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Blindsolving Failure Thread

Former Chess World Champion Michael Botvinnik taught how to practise under severe conditions. He let people smoke over the board and make loud background noise and so on.

So today I first ran 6-7 km and then went to the sauna. In the more than normal hot and steamy sauna I tried 10 blindfold in a row. And got 8 dnf:s ;) Actually I am surprised that I had two ok. The brain felt like jelly afterwards :D

Edit: But maybe I should not invent this event, Sauna Blindfolded.
Ville does not need any more favours while competing :)
 
Last edited:
MatsBergsten said:
trying-to-speedcube... said:
4x4 BLD: DNF in about 40 minutes... Off by 8 centers, 7 edges and 2 corners -.- Turns out I skipped one of my images...
Click to expand...

But now this DNF makes the success later all the more sweet :)

Edit: and why don't you do 4x4BLD in the weekly contest?
Click to expand...

I don't know... I practised a lot, so I can try... With my sub-20 already in the pocket :D
 
Mike Hughey said:
8 5x5x5 BLD DNFs in a row (including 2 in official competition). :(

And counting. :( :( :(

How bad will it get????
Click to expand...

Using your current new success rate (official only since gathering this data is much quicker than going through your old weekly competitions here) I find that your success rate is 4/9 or 44.4%. (I know it's better than that in practice, again ease of calculation).

This means the expected number of solves for you to get a success is 2.25 solves. The standard deviation is sqrt[(5/9) / (4/9)^2] = 3sqrt(5)/4 = 1.68 solves.

To get 8 solves unsuccessfully has a probability of (5/9)^8 = 0.9% :-(

As for your next success I predict you will get one in the next 2.25 +- 1.68 solves, but to go more in depth it will be:

chance it's your next solve: 4/9 = 44.4%
chance it's at least one of your next 2 solves: 1 - (5/9)^2 = 69.1%
chance it's at least one of your next 3 solves: 1 - (5/9)^3 = 82.9%
number of solves you have to complete from today to have at least a 99% chance of having at least one success: 7.83 solves.

At the risk of being the math nerd, you did ask the question "how bad will it get?" ;-)

Chris
 
cmhardw said:
Mike Hughey said:
8 5x5x5 BLD DNFs in a row (including 2 in official competition). :(

And counting. :( :( :(

How bad will it get????
Click to expand...

Using your current new success rate (official only since gathering this data is much quicker than going through your old weekly competitions here) I find that your success rate is 4/9 or 44.4%. (I know it's better than that in practice, again ease of calculation).

This means the expected number of solves for you to get a success is 2.25 solves. The standard deviation is sqrt[(5/9) / (4/9)^2] = 3sqrt(5)/4 = 1.68 solves.

To get 8 solves unsuccessfully has a probability of (5/9)^8 = 0.9% :-(

As for your next success I predict you will get one in the next 2.25 +- 1.68 solves, but to go more in depth it will be:

chance it's your next solve: 4/9 = 44.4%
chance it's at least one of your next 2 solves: 1 - (5/9)^2 = 69.1%
chance it's at least one of your next 3 solves: 1 - (5/9)^3 = 82.9%
number of solves you have to complete from today to have at least a 99% chance of having at least one success: 7.83 solves.

At the risk of being the math nerd, you did ask the question "how bad will it get?" ;-)

Chris
Click to expand...

Mathematical ownage :D
 
cmhardw said:
Mike Hughey said:
8 5x5x5 BLD DNFs in a row (including 2 in official competition). :(

And counting. :( :( :(

How bad will it get????
Click to expand...

Using your current new success rate (official only since gathering this data is much quicker than going through your old weekly competitions here) I find that your success rate is 4/9 or 44.4%. (I know it's better than that in practice, again ease of calculation).

This means the expected number of solves for you to get a success is 2.25 solves. The standard deviation is sqrt[(5/9) / (4/9)^2] = 3sqrt(5)/4 = 1.68 solves.

To get 8 solves unsuccessfully has a probability of (5/9)^8 = 0.9% :-(

As for your next success I predict you will get one in the next 2.25 +- 1.68 solves, but to go more in depth it will be:

chance it's your next solve: 4/9 = 44.4%
chance it's at least one of your next 2 solves: 1 - (5/9)^2 = 69.1%
chance it's at least one of your next 3 solves: 1 - (5/9)^3 = 82.9%
number of solves you have to complete from today to have at least a 99% chance of having at least one success: 7.83 solves.

At the risk of being the math nerd, you did ask the question "how bad will it get?" ;-)

Chris
Click to expand...
Actually, thank you! That's much appreciated - now I can watch as I go by plugging numbers in and see just how ridiculous this gets.

By the way, I'm now at 9 in a row and counting. :( (Last one was off by just 3 X centers - I was in the middle of it when you posted this.)

McWizzle94 said:
Mathematical ownage :D
Click to expand...
So true, so true!
 
Mike Hughey said:
Actually, thank you! That's much appreciated - now I can watch as I go by plugging numbers in and see just how ridiculous this gets.

By the way, I'm now at 9 in a row and counting. :( (Last one was off by just 3 X centers - I was in the middle of it when you posted this.)

McWizzle94 said:
Mathematical ownage :D
Click to expand...
So true, so true!
Click to expand...

Haha Mike, I just meant to answer your question with the actual probabilities, in case you were interested. And to McWizzle, I in no way intended that to be mathematical ownage - I was also curious myself (since I think everyone has DNF streaks sometimes).

Chris
 
Mike Hughey said:
cmhardw said:
Mike Hughey said:
8 5x5x5 BLD DNFs in a row (including 2 in official competition). :(
Click to expand...

To add a little something I tried a better estimate. This year Mike has done 36 attempts at 5x5BLD in the weekly Forum contests the first 12 weeks. Of those only 15 were DNF:s, so the 44% would be better estimated by (21+4) / (36+9) = 25 / 45 = 0.55...

But then on the other hand several BLD attempts are not independent trials of the same stochastic procedure. And the probability for success depends on "daily form" which also can be formulated as tiredness or disturbing environment etc. etc.

McWizzle94 said:
Mathematical ownage :D
Click to expand...

What is meant by this? (I don't understand the English meaning)


Finally (Mike), while I was counting your success rate in the result files I happened to look at mine. And I had to accept a really frightening fact, I have done 27 attempts at 4x4BLD in the weekly contests. 23 DNF;s. That I call BLD Failure.

I have mentally hidden this fact from myself behind excuses like "I'm going for speed* or "If I really concentrated I would have made it".
Click to expand...
Click to expand...
 
MatsBergsten said:
But then on the other hand several BLD attempts are not independent trials of the same stochastic procedure. And the probability for success depends on "daily form" which also can be formulated as tiredness or disturbing environment etc. etc.
Click to expand...

Hi Mats,

I think this is interesting, that's true that success rate would depend on factors related to the day as well. I also think that success rate depends on the number of solves done in the person's lifetime up to that point (assuming you are not just trying to go super super fast on that one particular solve, which will lower the success rate). I feel like it might be logarithmic or logisitc in some way. Say everytime you increase the number of solves you have done in your lifetime by a factor of X (X is a positive number greater than 1, probably between 1 and 2 but maybe larger), you increase your success rate by 0.01% or something to that affect. Or maybe it's logisitc somehow, approaching 99% or 100% or something. These are of course guesses based on a qualitative analysis of people's solves haha, but it's an idea at least.

Chris

P.S. I didn't know you were an actuary! I passed my exam 1, probability for the SOA and am studying for exam 2, Financial Math which I hope to take in August.
 
cmhardw said:
P.S. I didn't know you were an actuary! I passed my exam 1, probability for the SOA and am studying for exam 2, Financial Math which I hope to take in August.
Click to expand...

Well how would you know:) (I realize it's in my profile)

I'm probably the worlds oldest blindsolving pi juggling actuary. Probability close to 1.
I'm so old so in my days Financial Math was almost not invented :D Very good luck with your exams!

As for the success probability, of course it changes over time. But it seems to me it can be decreasing also, e.g. if you don't practise any longer. Otherwise I mostly agree with you.

And again, with English not my native tongue, please explain "Mathematical ownage".
 
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