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<blockquote data-quote="EzekielRaiden" data-source="post: 6876602" data-attributes="member: 6790260">Which is why I did things like presenting the 4d5k3+3 as opposed to 4d6k3 reroll 1s. Because the d5-based set cannot possibly be the same as the unmodified d6 set--they're physically and numerically different generators, with no rejection rules involved. Yet the probability distribution for 4d5k3+3 is, in absolutely every sense, precisely identical to that of 4d6k3 reroll 1s. Every probability is identical, and there is no difference in the possible results that you can get.Therefore, rigorously applying selection rules, and looking only at those things that come out after the selection rules have been applied, must be (in some cases) equivalent to changing the type of generator used, which *means* changing the distribution.Or, another way to put it,&nbsp; [MENTION=6799649]Arial Black[/MENTION]: you seem to be focusing on P(x), where x is &quot;a result that can be produced from rolling 4 six-sided dice and discarding the lowest of them.&quot; But none of us are focusing on that. We are focusing on P(x|C), where &quot;x|C&quot; is &quot;result x as defined before, given that condition C has been met.&quot;It is not the gambler's fallacy to assert that P(x) =/= P(x|C). It's <a href="https://en.wikipedia.org/wiki/Bayes'_theorem" target="_blank">Bayes' theorem.</a> P(x|C) = P(C|x)*P(x)/P(C). These are not independent events. Whether a die, or set of dice, is rerolled does depend on the current state thereof.</blockquote>

[QUOTE="EzekielRaiden, post: 6876602, member: 6790260"] Which is why I did things like presenting the 4d5k3+3 as opposed to 4d6k3 reroll 1s. Because the d5-based set [I]cannot possibly[/I] be the same as the unmodified d6 set--they're physically and numerically different generators, with no rejection rules involved. Yet the probability distribution for 4d5k3+3 is, in absolutely every sense, precisely identical to that of 4d6k3 reroll 1s. Every probability is identical, and there is no difference in the possible results that you can get. Therefore, [I]rigorously[/I] applying selection rules, and looking [I]only[/I] at those things that come out after the selection rules have been applied, must be (in some cases) equivalent to changing the type of generator used, which *means* changing the distribution. Or, another way to put it, [MENTION=6799649]Arial Black[/MENTION]: you seem to be focusing on P(x), where x is "a result that can be produced from rolling 4 six-sided dice and discarding the lowest of them." But [I]none of us[/I] are focusing on that. We are focusing on P(x|C), where "x|C" is "result x as defined before, [I]given[/I] that condition C has been met." It is not the gambler's fallacy to assert that P(x) =/= P(x|C). It's [URL="https://en.wikipedia.org/wiki/Bayes'_theorem"][B]Bayes' theorem.[/B][/URL] P(x|C) = P(C|x)*P(x)/P(C). These are [I]not[/I] independent events. Whether a die, or set of dice, is rerolled [I]does[/I] depend on the current state thereof. [/QUOTE]

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