have been sold worldwide since it's introduction.

Kociemba has got some numbers here.

20 face turn metric moves (MOST ULTRA VERY LIKELY PROBABLY BUT NOT DEFINITELY) suffice to reach any state on the cube.

The WCA ranking and statistics pages are not fully working right now, but I am estimating that:
5000 people have any recorded average of 5 for 3x3x3
1500 people have a sub-20 average of 5 for 3x3x3
(will change if someone tells me the correct numbers)

My estimate:

Let us treat non-cubers, sup-20, and sub-20 cubers as three distinct subgroups of move-generating apparati:

Assume that
non-cubers (or to be more precise, single cubes that pass around between multiple non-cubers) number 3.5 x 10^8
sup-20 cubers number 10000
sub-20 cubers number 2000

For the first subgroup, assume that the cubes net an average of 10 hours mindless turn time at 20 turns a minute, making for a total cubespace exploration of 10 hours x 20x60 turns an hour x 3.5x10^8 cubes = 4.2 x 10^12 cubestates generated.

For the second subgroup, assume to reach a sup-20 average requires an average of 250 hours of cubing. Assume that this cubing time averages 2 TPS. This would give us 250 hours x 2x60x60 turns per hour x 10000 sup-20 cubers = 1.8 x 10^10 cubestates generated.

For the third subgroup, assume to reach a sub-20 average requires an average of 5000 hours to reach sub-20. Assume that this cubing time averages 3 TPS. This would give us 5000 hours x 3x60x60 turns per hour x 2000 sub-20 cubers = 1.1 x 10^11 cubestates generated.

The latter two subgroups (somewhat predictably) contribute negligibly to the total cubestate count according to my calculations, but I am leaving them there in the case that someone makes a comment / suggests different numbers that would make them matter.

Therefore we have 4.211 x 10^12 / 4.3252 x 10^19 =~ 9.73 x 10^-7 = .0000973 (or around one ten thousandth of one) percent of possible cube states have been explored in the history of the Rubik's cube's existence.

MAIN PROBLEMS WITH THIS METHOD (as I see it):

-> Assuming total turns = cubestates generated. I tried to make the numbers fairly conservative to "compensate" for this.
-> Actual numbers plugged in
 
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Assuming total turns = cubestates generated

Exactly. Some cubestates are much more common than others, such as the solved state, and all LL cases, even some f2l cases. So this may make the total number even less because of the sheer number of turns that have been made to get cubes into such states by cubers.
 
I would like to think ~4.3252 x 10^19. I haven't done any math so this may be a ridiculous guess. I just think that cubers go through a lot of them and have probably done all of the close to solved ones, and noncubers have done a lot of scrambling. :) I doubt I'm right, there are probably a lot of fully scrambled combinations that have never been done.
 
I would like to think ~4.3252 x 10^19. I haven't done any math so this may be a ridiculous guess. I just think that cubers go through a lot of them and have probably done all of the close to solved ones, and noncubers have done a lot of scrambling. :) I doubt I'm right, there are probably a lot of fully scrambled combinations that have never been done.
SO WHY POST IT?
Oh, and: you're way off.
 
id probably just get 45 quintillion and divide it by 350 million, times the number of people who actually play with there cube (attainable by a survey). that could give you a VERY VERY rough estimate i think.

The units of that answer are:
(possible permutations of a cube) * (people who have solved the cube) / (number of cubes that have existed)

The units of the answer asked for are: (possible positions that have been reached)

Basically, your answer should be a number of cube states. Your answer via that calculation is in positions-solvers per cube that has existed, which makes no sense.

I'm not trying to be rude, but please try again,
Chris
 
ok so...
(43 quintillion)(~1 billion)(~350 million) i say 1 billion solved becasue of speedsolvers and people solving more than once really adds up
i came out to the answer

~12,285,714,260

am i close?

EDIT: i give up
 
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ok so...
(43 quintillion)(~1 billion)(~350 million) i say 1 billion solved becasue of speedsolvers and people solving more than once really adds up
i came out to the answer

~12,285,714,260

am i close?

Unless each cube sold has seen less than 36 possibilities, no.
 
I can safely say that all 4.32E19 states of the 3x3x3 have not been seen yet by all the people of this planet. But I can be sure of this.

All 3.67E6 states of the 2x2x2 cube have been seen by now. And all 1 states of the 1x1x1 have bee seen by now. And all 933000 states of the pyraminx have been seen by now too.

And I'm sure then a computer fast enough to generate and solve all 4.32E19 states of the 3x3x3 cube comes along, we can say we've seen all the states of the 3x3x3.

So far, a computer has scanned abut 1E16 states of the 3x3x3 so far (or about 0.023% of total combinations).
 
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