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Probability Thread

I think it is much higher than that.
Yeah, just googling (without quotes):
how many people know how to solve the rubik's cube

The first search result states:
It is estimated that only 5.8% of the total population can solve the Rubik's cube. So if you are among the elite ones, you deserve to take pride in the feat achieved. While learning to solve the Rubik's cube the mind multi-tasks.

So according to that, P(someone can solve the 3x3x3 cube) is approximately 29/500.

(There are A LOT of people who can solve the cube who never participated in a competition, and I'm guessing that percentage is going to increase over time as the thresholds for being able to enter become more strict... as the overall times continue to decrease and cubing tutorials get better and better.)
 
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and I think 10x isn’t TOO far off
I disagree.
The Rubiks Cube had it's biggest spike in popularity in the early 80s where it was THE popular thing to have and booklets about how to do it were cheap.
This also means that a silent majority from then never went to a competition and those that did were largely not documented.
 
I disagree.
The Rubiks Cube had it's biggest spike in popularity in the early 80s where it was THE popular thing to have and booklets about how to do it were cheap.
This also means that a silent majority from then never went to a competition and those that did were largely not documented.
That’s a fair assessment, I completely forgot about the 80’s.
 
Does not seem to add up.

If there was an event that had a 1/2 probability (heads up coin) and you did 5 tries, what are the chances at least 2 are heads up? Using your equation, (2^2)/5 would give 1/0.8, which does not make much sense.
 
If an event has a 1/54 chance of happening in a solve, in an Ao5, what is the chance the event happening at least twice in the average (a counting solve with the event)
Well, the probability of it happening twice in a given set of 2 solves is 1/54^2, or 1/2916. Since there are 5!/3!2! = 10 possible sets of 2 solves in an ao5 (also denoted as 5C2, pronounced 5 choose 2), the total probability is 10/2916

As always, correct me if I'm wrong
 
What is the chance for 2 5 movers in a row on 2x2?

I just got it so Im asking
That would be \( (9992/3673775)^2 \) , which is about 0.0007397% (I think, I might've placed the decimal point wrong)
It's easy to misinterpret these kinds of probabilities, I'm lazy so I'll just quote someone else's post.
It is indeed easy to misrepresent probabilities. What you are asking is 'what is the probability of getting a LL skip and PLL skip in a given mean of 3?' However, that probability is obviously a lot lower that the probability of someone getting a mo3 with both skips at some point in his life.

This seems like a better representation:
Pretty much every somewhat serious cuber gets a LL skip at some point. So, assume someone gets a LL skip. What is the probability there is a PLL skip on a different solve in the same mean? Well, there are 4 solves, 2 before and 2 after the LL skip, for which a PLL skip would satisfy. So, the probability becomes 4/72, which is 5.6%.

The probability of this happening at some point to a given person is even higher, because the number of LL skips you get in your life isn't quite limited to 1.
 
I just got 14 solves in a row on 4x4 with OLL or PLL Parity, or both. (One solve which cost me PB). Out of those 14 solves, 12 were OLL parity. I also got a streak of 9 OLL parities in a row. Would be cool if someone can calculate the percentage of all this happening.
 
probability of a 2 mover on 3x3? idk if someone asked this before "two" is too short to look up on the site and there's so many posts in this thread it would take forever to find this specific question.
How many 2 move states exist:
18 possible 1st moves, 18 possible second moves. If the moves are parallel, the order doesn't matter. We eliminate all scrambles that start with a U, F or L move and are followed by any of the 3 parallel moves, so that's 3*3*3 scrambles states eliminated. Therefore, there are 18^2 - 27 2 move states.

Divide this by the total number of states and simplify and you get a probability of 1 in 14562964065484800

If you want the probability of getting it in competition, there are 19 fewer total possible states so the probability is 297 in 43252003274489855981, non-simplifyable and basically the same probability, but since I accounted for the 27 duplicate 2 move states I couldn't just ignore 19 illegal states I suppose (although they're waaaayyy less significant than the 27 duplicate states considering their respective proportion to the the total they belong to)
 
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