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Proficiency vs Non-Proficiency
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<blockquote data-quote="Ovinomancer" data-source="post: 7597280" data-attributes="member: 16814"><p>But, have you done the right math? I have to ask because your poll doesn't make sense at all given the actual math involved in the question. Not that your options are off, but that they don't even make any sense. If you had asked, "on average, how many times..." that would make sense with your options, but you asked "out of 20 rolls" which does not. This is because "on average" has certain assumptions that make a "number of times" questions make sense, or, at least, more sense (you'd need to change your wording to "assuming averages, how many times out of 20 would..." As it is, though, the math to answer your question is combinatorics and would make more sense as "what are the odds that out of 20 attempts the unskilled would win X times?"</p><p></p><p></p><p>On the math front, [MENTION=59057]UngeheuerLich[/MENTION] has the chance of losing (ties not counted as loses) for +6 vs +0 as 105/400 or 26.25% chance of losing. It's 30% even if you count ties as losing, and that's an easier number to work with, so let's go with 30% and say being tied by the lesser skilled is as good as a loss. So, the lesser skilled has a 30% chance of winning and the greater skilled has a 70% chance of winning on every try.</p><p></p><p>However, to address the odds of a given number of wins in a given number of trials, we have to invoke the binomial distribution, which is a means to determine the chance of n successes in k trials given a p probability of success. This will generate a probability density function which, in this case, is pretty much bell shaped with a mean of 6 and a SD of ~2, so the most likely results would be that the lesser skilled competitor wins or ties between 2 and 10 times in 20 tries approx 95% of the trials run. This doesn't line up with your choices very well at all.</p><p></p><p>With expertise, this changes to be most likely 1-2 wins (better than 50% of the time) and almost always between 0 and 4 wins.</p><p></p><p></p><p>Now, all that said about the math, what is it you're hoping to illustrate with this? That bounded accuracy is bounded?</p></blockquote><p></p>
[QUOTE="Ovinomancer, post: 7597280, member: 16814"] But, have you done the right math? I have to ask because your poll doesn't make sense at all given the actual math involved in the question. Not that your options are off, but that they don't even make any sense. If you had asked, "on average, how many times..." that would make sense with your options, but you asked "out of 20 rolls" which does not. This is because "on average" has certain assumptions that make a "number of times" questions make sense, or, at least, more sense (you'd need to change your wording to "assuming averages, how many times out of 20 would..." As it is, though, the math to answer your question is combinatorics and would make more sense as "what are the odds that out of 20 attempts the unskilled would win X times?" On the math front, [MENTION=59057]UngeheuerLich[/MENTION] has the chance of losing (ties not counted as loses) for +6 vs +0 as 105/400 or 26.25% chance of losing. It's 30% even if you count ties as losing, and that's an easier number to work with, so let's go with 30% and say being tied by the lesser skilled is as good as a loss. So, the lesser skilled has a 30% chance of winning and the greater skilled has a 70% chance of winning on every try. However, to address the odds of a given number of wins in a given number of trials, we have to invoke the binomial distribution, which is a means to determine the chance of n successes in k trials given a p probability of success. This will generate a probability density function which, in this case, is pretty much bell shaped with a mean of 6 and a SD of ~2, so the most likely results would be that the lesser skilled competitor wins or ties between 2 and 10 times in 20 tries approx 95% of the trials run. This doesn't line up with your choices very well at all. With expertise, this changes to be most likely 1-2 wins (better than 50% of the time) and almost always between 0 and 4 wins. Now, all that said about the math, what is it you're hoping to illustrate with this? That bounded accuracy is bounded? [/QUOTE]
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