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<blockquote data-quote="ECMO3" data-source="post: 8486043" data-attributes="member: 7030563">This is a false assumption.&nbsp; It is actually the opposite. If you have already suceeded once, then statistically the data indicates the chance of success is higher than the chance of failure.If he saved once, statistically his estimated chance of successfully saving again is GREATER than 50%. If your only data are he saved once then you assume that success was an average roll AND you know that roll saved.&nbsp; That means by definition the population estimate from the single sample indicates success is more common than failure if the observed roll suceeded.Now if you know the roll and know it was high, or you know the target number to save through metagaming this might not be the case, but in a blind roll the assumption is the first observed result is the median result.When you use Barbs you are depending on luck, not statistics unless you have reason to believe the observed value (success) is not representative of the population.&nbsp; If you have seen more than one save (for example he made 1 and failed 2 in the previous three rounds), then this changes the math.&nbsp; But your assumption does not hold up unless you have seen that, or have reason to estimate the target roll to save is not low.The rest of your argument relies on this statistical fallacy.&nbsp; Redo the analysis mathematically with the assumption that this is a binomial population (success or failure) and the median roll is a &quot;success&quot; with a standard deviation of 0 and you will see what I am talking about.</blockquote>

[QUOTE="ECMO3, post: 8486043, member: 7030563"] This is a false assumption. It is actually the opposite. If you have already suceeded once, then statistically the data indicates the chance of success is higher than the chance of failure. If he saved once, statistically his estimated chance of successfully saving again is GREATER than 50%. If your only data are he saved once then you [I]assume[/I] that success was an average roll AND you know that roll saved. That means by definition the population estimate from the single sample indicates success is more common than failure if the observed roll suceeded. Now if you know the roll and know it was high, or you know the target number to save through metagaming this might not be the case, but in a blind roll the assumption is the first observed result is the median result. When you use Barbs you are depending on luck, not statistics unless you have reason to believe the observed value (success) is not representative of the population. If you have seen more than one save (for example he made 1 and failed 2 in the previous three rounds), then this changes the math. But your assumption does not hold up unless you have seen that, or have reason to estimate the target roll to save is not low. The rest of your argument relies on this statistical fallacy. Redo the analysis mathematically with the assumption that this is a binomial population (success or failure) and the median roll is a "success" with a standard deviation of 0 and you will see what I am talking about. [/QUOTE]

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