Robert-Y
Legend
You should race with Thom and see who can get a sub 50 avg of 12 first 
Follow along with the video below to see how to install our site as a web app on your home screen.
Note: This feature may not be available in some browsers.
So I guess this goes here.
In thirteen solves I got a LL skip. My first ever. Then a few later a PLL skip. Then a couple after that an OLL skip. What are the odds of getting ever possible type of lucky solve within fifteen solves? Btw that's rhetorical. Unless a cube theorist wants to crunch the numbers. Either way I'd be willing to doubt it's ever been done before. Oh and none of these were forced.
Every time I see you post in here I keep thinking you achieved the sub1 ao12PB Avg5 on the X-Cube 4x4
Pretty good average for me, I probably would have posted it even if it was with the Dayan (overall PB 58.57).
Average of 5: 1:00.121. 55.73 f2 D' r' D' r' R' F R' f2 F R' L F2 R2 D u2 F R' u R B' L2 R f2 R2 r2 L' D2 r' B D' f' U r U' r' U2 D' B2 r'
2. (53.98) L' B2 f' D' F2 r2 L B2 u' F2 f L R2 B' f D' u' r2 f u' L2 U R2 F2 f' r2 U2 r2 L f B2 D2 R' r U' R F' r' L2 D2
3. 1:03.04 F B f R' U R u' f2 D' r2 U2 D' R r2 L2 u L' B' D u r' u L2 u F' R U2 f' D R L2 F R2 F' R r2 F B L2 U
4. (1:14.68) D2 B2 D' F2 L' F2 u2 f2 U' r B' D2 L2 D2 U2 F' U' f' L2 B2 u U' B D f' r2 f' D2 r U' F2 f2 R F' D2 u f r U D2
5. 1:01.61 U' B2 L2 D2 U R B2 F' U' f' r' U' r' f u f' U f' B' D2 B D' R F R2 F f2 B2 U F2 D2 u2 r2 f' L2 B F U2 r2 f
What are the odds of getting every possible type of lucky solve within fifteen solves?
Closer: The probability is (number of ways to place the skips) * (probability of LL skip, OLL skip, PLL skip, and 12 nonskip solves in that order) ~= (15*14*13) * (1/15552 * 1/216 * 1/72 * 1^12) = .000011, which is slightly bigger than what you had.The probability of getting a LL skip on one of 13 solves (the other two being reserved for the PLL and OLL skip) is (1/15552)*13. Respectively we get (1/72)*13 and (1/216)*13. P(LL, OLL and PLL skip within 15 solves)=0.0009%. And the odds 111110 to 1.
zero
Closer: The probability is (number of ways to place the skips) * (probability of LL skip, OLL skip, PLL skip, and 12 nonskip solves in that order) ~= (15*14*13) * (1/15552 * 1/216 * 1/72 * 1^12) = .000011, which is slightly bigger than what you had.
Well, I guess if we wanted to be REALLY precise, we'd do something like this:
\( 15*14*13*(\frac{1}{216}\frac{1}{72} * \frac{1}{216}\frac{71}{72} * \frac{215}{216}\frac{1}{72} * (\frac{215}{216}\frac{71}{72})^{12}) \)
= \( 15*14*13 * (\frac{1^2 215^{13}}{216^{15}} \frac{1^2 71^{13}}{72^{15}}) = 0.00000885987... \)