# bloc 1x1x2 on corner probability



## deadalnix (Nov 15, 2010)

Basically, this is about getting an edge matched with an adjacent corner on a randomly scrambled cube.

I'm usually not so bad at proba, but I get really stick on that one. I don't even know where to start :/ Any ideo to start with will be really apreciated


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## Rinfiyks (Nov 15, 2010)

I've seen this question asked on this thread before.
http://www.speedsolving.com/forum/showthread.php?20384-Probability-Thread
I don't think anyone came up with an exact answer.
You could always get a computer to generate random cubes and find at estimate of the probability.


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## deadalnix (Nov 15, 2010)

Yes, I was considering this thread, but I think the problem is way too complex and would have messed up the thread.

I have noticed in my practice that it seems quite common, but not on enough scrambles to have an accurante result (I guess something like 0.4 is a realistic estimation).


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## irontwig (Nov 15, 2010)

The probability thread kinda sucks since it's too flooded with questions like "What's the probability for getting two PLL skips in a row?". I've tried to ask non-trivial questions, but to no avail. :/


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## Lucas Garron (Nov 15, 2010)

irontwig said:


> The probability thread kinda sucks since it's too flooded with questions like "What's the probability for getting two PLL skips in a row?". I've tried to ask non-trivial questions, but to no avail. :/


What about making a wiki page?

Perhaps a chart of probabilities, along with misc. facts, would be useful.


As for the original question in this thread:

1/24 for a pair to match, 24 possible pairs. So, about 1-(23/24)^24 ≈ 0.64


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## deadalnix (Nov 15, 2010)

Ok, I think you took the right direction lucas ! However, pairs are not independant, so your result isn't correct.


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## Lucas Garron (Nov 16, 2010)

deadalnix said:


> Ok, I think you took the right direction lucas ! However, pairs are not independant, so your result isn't correct.


Yeah, of course I know that, hence the "about" and squiggle marks. I was posting it because I thought this calculation should have been an obvious first step, but no one has done it in this thread yet.


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## cmhardw (Nov 16, 2010)

I tried looking at it via inclusion-exclusion a previous time it was brought up. It got quite difficult quite quickly. How is a 1x1x3 to be handled? Is this two separate 1x1x2 pairs? How about a 2x2x2? Is this 3 separate 1x1x2 pairs? Those were the issues I ran into with such an approach.


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