Probability Thread

[Stefan]

It does depend on it

No it doesn't.

Imagine there were just two OLL cases, one with 40% chance and one with 60% chance. Now what's the probability of getting the same case in two attempts?

The answer is 52%.

There. A perfectly fine answer to a perfectly fine question. Doesn't depend on the OLL case. Also note it's not 50%.

 

[guysensei1]

Umm... Was my question unclear? I'm not understanding this debate.

 

[Berd]

Umm... Was my question unclear? I'm not understanding this debate.

Just let it happen bebe

 

[Stefan]

Umm... Was my question unclear? I'm not understanding this debate.

Your original question was alright (the only problem I see is that you didn't specify whether you're just doing two solves, though that's a reasonable assumption). I'm not sure about your addition of "The average probability?".

 

[guysensei1]

I'm not sure about your addition of "The average probability?".

I'm not sure either. I was slightly confused at obelisk's response. I assumed he was talking about the probability of a single OLL and PLL case so I asked for the average of all of the probabilities but now that I think about it I make no sense

 

[obelisk477]

I'm not sure either. I was slightly confused at obelisk's response. I assumed he was talking about the probability of a single OLL and PLL case so I asked for the average of all of the probabilities but now that I think about it I make no sense

Yeah, the average of all probabilities is in fact 1/58 and 1/22 for OLL and PLL respectively, but as it turns out those don't really matter for the calculation.

EDIT: Would it still give you a good approximation to do it the way i tried? instead of summing all 57 OLLs or whatever

 

[Stefan]

Ugh, my edit was too slow, so let me move it here:

I think what you wanted is like the 52% of my example. If you understand that calculation, you can do it for real OLL with the probabilities from here. I'm too lazy, plus I'd rather have people do such stuff themselves if I believe they can.

 

[guysensei1]

Let me confirm this...

I can get the answer by doing
(Probability of OLL1)^2+(probability of OLL2)^2..........+(Probability of OLL57)^2+(probability of OLL skip)^2

And multiply that by

(Probability of PLL1)^2+...+(Probabilty of PLL21)^2+(Probabilty of PLL skip)^2 ?

 

[cmhardw]

Yeah, the average of all probabilities is in fact 1/58 and 1/22 for OLL and PLL respectively, but as it turns out those don't really matter for the calculation.

Oh wow! I never realized this! So:
(P(OLL1)+P(OLL2)+...+P(OLL58))/58 = 1/58

This is true because:

Spoiler

Given a universe with events a_1, a_2, ... , a_n that form a partition of the total space of outcomes then:

1/n * (P(a_1)+P(a_2)+...+P(a_n)) = 1/n

Because P(a_1)+P(a_2)+...+P(a_n)=1 since these events form a partition of the total possible outcomes.

Very cool! I'm excited to have learned something!

Let me confirm this...

I can get the answer by doing
(Probability of OLL1)^2+(probability of OLL2)^2..........+(Probability of OLL57)^2+(probability of OLL skip)^2

And multiply that by

(Probability of PLL1)^2+...+(Probabilty of PLL21)^2+(Probabilty of PLL skip)^2 ?

define event e_1 to be OLL1 followed by PLL1
define event e_2 to be OLL1 followed by PLL2
...
define event e_22 to be OLL1 followed by PLL22
define event e_23 to be OLL2 followed by PLL1
etc.

What is the probability of getting event e_1 twice in a row? Of getting event e_2 twice in a row? Of getting event e_3 twice in a row?

Now what is the probability of getting the same OLL and PLL pair twice in a row?

--edit--

Looking at your response I think what you and I have said is equivalent

 

[Stefan]

Let me confirm this...

I can get the answer by doing
(Probability of OLL1)^2+(probability of OLL2)^2..........+(Probability of OLL57)^2+(probability of OLL skip)^2

And multiply that by

(Probability of PLL1)^2+...+(Probabilty of PLL21)^2+(Probabilty of PLL skip)^2 ?

Yes.

 

[Stefan]

Oh wow! I never realized this! So:
(P(OLL1)+P(OLL2)+...+P(OLL58))/58 = 1/58

I might have never thought about it, either. I did find it interesting as well. Then again, I think it's only useful if you want to lie with statistics (if you say "average probability" instead of "probability", I suspect most people will ignore the "average", and you can legitimately change the answer that way if you like it better than the real answer)

Looking at your response I think what you and I have said is equivalent

Same idea, though his way is easier since OLL and PLL are independent and he exploits that (especially if you also exploit that many cases share the same probability).

 

[Antonie faz fan]

which oll and pll have the biggest propability to occure?? and wich ones the least? ( all edges down all corners up and E perm ?)

 

[Hssandwich]

which oll and pll have the biggest propability to occure?? and wich ones the least? ( all edges down all corners up and E perm ?)

H-perm?

 

[cmhardw]

which oll and pll have the biggest propability to occure?? and wich ones the least? ( all edges down all corners up and E perm ?)

OLL
PLL

 

[JustinTimeCuber]

OLL-20: 1/216 = 0.046%

PLL-H: 1/72 = 1.389%


OLL/PLL skips have equally low probabilities to the ones listed.

 

[Suzuha]

OLL-20: 1/216 = 0.046%

PLL-H: 1/72 = 1.389%


OLL/PLL skips have equally low probabilities to the ones listed.

That's interesting. I got that OLL like 6 times but like 2 OLL skips.

 

[cmhardw]

That's interesting. I got that OLL like 6 times but like 2 OLL skips.

Your distribution of X-OLLs and OLL skips is not as rare as it might seem

The probability above is the probability of two or fewer OLL skips in 8 "significant outcomes" where each "significant outcome" is a solve that has either an X-OLL or an OLL skip.

It is often the case that what seems at first to be a strange outcome is more common than you would expect.

 

[unsolved]

OLL-20: 1/216 = 0.046%

PLL-H: 1/72 = 1.389%


OLL/PLL skips have equally low probabilities to the ones listed.

So what are the solutions to each of those cubes that are the fewest number of moves? In the case of the top one, the last-layer-superflip is the case I am interested in.

 

[Lucas Garron]

So what are the solutions to each of those cubes that are the fewest number of moves? In the case of the top one, the last-layer-superflip is the case I am interested in.

Visit ACube.js, twist every LL edge, set "ignore position" for every LL piece, then "solve graphical input":
Gets you: F B' R2 D2 R F2 B2 L F B' . U2 R2 // 18q, 12f, 9s

EDIT: Simplified using slice moves: S' U2 R2 U S2 U S' U2 R2

 

[TheCoolMinxer]

What is the chance of getting a l3e skip on 7x7, being left with inner parity on the last edge, after realigning the centers?
Happened to me last weekend at a comp, overall PB by 25 sec! (3:58)