Turns per solve

AJT17

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I have done over 3000 solves on my GAN 11 M Pro, and I do f2l and I know almost full PLL, and only know 2 Look OLL
How many turns would I average per solve, so I can find roughly how many turns I have done on my cube.
 
Roughly 215000~ on your cube ever.
I believe this is too low - you did not account for scrambling.

(~70 moves per solve + ~20 moves per scramble) 3000 solves = 270,000 total turns

I feel this may still be a bit low as it does not include practicing algorithms, untimed slow solves, or just fidgeting around with things like sexy move or U2 U2.
 
I believe this is too low - you did not account for scrambling.

(~70 moves per solve + ~20 moves per scramble) 3000 solves = 270,000 total turns

I feel this may still be a bit low as it does not include practicing algorithms, untimed slow solves, or just fidgeting around with things like sexy move or U2 U2.
Well, maybe it would be somewhere between 300,000 and 350,000 turns, which is a lot.
 
A bit of off-shoot question that has been bugging me. How many turns are done in a single 60 cube MBLD attempt?
I don't know enough about the BLD meta to give you a definitive answer, but I would simply calculate it as such:
((average # edge comms * average length of edge comm) + (average # corner comms * average length of corner comm) + (parity))*60

iirc there are on average 11 (6 comms) edge targets and 7 (4 comms) corner targets with parity (OP parity is 16 moves and that's probably a little high but whatever). This obviously changes as sometimes you have cycle breaks or flipped/solved pieces, but lets just stick with it there. I don't know how long the average comm is but I've heard 8 movers being optimal so I'll assume 9 is speed-optimal.

Plugging in all the numbers you get:
(( 6 * 9 ) + ( 4 * 9 ) + ( 16 )) = 106 average BLD solve
106 * 60 = 6360

EDIT: previously assumed 1 target needed one comm and ended up with 160 move solutions. This has been corrected.

So if you have a sheet with your list of comms you can calculate the average length and I'm sure there's a distribution somewhere for the average number of targets.
 
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I don't know enough about the BLD meta to give you a definitive answer, but I would simply calculate it as such:
((average # edge comms * average length of edge comm) + (average # corner comms * average length of corner comm) + (parity))*60

iirc there are on average 11 (6 comms) edge targets and 7 (4 comms) corner targets with parity (OP parity is 16 moves and that's probably a little high but whatever). This obviously changes as sometimes you have cycle breaks or flipped/solved pieces, but lets just stick with it there. I don't know how long the average comm is but I've heard 8 movers being optimal so I'll assume 9 is speed-optimal.

Plugging in all the numbers you get:
(( 6 * 9 ) + ( 4 * 9 ) + ( 16 )) = 106 average BLD solve
106 * 60 = 6360

EDIT: previously assumed 1 target needed one comm and ended up with 160 move solutions. This has been corrected.

So if you have a sheet with your list of comms you can calculate the average length and I'm sure there's a distribution somewhere for the average number of targets.
The average number of 3-cycles for a large sample size, is ~10.4 algs per cube.

Now the tough part is to distribute this into corner and edge average alg length in correct ratio. Corner/edge alg ratio varies a lot according to various scrambles.

The average edge comm length is 8.6, so a ball park calculation is 10.4x8.6 = ~89 turns (STM) using 3-style.

The average number of 5-cycles for a large sample size of edges of a 3x3 is ~2.5 algs per cube and about 4.5 algs for 3-cycle of corners.

The average edge comm length using 5-cycles is 10.6, so the solve length would roughly look like, 10.6x2.5 +8.6x4.5 = 26.5 + 38.7 = ~65 turns

For a 60 cube attempt, it would be 65.2x60 = 3912 turns
 
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