guythatlikesOH
Member
I just got three PLL skips in a row (Which also gave me my best single time, and average). Probability: 1/373248.
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?
If you get a PLL skip, the probability of the next two solves also being PLL skips is 1/5184I 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
*1/72 for the original pll skip = 1/373248
Edit:
Okay, technically, the "probability of having 3 PLL-skip in a row" has an important question: out of how many solves?
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?
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?
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.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.
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.
What is the probability of having a move move solve for a cross?
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