Probability Thread

kcl · Feb 24, 2014

Odds of a complete cross skip?

Jaycee · Feb 24, 2014

kclejeune said:
Odds of a complete cross skip?

(1/12)*(1/11)*(1/10)*(1/9)*((1/2)^4) ------- that last term is really 1/16 so it can be written as (1/(12*11*10*9*16))

1/190080.

Then again it's 1 AM and I'm tired so I could be wrong.

kcl · Feb 24, 2014

Jaycee said:
(1/12)*(1/11)*(1/10)*(1/9)*((1/2)^4) ------- that last term is really 1/16 so it can be written as (1/(12*11*10*9*16))

1/190080.

Then again it's 1 AM and I'm tired so I could be wrong.

Does that change if I say odds of a complete cross skip on white? I suck at probability lol.

Jaycee · Feb 24, 2014

kclejeune said:
Does that change if I say odds of a complete cross skip on white? I suck at probability lol.

I was just about to edit the post saying that I think I did the math assuming one cross color, so I don't think so. :3

Tim Major · Feb 24, 2014

Kit Clement said:
I'm going to assume you meant 15552, not 15522. Also assuming no edge/OLL control of any sort.

Simpler question - what's the chance of none happening? In one solve, 15551/15552. Take that to the 15552nd power to get the probability of no LL skips in that many solves.

(15551/15552)^15552 = .368 or 36.8%

Thus, the probability of at least one happening is the complement of that.

1 - .368 = .632 or 63.2%

.501 = (15551/15552)^x

x = 10780

So if one was to do 10780 solves, it is more likely for them to have had an LL skip, than for them to have not.

And considering most people will atleast use SOME basic edge control, in an average of 10000 it is likely one will have an LL skip.

On an unrelated note, this reminded me of the birthday pairs problem, for those who haven't seen it;

how many people would you need to randomly pick out of a crowd before it is more likely than not (>50% chance) of two of the subjects sharing a birthday (as in, 24th of Feb, not 24th of Feb in a certain year)

TDM · Feb 24, 2014

Tim Major said:
On an unrelated note, this reminded me of the birthday pairs problem, for those who haven't seen it;

how many people would you need to randomly pick out of a crowd before it is more likely than not (>50% chance) of two of the subjects sharing a birthday (as in, 24th of Feb, not 24th of Feb in a certain year)

I've seen it before.

Isn't it 23 or something? I might have remembered it wrong, but I'm 99% sure that's it. Well, I'm at least 50% sure anyway

I was explaining this to my brother the other day and he didn't believe me, neither did anyone else who heard, so idk if it is actually 23.

CyanSandwich · Feb 25, 2014

TDM said:
I've seen it before.

Isn't it 23 or something? I might have remembered it wrong, but I'm 99% sure that's it. Well, I'm at least 50% sure anyway I was explaining this to my brother the other day and he didn't believe me, neither did anyone else who heard, so idk if it is actually 23.

http://en.wikipedia.org/wiki/Birthday_problem

Logiqx · Feb 25, 2014

kclejeune said:
Odds of a complete cross skip?

http://www.speedsolving.com/forum/s...ability-Thread&p=951837&viewfull=1#post951837

Explanations are in following posts.

Logiqx · Feb 25, 2014

Tim Major said:
On an unrelated note, this reminded me of the birthday pairs problem, for those who haven't seen it;

how many people would you need to randomly pick out of a crowd before it is more likely than not (>50% chance) of two of the subjects sharing a birthday (as in, 24th of Feb, not 24th of Feb in a certain year)

TDM said:
I've seen it before.

Me too... I often consider it during my life as a software developer.

Applied to cubing: If you take 6 random solves there's >50% chance of getting the same PLL twice.

You can illustrate this with six random numbers between 1 and 18:
http://www.random.org/integers/?num=6&min=1&max=18&col=6&base=10&format=html&rnd=new

The odds of at least 2 identical numbers are around 55%. This is an over-simplification because some PLL cases aren't 1/18 chance but it is close enough!

ChickenWrap · Feb 25, 2014

Logiqx said:
Me too... I often consider it during my life as a software developer.

Applied to cubing: If you take 6 random solves there's >50% chance of getting the same PLL twice.

You can illustrate this with six random numbers between 1 and 18:
http://www.random.org/integers/?num=6&min=1&max=18&col=6&base=10&format=html&rnd=new

The odds of at least 2 identical numbers are around 55%. This is an over-simplification because some PLL cases aren't 1/18 chance but it is close enough!

Aren't there 21 plls?

AmazingCuber · Feb 25, 2014

ChickenWrap said:
Aren't there 21 plls?

But most cases have a probability of 1/18. Check badmephisto's PLL page for more info.

Carrot · Feb 25, 2014

TDM said:
I've seen it before.

Isn't it 23 or something? I might have remembered it wrong, but I'm 99% sure that's it. Well, I'm at least 50% sure anyway I was explaining this to my brother the other day and he didn't believe me, neither did anyone else who heard, so idk if it is actually 23.

It's 23.

ceil(sqrt(2*n*log(2))+(3-2*log(2))/6) = birthday problem for years with n days.
put 365 in it and you get 23 ;-) (source: David Brink: "A (probably) exact solution to the Birthday Problem" it's an interesting paper)

for the how many times should I solve a cube before I have more than 50% chance of getting a PLL/OLL/LL skip: ceil(log(50%)/log((#-1)/#)) where #=15552 for LL skip, figure out how to use it

TDM · Feb 25, 2014

CyanSandwich said:
http://en.wikipedia.org/wiki/Birthday_problem

Oh cool, didn't know there was an entire Wikipedia page on it. ty.

Carrot said:
It's 23.

ceil(sqrt(2*n*log(2))+(3-2*log(2))/6) = birthday problem for years with n days.
put 365 in it and you get 23 ;-) (source: David Brink: "A (probably) exact solution to the Birthday Problem" it's an interesting paper)

for the how many times should I solve a cube before I have more than 50% chance of getting a PLL/OLL/LL skip: ceil(log(50%)/log((#-1)/#)) where #=15552 for LL skip, figure out how to use it

I have no idea of what ceil or log means, but if #=15552 for LL skip, I don't think I need to

The only problem is that it doesn't work in Google, which I usually use as a calculator, and I don't think my calculator has ceil. Do you know an online calculator that has it/if it's shown as something else on other calculators?

cmhardw · Feb 25, 2014

TDM said:
I have no idea of what ceil or log means...

ceil and log are functions. The ceiling function is often used to help with precisely calculating an integer answer when your intermediate results are decimal numbers. The Logarithm function is essential when working with exponents.

TDM · Feb 25, 2014

cmhardw said:
ceil and log are functions. The ceiling function is often used to help with precisely calculating an integer answer when your intermediate results are decimal numbers. The Logarithm function is essential when working with exponents.

Oh cool, there's not actually that complicated. ty!

Logiqx · Feb 25, 2014

Carrot said:
It's 23.
(source: David Brink: "A (probably) exact solution to the Birthday Problem" it's an interesting paper)

There are several approximations for the birthday problem but I like the two below. The second one is pretty easy to remember:

0.5 + sqrt(0.25 - 2 * N * log(0.5))

0.5 + 1.17741 * sqrt(N)

Here's a little Python script with a selection of values for N (re: cubing, computing and birthday):

Code:

import math
for n in [18, 54, 256, 365, 15552, 65536, 4294967296]:
    approx1 = 0.5 + 1.17741 * math.sqrt(n)
    approx2 = 0.5 + math.sqrt(0.25 - 2 * n * math.log(0.5))
    approx3 = math.sqrt(2 * n * math.log(2)) + (3 - 2 * math.log(2)) / 6
    print n, approx1, approx2, approx3

Results:

18 5.49532757142 5.52028868693 5.26427860676
54 9.15216115415 9.166596535 8.92111225942
256 19.33856 19.3451945187 19.1075113001
365 22.9943864654 22.9999431512 22.7633378354
15552 147.332043776 147.332897895 147.100997524
65536 301.91696 301.917380472 301.685916704
4294967296 77163.24176 77163.2432372 77163.0121865

cuber952 · Feb 26, 2014

This might be stupid but, do some OLLs stand a better chance of getting a PLL skip than others? If so, which OLLs give PLL skips most often?

- fedora

Jaycee · Feb 26, 2014

I'd think not, because AFAIK orientation and permutation are independent of each other.

Stefan · Feb 26, 2014

cuber952 said:
This might be stupid but, do some OLLs stand a better chance of getting a PLL skip than others? If so, which OLLs give PLL skips most often?

- fedora

Who is fedora?

guysensei1 · Feb 26, 2014

What is the probability of getting 1 ortega face solved on 2x2? What about a LBL type face?

I guess this isn't probability, but what are the god's numbers for ortega faces and LBL faces?

Thread starter	Similar threads	Forum	Replies	Date
	Cube probability and coin flips	Puzzle Theory	2	Jul 25, 2017
	Solvable cube if stickers randomly applied? What's the chance?	Puzzle Theory	12	Apr 1, 2016
	Branching OLL to PLL Cases?	General Speedcubing Discussion	9	Nov 30, 2014
U	Is an infinite scramble really "random"?	Puzzle Theory	22	Jul 3, 2014
	Probability of having a pre-made pair in a ZZ solve?	Puzzle Theory	11	Dec 30, 2010

Probability Thread

Member

Member

Member

Member

Platinum Member

Member

Member

Member

Member

Member

Member

Member

Member

Premium Member

Member

Member

Member

Member

Member

Member