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Probability Thread

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%
Click to expand...

.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)
 
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)
Click to expand...
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 :p 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.
 
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 :p 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.
Click to expand...
http://en.wikipedia.org/wiki/Birthday_problem
 
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)
Click to expand...

TDM said:
I've seen it before.
Click to expand...

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!
 
Last edited:
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!
Click to expand...

Aren't there 21 plls?
 
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 :p 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.
Click to expand...

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 :)
 
Last edited:
CyanSandwich said:
http://en.wikipedia.org/wiki/Birthday_problem
Click to expand...
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 :)
Click to expand...
I have no idea of what ceil or log means, but if #=15552 for LL skip, I don't think I need to :p 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?
 
Carrot said:
It's 23.
(source: David Brink: "A (probably) exact solution to the Birthday Problem" it's an interesting paper)
Click to expand...

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
 
Last edited:
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?
 
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