Probability Thread

i just did an ao12 where i got 2 h perms followed by three y perms all in a row

What's the chance of getting the same pll 2 times in a row followed by another pll 3 times in a row?

It depends which plls you have twice.. H perm is 1/72 and Y perm is 4/72 (correct if I'm wrong)
So first you get your H perm
Than there's 1/72 chance that you get it once again, then you get another pll, then again so multiply the first 1/72 by 4/72, and again because you got the y perm three times!
So prob(first pll)xprob(second pll)^2
It gives 1/72x4/72x4/72, which is 1/23328

Probability of getting the same PLL on 4x4 3 times in a row. It was the pure diag corner alg with half of yperm, pll parity, rest of y perm. I dunno if the chance of getting the same PLLs on a 4x4 is lower than on a 3x3...

Idk what you're talkin about, but the probability of a pll on 4^3 is twice lower than on 3^3, because pll parity is 1/2
But if you don't use reduction, like I do sometimes, you can get a looot more differents pll's, that's another story^^

If I well understood the question (I did not read old posts)
I think that the probability of 5 pll in a row is just
p(1 pll)*p(2 pll)*p(3 pll)*p(4 pll)*p(5 pll) , where p(x pll) is the probability of specific pll for instance p(H)=1/72 (badmephisto.com)
regardless if some pll are the same.
having H cannot influence next pll then there is not a p(H "after a previous H")

That's right and wrong
If i say I'll get 5 h perms in the 5 next cubes, the probability is p(H)^5, which is ridiculously low :3
But if I say i'll get 5 times the same pll, but without saying which one it'll be, the probability is just p(first pll I'll get)^4 ! It depends on the pll you get first, you can get a rough estimate but no more until you know which pll is your first one. The second probability is obviously higher than the first one.
(The last thing you said is obviously right, im talking about your algorithm to calculate the prob of 5 plls)

uhm,
some time passed from my studies, but if I well remember
if you normalize to 72 (p(T)=4/72), 5 pll can form 72^5 combinations.
The event "single pll" is indipendent from the other, then if I want to guess a 5pll combination
and I say UaTTHJb then p(UaTTHJb) should be p(Ua)*p(T)*p(T)*p(H)*p(Jb)
that is also equal to p(TTTTJb) because p(T)=p(Ua)=p(Jb)=4/72 (badmephisto.com)
and is also equal to p(TJbTTT)
Is that right ? Help me remember such rules!!

That's right, except that you clearly have not 72^5 combinations of 5 plls! You do if you count AUFs, but actually nobody says "i got a Jb perm with -1/4 auf"
21 plls equal 21^5 different combinations
Edit: i love editing
P(Ua,T,T,H,Jb) is not equal to P(TTTTT), because of H perm! The plls you talked about have 1/18 chance to happen, but the H perm has 1/72

Right,
otherwise also p(T) should be 1/72 ...

I never say that

The other day I got a OLL+PLL skip on 4x4, but had double parity. Anyone know the probability for that?

Just had 4 consecutive PLL skips occur during a 4x4x4 session.

Usually that's 144^4 = a 1 in 429,981,696 chance, but there were 2 OLLCPs, 1 1LLL case and a normal skip, which a few of have worked out to be 1/82,944. Is this correct?

Few days ago I got 2 LL skips in a row on 3x3.
What is propabilty of that? Also I was playing with some OLL skip techniques so, wasn't it double PLL skip?
What I did:
1st: RU'R' F'UF
2nd: case when evrything oriented, skip after using z rotation and J perm insertion.

Kudz said:

Few days ago I got 2 LL skips in a row on 3x3.
What is propabilty of that? Also I was playing with some OLL skip techniques so, wasn't it double PLL skip?
What I did:
1st: RU'R' F'UF
2nd: case when evrything oriented, skip after using z rotation and J perm insertion.

Click to expand...

Neither of them are LL skips.

What are the chances of getting L4E skip in 4x4+ cubes? I have got only like twice it on 4x4, and never on bigger cubes. So what is the probability for that?

joshsailscga said:

The other day I got a OLL+PLL skip on 4x4, but had double parity. Anyone know the probability for that?

Click to expand...

I got OLL+PLL skip yesterday, but got PLL parity. For getting OLL+PLL skip chance are of course the same as getting LL skip, but if you mean with parity, Idk then

joshsailscga said:

The other day I got a OLL+PLL skip on 4x4, but had double parity. Anyone know the probability for that?

Click to expand...

isn't it just chance of an LL skip*parity*parity ((2^3*3!*3^3*3!)*2*2)?

shadowslice e said:

isn't it just chance of an LL skip*parity*parity ((2^3*3!*3^3*3!)*2*2)?

Click to expand...

The total number of positions of the 4x4x4 last layer with paired dedges is:
If he is talking about that he got the following case, then there are 2 possible ways he could get it (it or its inverse) with regards to AUF.

Therefore, the final number is 2/(62,208) = 1/(31,104), which is equivalent to your answer.

Christopher Mowla said:

The total number of positions of the 4x4x4 last layer with paired dedges is:

http://i.imgur.com/BLEjTu3.gif
​

If he is talking about that he got the following case, then there are 2 possible ways he could get it (it or its inverse) with regards to AUF.

https://www.speedsolving.com/wiki/images/9/95/DblParity.png
​

Therefore, the final number is 2/(62,208) = 1/(31,104), which is equivalent to your answer.

Click to expand...

Sounds cool, thanks for the input. I don't know if it would change any of the numbers but the OLL parity was the speedsolving alg that flips one dedge and swaps two more, as well as flipping + swapping two corners.

joshsailscga said:

Sounds cool, thanks for the input. I don't know if it would change any of the numbers but the OLL parity was the speedsolving alg that flips one dedge and swaps two more, as well as flipping + swapping two corners.

Click to expand...

It sounds like you're describing the following case:


If this is so, this is not double parity, as the double parity case which looks a lot like this does not switch two dedges in addition to flipping one.

So, with this being the case, then since this case is equal to its inverse, there is only one version of it (with respect to AUF). We therefore have 1/(62,208).

What is the probability of getting a 3BLD Scramble with no cycle breaks? My Monty Carlo simulations give 736/100130 or roughly 0.74%.