Probability Thread

[guythatlikesOH]

I just got three PLL skips in a row (Which also gave me my best single time, and average). Probability: 1/373248.

 

[~Adam~]

If anyone fancies doing the math I recently had a 4 move last 4 centres while solving a 4x4 using Yau.

 

[Bestsimple]

I just got three PLL skips in a row (Which also gave me my best single time, and average). Probability: 1/373248.

May i ask how you got that?

 

[ben1996123]

May i ask how you got that?

luck, of course

 

[mark49152]

I just got three PLL skips in a row (Which also gave me my best single time, and average). Probability: 1/373248.

If you get a PLL skip, the probability of the next two solves also being PLL skips is 1/5184

 

[ben1996123]

If you get a PLL skip, the probability of the next two solves also being PLL skips is 1/5184

*1/72 for the original pll skip = 1/373248

 

[Renslay]

*1/72 for the original pll skip = 1/373248

That is the probability for that particular three solves. But if you have a PLL skip, then the probability of having three in a row PLL skip (i.e., two more), is (1/72)^2. Conditional probability.

Edit:
Okay, technically, the "probability of having 3 PLL-skip in a row" has an important question: out of how many solves?

 

[Ollie]

Edit:
Okay, technically, the "probability of having 3 PLL-skip in a row" has an important question: out of how many solves?

Why? The number of solves he did prior to the PLL skips has no effect on the probability of getting three in a row.

 

[Renslay]

Let's say I do N solves today. What is the probability of P = I have at least 3 PLL skips in a row?
Let p denote the probability of a single PLL skip.

N = 1 -> P = 0
N = 2 -> P = 0
N = 3 -> P = p^3
N = 4 -> P = (p^3)(1-p) + (1-p)(p^3) + p^4 (first 3 solves with PLL skip then no PLL skip -OR- first solve with no PLL skip then 3 PLL skip -OR- 4 PLL skips). Note that this is higher than p^3
...
N = k -> (okay, I'm too lazy for that)
...
N = inf -> P = 1

Or question number 2: what is the flaw in the reasoning above?

 

[Ollie]

Let's say I do N solves today. What is the probability of P = I have at least 3 PLL skips in a row?
Let p denote the probability of a single PLL skip.

N = 1 -> P = 0
N = 2 -> P = 0
N = 3 -> P = p^3
N = 4 -> P = (p^3)(1-p) + (1-p)(p^3) + p^4 (first 3 solves with PLL skip then no PLL skip -OR- first solve with no PLL skip then 3 PLL skip -OR- 4 PLL skips). Note that this is higher than p^3
...
N = k -> (okay, I'm too lazy for that)
...
N = inf -> P = 1

Or question number 2: what is the flaw in the reasoning above?

Not sure if you're deliberately trying to sound condescending, or what you're trying to prove, but the probability is the same regardless of how many solves you do in a day. If you flip a coin 15 times and it comes up heads on all of those, it does not make it more likely that you'll get a tails on the 16th time.

 

[CubeRoots]

That is the probability for that particular three solves. But if you have a PLL skip, then the probability of having three in a row PLL skip (i.e., two more), is (1/72)^2. Conditional probability.

Edit:
Okay, technically, the "probability of having 3 PLL-skip in a row" has an important question: out of how many solves?

he said, I just got 3 in a row. The probability that he would just get 3 in a row is 1/373248 as you know.

Your question is interesting but it doesn't change anything about the calculations that were made

 

[mark49152]

he said, I just got 3 in a row. The probability that he would just get 3 in a row is 1/373248 as you know.

Wrong. That is the probability for any three particular solves. The probability of having PLL skips on the first three solves of the day is 1/373248. The probability of having 3 in a row at some unspecified point in a session of many solves is obviously higher.

 

[Ollie]

Wrong. That is the probability for any three particular solves. The probability of having PLL skips on the first three solves of the day is 1/373248. The probability of having 3 in a row at some unspecified point in a session of many solves is obviously higher.

Irrelevant.

I just got three PLL skips in a row (Which also gave me my best single time, and average). Probability: 1/373248.

No mention of the average occurring at any specific point during a session, no mention that it was done in the first solves of the day and no mention of the number of solves done in the session. Therefore his calculation is correct. Stop over-complicating it.

 

[guythatlikesOH]

No mention of the average occurring at any specific point during a session, no mention that it was done in the first solves of the day and no mention of the number of solves done in the session. Therefore his calculation is correct. Stop over-complicating it.

Thank you for clarifying. Three successive solves in a row are an independent event, and therefore the probability of three solves in a row being PLL skips is not affected by any prior or following solves.

 

[Renslay]

There is a difference between "the probability of having the same outcome three times in a row for the next three cases" (dices, coins, PLL skip, etc), and "there will be three identical outcomes in a row in the next K cases" The first one is clearly independent as you say, P = (1/p)^3, however, the latter has lim P = 1 as K -> inf, because it is not specifyed where will be that row in the series. Of course no matter how many cubes I already solved, a chance for a PLL skip (or 3 in a row) is still the same. But that is because we're talking about the next case. But if I'm talking about "what is the chance that I will have a PLL skip today after ten thousand solves", that is pretty high, because I did not specify which solve contains the skip.

That's why I asked how many solves was that day, because I was wondering the chance of having 3 PLL skips in a row anywhere during the solving.

Again, one who has cubing for years has much more chance that he had 3 PLL skips in a row than a fresh cuber. But speaking of the chance that someone will have 3 PLL skips in a row is the same.

Aaaand yes, I'm starting to over-complicate it. Sorry for that.

 

[Crowned xerxes]

What is the probability of having a move move solve for a cross?

 

[Renslay]

What is the probability of having a move move solve for a cross?

You mean a one move solve? Depends on. For one particular color, or for any cross?
Anyway, this has the answer:
http://www.cubezone.be/crossstudy.html

 

[CubeRoots]

Right, it's like this:

The event is: I just got 3 PLL skips in a row. In other words: the last 3 solves I did were PLL skips.

The probability of this event is 1/373248

 

[ryanj92]

Surely the phrase 'if I do N solves what is the probability of getting 3 PLL skips in a row' has the following solution:

(N-2)*[(1/72)^3]*[(71/72)^N-3]
for N >= 3
The N-2 factor comes from the fact that the three PLL skips can be solves 1, 2, and 3, or 2, 3, and 4, right up to N-2, N-1, and N. This gives N-2 possibilites. Then the individual probability (in the knowledge that each solve is independent of the last solve) is the probability of one PLL skip cubed, multiplied by the probability that all other solves are not PLL skips.

EDIT: I've seen my own flaw already - my model assumes that ALL other solves have to not have a PLL skip, whereas you should include the other cases really (ie in 5 solves, solves 1-3 are skip, 4 is not skip and 5 is skip). in reality this diifference should be very small as probability of PLL skip < probability of non-PLL skip) so the above should give a good estimation?

 

[PedroSabioni]

Last layer skip probability?

What is the probability of having a LL skip after finishing the F2L?