Probability Thread

[Sa967St]

I'm sorry, but that's not even close.

Odds of getting a skip : 1/72
Odds of not getting a skip : 71/72
Solves with a skip : 3/5

1/72^3 * 71/72^2 * 5/3 = 1/226,845

I would think that's how you would do it, correct?

You got ninja'd, but I got a different final answer. Did you mean 5C3 instead of 5/3? I have no idea where you got the 226 845 from.

Also odds aren't the same thing as probability.
The odds of a PLL skip are 1:71 whereas the probability/likelihood/chance of a PLL skip is 1/72.

 

[cubernya]

You got ninja'd, but I got a different final answer. Did you mean 5C3 instead of 5/3? I have no idea where you got the 226 845 from.

Also odds aren't the same thing as probability.
The odds of a PLL skip are 1:71 whereas the probability/likelihood/chance of a PLL skip is 1/72.

Yeah I was thinking 5C3 but for whatever reason put down 5/3. Not sure why

 

[avgdi]

This isn't cubing related but I wanted to know if someone could figure out the probability of something for me.

A couple of years ago I randomly picked up a deck of cards. After shuffling it multiple times I guessed what color the top card would be before flipping it over. I got it right, so next I guessed which suit the second card would be. After getting that right I guessed which number the third card was going to be. I got that one right too and was rather surprised, so then I guessed what the exact card would be. Some how I got that one right too.

It was pretty crazy and me and my friends spent the next hour trying to do it again, but no one ever got it.

I was wondering what the probability of that happening is.

 

[qqwref]

First card's color: 1/2
Second card's suit: 1/4
Third card's number: 1/13
Fourth card exactly: 1/52
Since they are all independent events, you can get the probability by just multiplying everything together: 1/2 * 1/4 * 1/13 * 1/52 = 1/5408.

(This is approximate, though, because each card you already saw does influence the information of the next cards. For instance, if the first card was red, the second card would be more likely to be a spade or club than a heart or diamond. And the fourth card would actually be chosen out of 49 cards since you already know three cards it can't be. However, since I don't know what your actual guesses were, these approximate numbers give a pretty good estimate.)

 

[StachuK1992]

Not really probability, but how many distinct cases are there for the last 2 slots? (last 2 slot multislot.)

 

[jonlin]

Could you do the math for me?
There are 80 competitors at a competition.
What are the chances that all OLLs appear during that competition during all three rounds.
Assume that 24 get into the 2nd round, 12 get into the finals.
Also, for the 2x2 round, assuming there are 40 competitors, and 2 rounds, assuming 12 get into the finals, what is the chance that you get all CLL/EG cases on the last layer?
They don't have to be using CLL or EG.

 

[Sa967St]

Could someone please explain to me how 1/72 and 1/216 (?) are the probabilities of getting a PLL/OLL skip?

OLL skip:
P(EO skip) = 1/(2^3) = 1/8
P(CO skip) = 1/(3^3) = 1/27
P(EO+CO skip) = 1/(8*27) = 1/216

PLL skip:
P(CP skip, regarding AUF) = 1/6
P(EP skip | corners are solved) = 1/12
P(CP+EP skip) = 1/(12*6) =72

 

[Stefan]

P(CP skip) = 1/(4!/2) = 1/12
P(EP skip) = 1/3! = 1/6

CP and EP both permute four pieces.
P(CP skip) and P(EP skip) should be the same.

 

[Sa967St]

CP and EP both permute four pieces.
P(CP skip) and P(EP skip) should be the same.

For CP I ignored EP, and for EP I took CP in consideration. I know independently that they're both 1 in 12, but together you have to half it because of P parity. I updated my post.

 

[Stefan]

For CP I ignored EP

But for CP alone (ignoring EP) it's either 1/24 (if you don't allow AUF) or 1/6 (if you do allow AUF). I don't see how it's 1/12.

 

[Sa967St]

But for CP alone (ignoring EP) it's either 1/24 (if you don't allow AUF) or 1/6 (if you do allow AUF). I don't see how it's 1/12.

I don't know, I thought it was 1/12 not allowing AUF, I guess I'm wrong.
How is a PLL skip 1/72 then?

 

[Stefan]

There are 4!*4!/2=288 permutations (EP*CP/parity), of which 4 are skips (because of AUF). And 4/288 = 1/72.

 

[Christopher Mowla]

@Sarah, just use the number of permutations formula for the LL for these. For any of these skips, just start with the fraction:
\( \frac{\frac{1}{2}\left( \overset{\text{Edges}}{\mathop{\left( 4! \right)\left[ 2^{3} \right]}}\, \right)\left( \overset{\text{Corners}}{\mathop{\left( 4! \right)\left[ 3^{3} \right]}}\, \right)}{\frac{1}{2}\left( \overset{\text{Edges}}{\mathop{\left( 4! \right)\left[ 2^{3} \right]}}\, \right)\left( \overset{\text{Corners}}{\mathop{\left( 4! \right)\left[ 3^{3} \right]}}\, \right)} \)

Now, whatever you wish to find the probability of, remove it from the numerator (never touch the denominator). The only time you remove the 1/2 from the numerator is if both EP and CP are removed (because the odd permutation exists only if EP, CP, or both still need to be solved, not if both do not need to be solved). And, if AUF is to be considered, just multiply your answer by 4 (because 1/4 of the permutations can yield the same result. Because that's true and because the number of permutations are in the denominator, you put the 4 in the numerator, i.e., multiply by 4).

 

[Czery]

While solving my Square 1, I got an edge permutation skip!
Assuming I use full Vandenburgh, what is the probabilty that I will get this skip again?

 

[LouisCormier]

What are the odds of PLL skip on megaminx? OLL skip?

 

[TMOY]

PLL skip is 1/720, OLL skip is 1/1296.

 

[CubeRoots]

I don't know about OLL but PLL is definitely not p=1/720. Maybe you mean 1/72? there are 22 PLL cases (one of which is PLL skip), for almost every case there are 4 ways it could occur (just the way it is oriented), except the symetrical cases. The one that swaps opposite edges is an example, these can only occur in two ways because of the symmetry. I think there are 6 of these, So in total there are 15*4 + 6*2 = 72 cases + PLL skip so I guess it's roughly 1/72, 1/73 to be pedantic. Can anyone verify or fault my logic?

 

[Rpotts]

Can anyone verify or fault my logic?

What are the odds of PLL skip on megaminx? OLL skip?

There it is.

 

[Sa967St]

While solving my Square 1, I got an edge permutation skip!
Assuming I use full Vandenburgh, what is the probabilty that I will get this skip again?

I believe it's 1/576. 1/24 for the U edges and 1/24 for the D edges.

The one that swaps opposite edges is an example, these can only occur in two ways because of the symmetry. I think there are 6 of these, So in total there are 15*4 + 6*2 = 72 cases + PLL skip so I guess it's roughly 1/72, 1/73 to be pedantic. Can anyone verify or fault my logic?

Not quite.
Two PLLs are 1/36 (Z and E), three are 1/72 (H, Na, Nb), and the other 16 are 1/18. That leaves 1/72 for a PLL skip.

 

[TMOY]

I don't know about OLL but PLL is definitely not p=1/720. Maybe you mean 1/72? there are 22 PLL cases (one of which is PLL skip), for almost every case there are 4 ways it could occur (just the way it is oriented), except the symetrical cases. The one that swaps opposite edges is an example, these can only occur in two ways because of the symmetry. I think there are 6 of these, So in total there are 15*4 + 6*2 = 72 cases + PLL skip so I guess it's roughly 1/72, 1/73 to be pedantic. Can anyone verify or fault my logic?

Well, it's pretty easy. Assume there are actually 73 different PLL cases, counting different orientations separately. Since 73 is prime, this is only possible if there are 73 pieces of the same kind on the LL; this is obviously not the case, hence your logic must be wrong

Actually the correct result for the cube ls 1/72. But as Rpotts pointed out, we were talking about the megaminx, for which the correct result is 1/720. (There are 60 possible permutations of corners (5!/2, since only even permutations are possible), same for edges, hence 3600 possibilities disregarding AUF, hence the probability of a PLL skip is 1 out of 3600/5=720.)