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I’m working on trying to get the 60-second achievement in Hyper Hexagonest mode in Super Hexagon; for a while, I’ve had the feeling that 60 seconds is achievable with my current skill level (and that my current skill level isn’t increasing particularly quickly), I just need to have a lucky run where my guesses and reflexes hold up enough times in a row. (For what it’s worth, my current high score is 48.43 seconds; I’ve broken 40 seconds maybe 10 times?)

Which raises the question: how often would I expect to have such a lucky run? Unlike the other Super Hexagon modes, that question feels relatively tractable to calculate for Hyper Hexagonest. You get the same collection of patterns throughout that mode (at last as far I’ve pushed it, I assume that changes at 60 seconds), and it doesn’t seem to speed up at all (except maybe right at the beginning). There are one or two visual distractions it adds (e.g. sometimes the screen zooms in and out); I’m not sure if those happen at regular times or are part of the patterns, but I don’t think that happens more as the time progresses. So, for all I know, there’s a fixed percentage that I’ll pass each pattern, and (assuming we don’t believe in hot hands or nerves), we should be able to map my chances of reaching a given time with an exponential curve.

And, of course, the exponent matters: if, say, I reach 20 seconds in 1 out of 5 times, then I should reach 60 seconds in 1 out of 125 times (clearly not the case!), or if I reach 20 seconds in 1 out of 20 times, then I should reach 60 seconds in 1 out of 800 times (plausible, and holds out enough hope that I should keep going). But if I only reach 20 seconds in 1 out of 40 times, then I’d only reach 60 seconds in 1 out of 64,000 times, and I should just cut my losses now.

So I recorded my times on 200 tries. Here’s the raw data, in case anybody wants to play around with it; the numbers are my time in seconds, rounded down to the nearest second. If we group it in 10-second chunks (where 10–20 includes 10 seconds but doesn’t include 20), then we have:

  • 0–10 seconds: 110 runs
  • 10–20 seconds: 55 runs
  • 20–30 seconds: 22 runs
  • 30–40 seconds: 8 runs
  • 40–50 seconds: 5 runs
  • >50 seconds: no runs

It looks like the numbers are getting cut by a little more than half every 10 seconds; if we take the first 10 seconds as the best indicator of the decay factor, then my chances of reaching a given time get multiplied by .45 every 10 seconds. That would lead to a prediction of

  • >10 seconds: 90.0 runs predicted, 90 runs actual
  • >20 seconds: 40.5 runs predicted, 35 runs actual
  • >30 seconds: 18.2 runs predicted, 13 runs actual
  • >40 seconds: 8.2 runs predicted, 5 runs actual
  • >50 seconds: 3.7 runs predicted, 0 runs actual
  • >60 seconds: 1.7 runs predicted, 0 runs actual

That’s clearly not right: all the predictions except for the first one are high. So I guess that means that the jump from 0 to 10 seconds is easier than the subsequent jumps.

If I look at the other ratios to calculate how much my chances are of reaching 10 seconds, then I get actually a surprisingly consistent answer:

  • (>10 seconds) / (>0 seconds) = 90 / 200 = .45
  • (>20 seconds) / (>10 seconds) = 35 / 90 = .39
  • (>30 seconds) / (>20 seconds) = 13 / 35 = .37
  • (>40 seconds) / (>30 seconds) = 5 / 13 = .38
  • (>50 seconds) / (>40 seconds) = 0 / 5 = .0

The first of those is unexpectedly high and the last is really really low, but the middle are consistent. (Honestly, way too consistent given the sample sizes.) And maybe we can explain the last with nerves and the first with a slow ramp-up? If we assume that I’ll reach 10 seconds in 45% of my attempts and that I’ll reach the next 10 second mark in 38% of my attempts for times after 10 seconds, then that predicts that, out of 200 tries, I’ll reach:

  • >10 seconds: 90.0 runs predicted, 90 runs actual
  • >20 seconds: 34.2 runs predicted, 35 runs actual
  • >30 seconds: 13.0 runs predicted, 13 runs actual
  • >40 seconds: 4.9 runs predicted, 5 runs actual
  • >50 seconds: 1.9 runs predicted, 0 runs actual
  • >60 seconds: .7 runs predicted, 0 runs actual

Obviously I’m cooking the books somewhat to make those first four predictions look good. Still, this is enough to suggest to me that I should keep on plugging away: assuming that the game doesn’t suddenly get harder at 45 seconds (which isn’t beyond the realm of possibility given my observed performance), then I should reach 60 seconds after a few hundred more attempts. (One out of every 280 attempts, if the above formula holds.) And that I am willing to try; if the numbers were suggesting it would take a few tens of thousands of attempts, then I’d be a lot more dubious about the wisdom of that attempt.

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November 3, 2012 @ 15:49:11Current Revision
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Deleted: I'm working on trying to get the 60-second achievement in Hyper Hexagonest mode in <a href="http:// www.bactrian.org/~carlton/ dbcdb/1727/"><cite>Super Hexagon</cite></a>; for a while, I've had the feeling that 60 seconds is achievable with my current skill level (and that my current skill level isn't increasing particularly quickly), I just need to have a lucky run where my guesses and reflexes hold up enough times in a row. (For what it's worth, my current high score is 48.43 seconds; I've broken 40 seconds maybe 10 times?) Added: I'm working on trying to get the 60-second achievement for Hyper Hexagonest mode in <a href="http:// www.bactrian.org/~carlton/ dbcdb/1727/"><cite>Super Hexagon</cite></a>; for a while, I've had the feeling that 60 seconds is achievable with my current skill level (and that my current skill level isn't increasing particularly quickly), I just need to have a lucky run where my guesses and reflexes hold up enough times in a row. (For what it's worth, my current high score is 48.43 seconds; I've broken 40 seconds maybe 10 times?)
Unchanged: Which raises the question: how often would I expect to have such a lucky run? Unlike the other <cite>Super Hexagon</cite> modes, that question feels relatively tractable to calculate for Hyper Hexagonest. You get the same collection of patterns throughout that mode (at last as far I've pushed it, I assume that changes at 60 seconds), and it doesn't seem to speed up at all (except maybe right at the beginning). There are one or two visual distractions it adds (e.g. sometimes the screen zooms in and out); I'm not sure if those happen at regular times or are part of the patterns, but I don't think that happens more as the time progresses. So, for all I know, there's a fixed percentage that I'll pass each pattern, and (assuming we don't believe in hot hands or nerves), we should be able to map my chances of reaching a given time with an exponential curve.Unchanged: Which raises the question: how often would I expect to have such a lucky run? Unlike the other <cite>Super Hexagon</cite> modes, that question feels relatively tractable to calculate for Hyper Hexagonest. You get the same collection of patterns throughout that mode (at last as far I've pushed it, I assume that changes at 60 seconds), and it doesn't seem to speed up at all (except maybe right at the beginning). There are one or two visual distractions it adds (e.g. sometimes the screen zooms in and out); I'm not sure if those happen at regular times or are part of the patterns, but I don't think that happens more as the time progresses. So, for all I know, there's a fixed percentage that I'll pass each pattern, and (assuming we don't believe in hot hands or nerves), we should be able to map my chances of reaching a given time with an exponential curve.
Unchanged: And, of course, the exponent matters: if, say, I reach 20 seconds in 1 out of 5 times, then I should reach 60 seconds in 1 out of 125 times (clearly not the case!), or if I reach 20 seconds in 1 out of 20 times, then I should reach 60 seconds in 1 out of 800 times (plausible, and holds out enough hope that I should keep going). But if I only reach 20 seconds in 1 out of 40 times, then I'd only reach 60 seconds in 1 out of 64,000 times, and I should just cut my losses now.Unchanged: And, of course, the exponent matters: if, say, I reach 20 seconds in 1 out of 5 times, then I should reach 60 seconds in 1 out of 125 times (clearly not the case!), or if I reach 20 seconds in 1 out of 20 times, then I should reach 60 seconds in 1 out of 800 times (plausible, and holds out enough hope that I should keep going). But if I only reach 20 seconds in 1 out of 40 times, then I'd only reach 60 seconds in 1 out of 64,000 times, and I should just cut my losses now.
Unchanged: So I recorded my times on 200 tries. Here's the <a href="http:// malvasiabianca.org/wp-content/ uploads/2012/ 11/super-hexagon- times.txt">raw data</a>, in case anybody wants to play around with it; the numbers are my time in seconds, rounded down to the nearest second. If we group it in 10-second chunks (where 10&ndash;20 includes 10 seconds but doesn't include 20), then we have:Unchanged: So I recorded my times on 200 tries. Here's the <a href="http:// malvasiabianca.org/wp-content/ uploads/2012/ 11/super-hexagon- times.txt">raw data</a>, in case anybody wants to play around with it; the numbers are my time in seconds, rounded down to the nearest second. If we group it in 10-second chunks (where 10&ndash;20 includes 10 seconds but doesn't include 20), then we have:
Unchanged: <ul>Unchanged: <ul>
Unchanged: <li>0&ndash;10 seconds: 110 runs</li>Unchanged: <li>0&ndash;10 seconds: 110 runs</li>
Unchanged: <li>10&ndash;20 seconds: 55 runs</li>Unchanged: <li>10&ndash;20 seconds: 55 runs</li>
Unchanged: <li>20&ndash;30 seconds: 22 runs</li>Unchanged: <li>20&ndash;30 seconds: 22 runs</li>
Unchanged: <li>30&ndash;40 seconds: 8 runs</li>Unchanged: <li>30&ndash;40 seconds: 8 runs</li>
Unchanged: <li>40&ndash;50 seconds: 5 runs</li>Unchanged: <li>40&ndash;50 seconds: 5 runs</li>
Unchanged: <li>&gt;50 seconds: no runs</li>Unchanged: <li>&gt;50 seconds: no runs</li>
Unchanged: </ul>Unchanged: </ul>
Unchanged: It looks like the numbers are getting cut by a little more than half every 10 seconds; if we take the first 10 seconds as the best indicator of the decay factor, then my chances of reaching a given time get multiplied by .45 every 10 seconds. That would lead to a prediction ofUnchanged: It looks like the numbers are getting cut by a little more than half every 10 seconds; if we take the first 10 seconds as the best indicator of the decay factor, then my chances of reaching a given time get multiplied by .45 every 10 seconds. That would lead to a prediction of
Unchanged: <ul>Unchanged: <ul>
Unchanged: <li>&gt;10 seconds: 90.0 runs predicted, 90 runs actual</li>Unchanged: <li>&gt;10 seconds: 90.0 runs predicted, 90 runs actual</li>
Unchanged: <li>&gt;20 seconds: 40.5 runs predicted, 35 runs actual</li>Unchanged: <li>&gt;20 seconds: 40.5 runs predicted, 35 runs actual</li>
Unchanged: <li>&gt;30 seconds: 18.2 runs predicted, 13 runs actual</li>Unchanged: <li>&gt;30 seconds: 18.2 runs predicted, 13 runs actual</li>
Unchanged: <li>&gt;40 seconds: 8.2 runs predicted, 5 runs actual</li>Unchanged: <li>&gt;40 seconds: 8.2 runs predicted, 5 runs actual</li>
Unchanged: <li>&gt;50 seconds: 3.7 runs predicted, 0 runs actual</li>Unchanged: <li>&gt;50 seconds: 3.7 runs predicted, 0 runs actual</li>
Unchanged: <li>&gt;60 seconds: 1.7 runs predicted, 0 runs actual</li>Unchanged: <li>&gt;60 seconds: 1.7 runs predicted, 0 runs actual</li>
Unchanged: </ul>Unchanged: </ul>
Unchanged: That's clearly not right: all the predictions except for the first one are high. So I guess that means that the jump from 0 to 10 seconds is easier than the subsequent jumps.Unchanged: That's clearly not right: all the predictions except for the first one are high. So I guess that means that the jump from 0 to 10 seconds is easier than the subsequent jumps.
Unchanged: If I look at the other ratios to calculate how much my chances are of reaching 10 seconds, then I get actually a surprisingly consistent answer:Unchanged: If I look at the other ratios to calculate how much my chances are of reaching 10 seconds, then I get actually a surprisingly consistent answer:
Unchanged: <ul>Unchanged: <ul>
Unchanged: <li>(&gt;10 seconds) / (&gt;0 seconds) = 90 / 200 = .45</li>Unchanged: <li>(&gt;10 seconds) / (&gt;0 seconds) = 90 / 200 = .45</li>
Unchanged: <li>(&gt;20 seconds) / (&gt;10 seconds) = 35 / 90 = .39</li>Unchanged: <li>(&gt;20 seconds) / (&gt;10 seconds) = 35 / 90 = .39</li>
Unchanged: <li>(&gt;30 seconds) / (&gt;20 seconds) = 13 / 35 = .37</li>Unchanged: <li>(&gt;30 seconds) / (&gt;20 seconds) = 13 / 35 = .37</li>
Unchanged: <li>(&gt;40 seconds) / (&gt;30 seconds) = 5 / 13 = .38</li>Unchanged: <li>(&gt;40 seconds) / (&gt;30 seconds) = 5 / 13 = .38</li>
Unchanged: <li>(&gt;50 seconds) / (&gt;40 seconds) = 0 / 5 = .0</li>Unchanged: <li>(&gt;50 seconds) / (&gt;40 seconds) = 0 / 5 = .0</li>
Unchanged: </ul>Unchanged: </ul>
Unchanged: The first of those is unexpectedly high and the last is really really low, but the middle are consistent. (Honestly, way too consistent given the sample sizes.) And maybe we can explain the last with nerves and the first with a slow ramp-up? If we assume that I'll reach 10 seconds in 45% of my attempts and that I'll reach the next 10 second mark in 38% of my attempts for times after 10 seconds, then that predicts that, out of 200 tries, I'll reach:Unchanged: The first of those is unexpectedly high and the last is really really low, but the middle are consistent. (Honestly, way too consistent given the sample sizes.) And maybe we can explain the last with nerves and the first with a slow ramp-up? If we assume that I'll reach 10 seconds in 45% of my attempts and that I'll reach the next 10 second mark in 38% of my attempts for times after 10 seconds, then that predicts that, out of 200 tries, I'll reach:
Unchanged: <ul>Unchanged: <ul>
Unchanged: <li>&gt;10 seconds: 90.0 runs predicted, 90 runs actual</li>Unchanged: <li>&gt;10 seconds: 90.0 runs predicted, 90 runs actual</li>
Unchanged: <li>&gt;20 seconds: 34.2 runs predicted, 35 runs actual</li>Unchanged: <li>&gt;20 seconds: 34.2 runs predicted, 35 runs actual</li>
Unchanged: <li>&gt;30 seconds: 13.0 runs predicted, 13 runs actual</li>Unchanged: <li>&gt;30 seconds: 13.0 runs predicted, 13 runs actual</li>
Unchanged: <li>&gt;40 seconds: 4.9 runs predicted, 5 runs actual</li>Unchanged: <li>&gt;40 seconds: 4.9 runs predicted, 5 runs actual</li>
Unchanged: <li>&gt;50 seconds: 1.9 runs predicted, 0 runs actual</li>Unchanged: <li>&gt;50 seconds: 1.9 runs predicted, 0 runs actual</li>
Unchanged: <li>&gt;60 seconds: .7 runs predicted, 0 runs actual</li>Unchanged: <li>&gt;60 seconds: .7 runs predicted, 0 runs actual</li>
Unchanged: </ul>Unchanged: </ul>
Unchanged: Obviously I'm cooking the books somewhat to make those first four predictions look good. Still, this is enough to suggest to me that I should keep on plugging away: assuming that the game doesn't suddenly get harder at 45 seconds (which isn't beyond the realm of possibility given my observed performance), then I should reach 60 seconds after a few hundred more attempts. (One out of every 280 attempts, if the above formula holds.) And that I am willing to try; if the numbers were suggesting it would take a few tens of thousands of attempts, then I'd be a lot more dubious about the wisdom of that attempt.Unchanged: Obviously I'm cooking the books somewhat to make those first four predictions look good. Still, this is enough to suggest to me that I should keep on plugging away: assuming that the game doesn't suddenly get harder at 45 seconds (which isn't beyond the realm of possibility given my observed performance), then I should reach 60 seconds after a few hundred more attempts. (One out of every 280 attempts, if the above formula holds.) And that I am willing to try; if the numbers were suggesting it would take a few tens of thousands of attempts, then I'd be a lot more dubious about the wisdom of that attempt.
 Added: <strong>Update:</strong> And, of course, later on this afternoon, I did in fact break 60 seconds; I didn't count, but maybe it took another 30 tries or so? So my belief that that goal was in reach, and that it was mostly a matter of luck as to when I'd get there, turned out to be correct.

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