Probability Thread

[ruffleduck]

That is true. Okay so I guess I will only use ZBLL in this example. I am aware that depending on the probability the case appears, you may need to learn more algs to get 50% of ao5s full ZBLL. What is the min-max number of ZBLLs you would need to know to have a 50% chance of having full ZBLL during an ao5?

ZBLLprobability^5 = 0.5.
(ZBLLprobability^5)^(1/5) = (0.5)^(1/5)
ZBLLprobability = 0.871
You would need to know 87.1% of all ZBLL's to have a full ZBLL ao5 50% of the time.
0.871 * 493 = 430
87.1% of all ZBLL's, rounded up, is 430 algs.

 

[ruffleduck]

Does anyone know the odds of, after solving your third pair, you get a fourth pair skip into an OLL skip?

Assuming you use CFOP (the probability would be a lot lower if you use ZZ hehe)

4th pair skip probability = 1 / (2 * 3 * 5 * 5) = 1/150
OLL skip probability = 1 / (3^3 * 2^3) = 1/216

probability of both = (1/150) * (1/216) = 1/32400

 

[Skewby_Doo]

Anyone know the chances of a solved center in 4x4?

 

[Jorian Meeuse]

Just interested after seeing this on JDAW6's Progression, what's the chance of getting a LL Skip and a PLL skip in the same Mo3?

Assuming the two are in different solves, that would be 1/15552(LL skip)*1/72(PLL skip)*6, which is about
0.0005%.

It is indeed easy to misrepresent probabilities. What you are asking is 'what is the probability of getting a LL skip and PLL skip in a given mean of 3?' However, that probability is obviously a lot lower that the probability of someone getting a mo3 with both skips at some point in his life.

This seems like a better representation:
Pretty much every somewhat serious cuber gets a LL skip at some point. So, assume someone gets a LL skip. What is the probability there is a PLL skip on a different solve in the same mean? Well, there are 4 solves, 2 before and 2 after the LL skip, for which a PLL skip would satisfy. So, the probability becomes 4/72, which is 5.6%.

The probability of this happening at some point to a given person is even higher, because the number of LL skips you get in your life isn't quite limited to 1.

 

[cubenerd74]

Anyone know the chances of a solved center in 4x4?

That would be 24*6/24*23*22*21=1/1771=0.05647%.

 

[Davian]

Anyone know the chance of an cubeshape done scramble on square 1? I got this recently and I think its pretty rare.

 

[ProStar]

It is indeed easy to misrepresent probabilities. What you are asking is 'what is the probability of getting a LL skip and PLL skip in a given mean of 3?' However, that probability is obviously a lot lower that the probability of someone getting a mo3 with both skips at some point in his life.

This seems like a better representation:
Pretty much every somewhat serious cuber gets a LL skip at some point. So, assume someone gets a LL skip. What is the probability there is a PLL skip on a different solve in the same mean? Well, there are 4 solves, 2 before and 2 after the LL skip, for which a PLL skip would satisfy. So, the probability becomes 4/72, which is 5.6%.

The probability of this happening at some point to a given person is even higher, because the number of LL skips you get in your life isn't quite limited to 1.

Especially since JDAW an LL skip every week

 

[JDAW6]

No, I don’t get a ll skip every week. Please stop spread rumours and false info.

Especially since JDAW an LL skip every week

 

[SpeakCuba]

No, I don’t get an ll skip every week. Please stop spreading rumours and false info.

Please stop spreading…false info.

 

[JDAW6]

No, I don’t get a ll skip every week. Please stop spread rumours and false info.

Why are people mehing and giving bad reactions to this post? It is straight facts

 

[SpeakCuba]

Why are people mehing and giving bad reactions to this post? It is straight facts

We don’t want to get into this, but you say it is “straight facts” that you don’t get a LL skip every week, which would be true of all cubers. However, at the same time you claim that you get a LL skip every couple of days on your progression thread.

 

[JDAW6]

We don’t want to get into this, but you say it is “straight facts” that you don’t get a LL skip every week, which would be true of all cubers. However, at the same time you claim that you get a LL skip every couple of days on your progression thread.

I did that once, I only have 4 ll skips

 

[SpeakCuba]

I did that once, I only have 4 ll skips

Cool.

 

[OtaMota]

What the probability of a l4e skip on 4x4? Just got PB3 because I skipped L4E.

 

[OtaMota]

Still want to know how rare a 4x4 l4e skip is. (Reduction in case that affects anything)

 

[OtaMota]

Fine, I will do it myself, but would like some to fact check this.

We have a total of 8 wings, and we do not need to consider the orientation of them, just permutation. So do I just do 8! to do this? This is quite easy to calculate if that is the case.

1/40320

2.6 times rarer than a LL skip on 3x3.

 

[ruffleduck]

Fine, I will do it myself, but would like some to fact check this.

We have a total of 8 wings, and we do not need to consider the orientation of them, just permutation. So do I just do 8! to do this? This is quite easy to calculate if that is the case.

1/40320

2.6 times rarer than a LL skip on 3x3.

Hey, sorry for not trying to answer your question earlier. Your attempt is incorrect. It is correct that 8! represents the number of the wings' permutations, but there is more than 1 state of wings' permutations that classifies as a L4E skip. The number of states that would be a L4E skip would be the number of permutations and orientations 4 normal 3x3x3 edges could experience which would be 4!*2^4 = 384. Therefore, the probability of L4E skip is 384/40320 = ~0.01%.

Please correct me if I missed something

 

[OtaMota]

Hey, sorry for not trying to answer your question earlier. Your attempt is incorrect. It is correct that 8! represents the number of the wings' permutations, but there is more than 1 state of wings' permutations that classifies as a L4E skip. The number of states that would be a L4E skip would be the number of permutations and orientations 4 normal 3x3x3 edges could experience which would be 4!*2^4 = 384. Therefore, the probability of L4E skip is 384/40320 = ~0.01%.

Please correct me if I missed something

Your 384/40320 (1/105) and 0.01% do not add up.

 

[ruffleduck]

Your 384/40320 (1/105) and 0.01% do not add up.

Good point. I shouldn't have put the percentage symbol: the probability is ~0.01 = ~1%

 

[OtaMota]

Good point. I shouldn't have put the percentage symbol: the probability is ~0.01 = ~1%

Thank you for your help. I guess it is not as rare as I thought, but the wow reactions on my original post made it seem rare, as it might imply they might not have had l4e skip.