What are the odds of a skip on megaminx?

Really? I've gotten maybe 7 OLL skips and only a couple PLL skips.
720 and 933120 are numbers I have heard and unfortunately, I do not know how to prove them as I do not understand Permutations enough. I can prove OLL however.

In 3x3, CO and EO are separate things, they do not affect each other, so we can calculate each one and multiply at the end. If you have 4 edges, and each can be flipped in 2 ways, you can do 2^4 to find the number of EOs that can happen, which is 16. However, there is a parity, you can't have a single flipped edge, so we divide the 16 by 2, and we have a 1/8 chance for an EO skip. Corners use the same logic. There are 4 corners, each can be twisted in 3 ways, which gives us 3^4, 81. Divide by 3 because you can't have a single twisted corner, and you get 27. 27*8 is 216. 1/216 is the probability of an OLL skip on 3x3.

Let's translate this to Megaminx. You have 5 edges and corners this time. Instead of 2^4 and 3^4 for edges and corners respectively, it is now 2^5 and 3^5. This gives us 32 and 243. However, because of parities, you divide by 2 and 3 and you get 16 and 81. Multiply them and you get 1296.

I do hope that someone who understand's permutations more than me can show how 720 works. For LL skip, you multiply the chance of a PLL skip and the chance of an OLL skip. If a PLL skip is in fact 1/720, then a LL skip will be 1/933120.
 
For OLL: 2^4 * 3^4 = 1296

There are 2 ways to orient an edge and 3 ways to orient a corner. There are 5 corners and 5 edges, but we use 4 in the calculation because the orientation of the 5th piece is dictated by the others.

For PLL: 5! * 5! / 2 = 7200

Divide by 2 because half of the cases are parity and impossible to reach.

Entire LL: 1296 * 7200 = 9331200
 
Could you elaborate this? Doing 4! * 4! / 2 does not equal 72, which is 3x3 PLL skip so what makes Megaminx different?
That is a good point

3x3 PLL skip (with AUF) can be calculated by 4! * 4! / 2 / 4 = 72

Maybe to account for megaminx AUF in PLL you can divide by 5
 
i got 4 ep skips in a row, which is 1 / 12,950,000

also, its not like those solves were just 4 out of hundreds, i hadnt done mega in months and i sat down and did an ao5... and got 4 ep skips on the first 4 solves
 
????? how come nobody has given the correct calculations for permutations yet

(I was too lazy to respond when I first saw this thread but you guys really forced my hand here.)

Only even corner permutations and even edge permutations are possible on a megaminx, so there are 5!/2 = 60 corner permutations and 5!/2 = 60 edge permutations; all combinations are possible, so there are 3600 permutations in total. Of these, 5 count as PLL skips up to AUF, so the probability of getting a PLL skip is 5/3600 = 1/720.

If you use EP-CP order for PLL (as most people do), the probability of EP skip is 1/12, not 1/60, because AUF doesn't matter. Getting 4 EP skips out of 4 solves occurs with probability 1/12^4 = 1/20736. It still is rare, but you're off by three orders of magnitude if you don't correctly account for AUF.
 
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