kcl
Member
Odds of a complete cross skip?
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.
Does that change if I say odds of a complete cross skip on white? I suck at probability lol.
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%
I've seen it before.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.
Odds of a complete cross 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)
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!
Aren't there 21 plls?
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.
Oh cool, didn't know there was an entire Wikipedia page on it. ty.
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?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
It's 23.
(source: David Brink: "A (probably) exact solution to the Birthday Problem" it's an interesting paper)
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
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