#### riffz

##### Member
Strange. If that is indeed true, then either
1) I'm 1 in 512 lucky
2) I'm "doing it wrong"
3) My scrambling program (cubetimer.com) keeps giving me scrambles that don't require parity fix (Could this be the result of scrambles always having the same HTM/QTM lengths?)
Coincidence.

#### Cubenovice

##### Forever Slow
In my last ~10 attempts (some edges first, some corners first), I've only needed to use it once.
Wait, 10 attempts or 10 successes?

When you have a few DNF's they may have had parity without you identifying it!

#### Marcell

##### Member
Wait, 10 attempts or 10 successes?

When you have a few DNF's they may have had parity without you identifying it!
A blindfolded solve does not have to be successful in order to determine parity. In fact, you don't even have to start solving it. You can tell if you have parity by the memo itself.

#### Cubenovice

##### Forever Slow
Something tells me you didn't quite follow the discussion ;-)

#### riffz

##### Member
In my last ~10 attempts (some edges first, some corners first), I've only needed to use it once. I use computer generated scrambles. Is that about right?.
10*(1/2)^1*(1/2)^9 = 10*(1/2)^10 = 5/512

= 0.9765625%

probability that in 10 solves you will get parity only once.

This is of course assuming that you only do 10 solves.

#### riffz

##### Member
of 10 solves i get ~4-6 times parity.. at long time, it feels ligke nearly 50 percent.. i use this scrambler:
http://www.worldcubeassociation.org/regulations/scrambles/scramble_cube.htm?size=3&num=50&len=30&col=yobwrg&subbutton=Scramble!

what scrambler you use?
Think about it for a second. The probability that any one solve will have parity is 50%. So obviously 4/10, 5/10, and 6/10 are all very close to 50% and would seem to be common outcomes

The probability that in 10 solves you will get parity between 4 and 6 times is:

(10C4)*(1/2)^10 + (10C5)*(1/2)^10 + (10C6)*(1/2)^10

= (2(10C4) + (10C5)) * (1/2)^10

= 65.625%

#### Rune

Think about it for a second. The probability that any one solve will have parity is 50%. So obviously 4/10, 5/10, and 6/10 are all very close to 50% and would seem to be common outcomes

The probability that in 10 solves you will get parity between 4 and 6 times is:

(10C4)*(1/2)^10 + (10C5)*(1/2)^10 + (10C6)*(1/2)^10

= (2(10C4) + (10C5)) * (1/2)^10

= 65.625%
Really?

#### Rune

Yes:
(easy to see it's about 0.2+0.25+0.2, but can anyone make the "at most 6 heads" part work?)

Rune, what are you doing here? Are you blindcubing now? That would be cool, if you did it in competition you'd be the oldest by 20 years
You could as well ask me to start pen spinning. Maybe I would be the oldest there by 30 years!
No, Stefan, no! The memory - the little that was - has gone. That is, some "muscle memory" is still there. OK, you have Mats and Mike and others. I think they are just phenomens and exceptions from "normal aging".
You are asking me, what I´m doing here? My lot nowdays is to admire others´s progress. That´s all.
Shall I after all try your little problem? It seems to be the complement to "at least 7 heads, at most 10 heads". In that case 0.828125.

#### Stefan

##### Member
You are asking me, what I´m doing here? My lot nowdays is to admire others´s progress. That´s all.
Nah, with "here" I meant specifically this thread, because it's about blindcubing and I know you're always resisting to learn it. And it's hard for me to give up, because you're smart and it's so easy

I myself ignore most threads, don't have enough time/interest for all. So I thought you would ignore this thread for similar reasons. I actually am ignoring this thread, only saw your name and got my hopes up...

It seems to be the complement to "at least 7 heads, at most 10 heads". In that case 0.828125.
No, what we're after is 4-6 heads, the complement would be "at least 7 or at most 3". I have tried quite a few ways to ask for that, but Wolframalpha never understands me. I'm sure it's possible to ask with a mathematical expression, but I want it to understand it in English.

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#### cmhardw

Yes:
(easy to see it's about 0.2+0.25+0.2, but can anyone make the "at most 6 heads" part work?)
I've tried multiple ways to reword the sentence and none of them seem to accept the two inputs. I even tried a more expression like approach with:
Probability of 10 coin flips, 4 heads + Probability of 10 coin flips, 5 heads + Probability of 10 coin flips, 6 heads

This expression also did not work. Each individual part works correctly, but summing them does not.

#### Mike Hughey

Staff member
No, Stefan, no! The memory - the little that was - has gone. That is, some "muscle memory" is still there.
Aww, Rune - it would be so cool if you would learn! Are you sure? Have you tried some of the memory methods? The nice thing about the techniques is that your built-in memory doesn't have to be all that good - with those techniques, it's all about training, rather than natural ability. With a 3x3x3, there's really not all that much to memorize. Perhaps with a simple system, you could really retain it? I'd love it if you'd try!

#### Selkie

What would you recommend as a memorisation technique that would be the best to scale to larger cubes? I am itching to start 3x3 BLD but would like to learn techniques that are scalable even if they take more time investment initially.

I am sure this has been asked before but I couldn't find it quickly and thought my error could be hidden in the 1 answer thread

#### cmhardw

What would you recommend as a memorisation technique that would be the best to scale to larger cubes? I am itching to start 3x3 BLD but would like to learn techniques that are scalable even if they take more time investment initially.

I am sure this has been asked before but I couldn't find it quickly and thought my error could be hidden in the 1 answer thread
I would suggest to start with prepared lists. For right now something like:

A=Alligator
B=Banana
C=Cat

etc. Should you get interested in scaling this later then you can upgrade to Letter Pairs:

AB = Abacus
AC = Achilles

etc. You can just start by either saying the letters/words, or forming them into non-sense "words" if they happen to make sense together. After that you can extend to journeys or auditory loop if you get interested in either moving up to big cubes, or just in making the method a bit more efficient.

Another method I used to use that is an easy stepping stone toward journeys is to use your prepared lists (the top one) and use it in the linked lists system. Maybe use linked lists for corners and just saying the letters/sounds for edges.

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#### Selkie

Armed with this information I will start my initial letter list with a bottle of wine

EDIT: I guess if I choose my list with things that will extend easily, that will same a little time later:-

A - Aardvark (AA)
B - BlackBerry (BB)
C - Car Park (CP)

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#### Julian

##### Member
What algs do you recommend for the M-slice cases of M2? And what setups do you recommend for the 4-setup move cases? Also what are the 2 parity algs (depending on which piece type I start with)?
Thanks a bunch

#### ilikecubing

##### Member
What algs do you recommend for the M-slice cases of M2? And what setups do you recommend for the 4-setup move cases? Also what are the 2 parity algs (depending on which piece type I start with)?
Thanks a bunch
These

UB - M2

BU - F' D R' F D' M2 D F' R D' F

UF - U2 M' U2 M'

FU - F E R U R' E' R U' R' F' M2

DB - M U2 M U2

BD - M2 D R' U R' U' M' U R U' M R D'

Setups can be found here

http://www.stefan-pochmann.info/spocc/blindsolving/M2R2/

For pairty there is R perm and U' F2 U M2 U' F2 U assuming that you use old Pochmann for corners and you are doing corners first

If you are doing edges first,pairty is just U' F2 U M2 U' F2 U

#### JonnyWhoopes

These

UB - M2

BU - F' D R' F D' M2 D F' R D' F

UF - U2 M' U2 M'

FU - F E R U R' E' R U' R' F' M2

DB - M U2 M U2

BD - M2 D R' U R' U' M' U R U' M R D'

Setups can be found here

http://www.stefan-pochmann.info/spocc/blindsolving/M2R2/

For pairty there is R perm and U' F2 U M2 U' F2 U assuming that you use old Pochmann for corners and you are doing corners first

If you are doing edges first,pairty is just U' F2 U M2 U' F2 U
All this, however, for FU I suggest this alg.

D R' M' U R' U' M U R U' R D' M2