OtaMota
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Do you know how to calculate an ao5? What about BPA? How do you know if someone already locked PR after the 4th solve? How do you calculate what someone needs on the last solve to get a sub-5 average? All of these questions will be answered in this guide. I will be splitting the guide up into different sections, determined by how many solves have been completed in an Ao5 / Mo3 so far. (I will also spoiler block them as the sections are pretty long, and I think that it would be better for readability) Another definition that is important to know here is target, which in this case is whatever the solver wants to get with their average, whether it’s a PR, podium at a comp, or an average under a certain time.
Now that you know about these tools and how to calculate pace, a good way to practice calculating these is to use it in your own solves. Pick a random target (your global is a good example) and do an Ao5, and after every solve, try to calculate the pace, and what times you need to get an Ao5 under the target.
I hope you were able to find this helpful.
If all solves are completed, the only thing you can calculate is the final average. For Mo3, it’s as simple as taking the mean of all 3 solves. (A mean is calculated by summing all of the results, then dividing by the number of results)
For Ao5, in the WCA, what they do is taking a mean of the solves of the average that are not the best or worst of the average. An example we can use is Ruihang Xu’s 4.78 PR average. The best and worst solves of Ruihang’s PR are 4.12 and 6.33. The mean of the middle 3 solves is 4.776667. It is important to note that Ao5s and Mo3s in the WCA are rounded to the nearest centisecond (0.0x). So Ruihang’s Average ends up being 4.78.
One more thing to note is about DNFs. DNFs are any attempts that are not completed, mathematically, they can be thought of as taking an infinite amount of time. If there is a DNF in Mo3, the whole Mo3 is considered a DNF. For Ao5s, it is possible to get a DNF and still have a non-DNF average, as the worst solve, the DNF, will not be used for the calculation. However, if there are 2 DNFs in the average, the Ao5 will be a DNF, as one of the DNFs would have to be used for the calculation of the average, which is not possible to mathematically define.
For Ao5, in the WCA, what they do is taking a mean of the solves of the average that are not the best or worst of the average. An example we can use is Ruihang Xu’s 4.78 PR average. The best and worst solves of Ruihang’s PR are 4.12 and 6.33. The mean of the middle 3 solves is 4.776667. It is important to note that Ao5s and Mo3s in the WCA are rounded to the nearest centisecond (0.0x). So Ruihang’s Average ends up being 4.78.
One more thing to note is about DNFs. DNFs are any attempts that are not completed, mathematically, they can be thought of as taking an infinite amount of time. If there is a DNF in Mo3, the whole Mo3 is considered a DNF. For Ao5s, it is possible to get a DNF and still have a non-DNF average, as the worst solve, the DNF, will not be used for the calculation. However, if there are 2 DNFs in the average, the Ao5 will be a DNF, as one of the DNFs would have to be used for the calculation of the average, which is not possible to mathematically define.
When 4 solves are done, this is where BPA and WPA come into play. Because of how Ao5 works, even if the last solve is really good or really bad, because that solve won’t be part of the calculations toward the average. After the 4th solve is completed, you have a range of what the final Ao5 could be, regardless of the result of the 5th solve. A BPA is calculated by taking the 3 best times out of the 4 done so far. For Ruihang, that would be 4.12, 4.75, 4.48. BPA going into that last solve is 4.45. WPA calculated taking the 3 worst times. For Ruihang, that would be 6.33, 4.75, 4.48. The WPA would be 5.19. The range the final average could be in was 4.45 to 5.19. If the BPA is faster than the target, it means the target is attainable. If the WPA is faster than the target, it means the target is guaranteed, regardless of the 5th solve’s result (could even be a DNF, and the target is still achieved). An example of someone locking their target in only 4 solves is Matty Hiroto Inaba; for his 4.72 NAR average, his WPA (4.72) was under the previous NAR of 4.86, so even after getting a high 6 on his 5th solve, he still had NAR average.
Note: If there is a DNF in one of the first 4 solves, the WPA is automatically a DNF. If there are 2 DNFs in the first 4 solves, the BPA is automatically a DNF; it is not possible to attain a non-DNF average.
Going back to the Ruihang example, we determined that the BPA is faster than his previous PR his target, so it was possible for him to achieve PR. However, his WPA was slower than his target, so PR wasn’t guaranteed. How do we determine the exact time Ruihang needs on the 5th solve to get PR average? What you would do is take the target, and multiply by 3 (4.84 * 3 = 14.52). Then, subtract from the number the 2 middle times of the first 4 solves (14.52 – 4.75 – 4.48 = 5.29) That 5.29 is the time that would be needed on the last solve to get a 4.84 average. However, because of how the rounding works, a 4.843333 average would still be a 4.84 average, so add 0.01 to the 5.29 to get 5.30. 5.30 is in fact the slowest time Ruihang could get on the last solve, and still PR. His actual 5th solve of 5.11 was under 5.30, so he was able to get his target of PR.
Though BPAs and WPAs are not applicable to Mo3s, due to all of the solves being counted in the average, determining what time is needed for the target is still possible. In this case, multiply the target by 3, then subtract the previous 2 solves, and add 0.01.
Note: If there is a DNF in one of the first 4 solves, the WPA is automatically a DNF. If there are 2 DNFs in the first 4 solves, the BPA is automatically a DNF; it is not possible to attain a non-DNF average.
Going back to the Ruihang example, we determined that the BPA is faster than his previous PR his target, so it was possible for him to achieve PR. However, his WPA was slower than his target, so PR wasn’t guaranteed. How do we determine the exact time Ruihang needs on the 5th solve to get PR average? What you would do is take the target, and multiply by 3 (4.84 * 3 = 14.52). Then, subtract from the number the 2 middle times of the first 4 solves (14.52 – 4.75 – 4.48 = 5.29) That 5.29 is the time that would be needed on the last solve to get a 4.84 average. However, because of how the rounding works, a 4.843333 average would still be a 4.84 average, so add 0.01 to the 5.29 to get 5.30. 5.30 is in fact the slowest time Ruihang could get on the last solve, and still PR. His actual 5th solve of 5.11 was under 5.30, so he was able to get his target of PR.
Though BPAs and WPAs are not applicable to Mo3s, due to all of the solves being counted in the average, determining what time is needed for the target is still possible. In this case, multiply the target by 3, then subtract the previous 2 solves, and add 0.01.
Even with just 3 solves, there is a lot of information to can gather in terms of pace towards the target. On the 3rd solve, it is possible determine if you can lock the target on the 4th solve, and what time would be needed to do so. It is also possible to determine what time you need on 4th to stay in contention for the target.
To determine what time you need on 4th can stay in contention for the target, take the target times 3, and subtract from it the 2 fastest times so far. For Ruihang, this would be 5.66; if Ruihang got a solve slower than this on his 4th solve, he would not be able to get his target, no matter what he does on the 5th solve. Another case study to analyze is Teodor Zajder’s 0.97 ER 2x2 Average. Doing the calculation we did for Ruihang, if Teodor gets a solve slower than 1.18 on his 4th solve, he is out of contention for ER. However, he did get a slower time, a 2.30. How did he still get ER? If the number you get (here, 1.18) is slower than the slowest time so far (for Teodor it was 1.16), regardless of what happens on the 4th solve, he could still get the target on the 5th solve if it is fast enough.
How do you calculate the possibility of locking in the target, only doing 4 solves? One way to determine this is using the same thing we did for Teodor, if you can get the target no matter what you do on the 4th solve, if the 5th solve is good enough, the vice versa is also true; if your 4th solve is good enough, you can get the target no matter what you do on the 5th solve. Again, this only applies if the target * 3 - 2 fastest solves is slower than the slowest solve so far. Another way to determine the possibility, as well as what time would be needed to do so, would be to take the target times 3, then subtract from it the 2 slowest solves so far. For Matty’s NAR, this would be 5.06, then you would check if this is faster than the fastest solve so far. If it is, locking the NAR after only 4 solves is not possible, and the 5th solve needs to be good as well. However, 5.06 is faster than the 4.56 he got on the 3rd solve, so locking in NAR was possible for him. An example where it would not be possible would be Ruihang. If he got a 3.45 on the 4th solve, he would lock in PR… except that would be faster than the fastest solve he had so far (4.12), so no matter how fast he was on the 4th solve, he would not be able to lock in PR on the 4th solve. This would be different though if the 4.12 was instead a 3.24 for example, in that case, getting a 3.45 on the 4th solve would lock PR, as the 3.45 would be guaranteed to count.
Notes for DNFs: If there is a DNF on one of the first 3 solves, it is not possible to lock in the target only doing 4 solves. If there are 2 DNFs in the first 3 solves, it is not possible to get the target at all.
To determine what time you need on 4th can stay in contention for the target, take the target times 3, and subtract from it the 2 fastest times so far. For Ruihang, this would be 5.66; if Ruihang got a solve slower than this on his 4th solve, he would not be able to get his target, no matter what he does on the 5th solve. Another case study to analyze is Teodor Zajder’s 0.97 ER 2x2 Average. Doing the calculation we did for Ruihang, if Teodor gets a solve slower than 1.18 on his 4th solve, he is out of contention for ER. However, he did get a slower time, a 2.30. How did he still get ER? If the number you get (here, 1.18) is slower than the slowest time so far (for Teodor it was 1.16), regardless of what happens on the 4th solve, he could still get the target on the 5th solve if it is fast enough.
How do you calculate the possibility of locking in the target, only doing 4 solves? One way to determine this is using the same thing we did for Teodor, if you can get the target no matter what you do on the 4th solve, if the 5th solve is good enough, the vice versa is also true; if your 4th solve is good enough, you can get the target no matter what you do on the 5th solve. Again, this only applies if the target * 3 - 2 fastest solves is slower than the slowest solve so far. Another way to determine the possibility, as well as what time would be needed to do so, would be to take the target times 3, then subtract from it the 2 slowest solves so far. For Matty’s NAR, this would be 5.06, then you would check if this is faster than the fastest solve so far. If it is, locking the NAR after only 4 solves is not possible, and the 5th solve needs to be good as well. However, 5.06 is faster than the 4.56 he got on the 3rd solve, so locking in NAR was possible for him. An example where it would not be possible would be Ruihang. If he got a 3.45 on the 4th solve, he would lock in PR… except that would be faster than the fastest solve he had so far (4.12), so no matter how fast he was on the 4th solve, he would not be able to lock in PR on the 4th solve. This would be different though if the 4.12 was instead a 3.24 for example, in that case, getting a 3.45 on the 4th solve would lock PR, as the 3.45 would be guaranteed to count.
Notes for DNFs: If there is a DNF on one of the first 3 solves, it is not possible to lock in the target only doing 4 solves. If there are 2 DNFs in the first 3 solves, it is not possible to get the target at all.
Even with 2 solves, it is still possible to gather some knowledge of how the average could go. If you take the target times 3, and subtract the 2 times so far, you get the threshold needed to be done on solve 3, in order to be able to lock in the target on solve 4. For Ruihang, if he got a 4.08 on the 3rd solve, he could lock in PR on the 4th solve if the solve was fast enough. However, Ruihang was unable to do so. On the other hand, if Matty got a 5.06 on the 3rd solve, he would be able to lock in NAR on the 4th solve. His 3rd solve was a 4.56, which is faster than 5.06, which enabled the potential to lock in the target on the 4th solve.
DNFs: If there is a DNF on one of the first 2 solves, it is not possible to lock in the target only doing 4 solves. If there are 2 DNFS in the first 2 solves, it is not possible to get the target at all.
DNFs: If there is a DNF on one of the first 2 solves, it is not possible to lock in the target only doing 4 solves. If there are 2 DNFS in the first 2 solves, it is not possible to get the target at all.
If only 1 solve is done, there is not a lot of info to gather, except for DNFs I guess. If you have a DNF in the first solve of a Mo3, you already have a DNF Mo3. If you have a DNF in the first solve of an Ao5, you are not able to lock in the target until the last solve is completed.
Now that you know about these tools and how to calculate pace, a good way to practice calculating these is to use it in your own solves. Pick a random target (your global is a good example) and do an Ao5, and after every solve, try to calculate the pace, and what times you need to get an Ao5 under the target.
I hope you were able to find this helpful.
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