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Rolled character stats higher than point buy?
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<blockquote data-quote="EzekielRaiden" data-source="post: 6872465" data-attributes="member: 6790260"><p>Okay, so you're employing a semantic argument. When I say "roll higher stats," I mean "roll <em>and keep</em> higher stats." Because rolls you discard don't matter for actual play, by definition. For <em>generating</em> stats, yes, you're always going to have a percentage chance of getting a stat roll you can't keep--because the generator doesn't care what your manipulations are. But if you generate 10,000 <em>actually acceptable</em> characters, their average is different, and within that set, there IS a higher probability of having good stats when compared to a set with no rejection rules whatsoever.</p><p></p><p></p><p></p><p><em>Teeeeeechnically</em> speaking that's not accurate. The probabilities--again, EXCLUSIVELY FOR KEPT ROLLS--have to go up, because the sum of the probabilities must be 1. They go up by an amount proportional to the number of options removed. For communication's sake, instead of talking about "rolling" (which seems to be the core of the bugaboo), I'm going to talk about <em>recorded</em> stats--you never, ever "record" stats that don't match whatever rules you're using, so from the perspective of the "recorded" stats, it doesn't matter how many "rolls" you had to go through to generate a number, it only cares about those sets that play by da rules.</p><p></p><p>Like I said: by mathematical definition, if the average result goes up, EITHER the probabilities are different, or the possible results are different, or both. With your first example case, if the probabilities remained identical, the average result should be 2/6+3/6+4/6+5/6+6/6 = 20/6 = 20/6 = 3.333..., not 4 as you have claimed. The only way to get the 4 you have claimed is if we say that the uniform distribution is now p=0.2, not 0.166...; this gives us 2/5+3/5+4/5+5/5+6/5 = 20/5 = 4, just as you've said. <em>But that means the probabilities are now higher than they were before.</em> They are, in fact, higher by exactly (probability of number of options removed)/(number of options remaining) = (1/6)/(5) = 1/30. 1/6 + 1/30 = 5/30 + 1/30 = 6/30 = 1/5. For a non-uniform distribution, the changes are harder to calculate, but they still exist.</p><p></p><p>It's true that, for simple rejection of a particular result value, you have to truncate off the rejected part when plotting the "kept" ("recorded") distribution. But the recorded distribution will, as a result, be (overall) <em>taller</em> than the "rolled" distribution (given the same axes), as you cannot truly "remove" area from a complete histogram--that is, one where the probabilities sum to 1, which will always be the case for any dice-based distribution (since there is only a finite number of possible results, each with positive probability). This is the graphical analogue to the "probability increases from 1/6 to 1/5" statement. Some probabilities may go down, but the net effect will be most probabilities going up; a uniform distribution guarantees all probabilities go up.</p><p></p><p></p><p></p><p><a href="http://anydice.com/program/8169" target="_blank">The data seems to say otherwise.</a></p><p></p><p>The probability of 6 on 4d6k3r1 is <em>much</em> lower than on just 4d6k3--reduced by an order of magnitude, in fact. Your probability of <em>recording</em> 17-or-higher has almost doubled. That sounds like a change to the math to me! Of course, if you aren't using a computer to automatically get rid of values you won't accept, you'll spend extra time, because you'll need to reroll all the ones that show up, but that's not the same thing. (Technically, you could just use d10s, dividing even numbers in half and adding 1 to each die--that would produce the same distribution WITHOUT any "rules for removal" parts.) That is: 4d6k3r1 is precisely equivalent to 4d5k3+3--yet with the former I have "rejection rules," while with the latter I do not. Seems hard to argue with that...</p><p></p><p></p><p></p><p>I haven't been trying to say that it does, and given my repeated statements to that fact, I'm getting a little annoyed here. Rolling 4 dice--whether you keep 3 or not, whether you ignore 1s or not--has a particular distribution. Now, in my opinion, you are being <em>fantastically</em> pedantic, since as I have repeatedly said, results you <em>could roll</em> but which you won't (using my new terms) <em>record</em> don't, and CAN'T, matter. So the difference between "recorded" and "rolled" is, essentially, semantics--characters end up having better stats, on the average, if you remove the possibility of characters having <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":)" title="Smile :)" data-smilie="1"data-shortname=":)" /><img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":)" title="Smile :)" data-smilie="1"data-shortname=":)" /><img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":)" title="Smile :)" data-smilie="1"data-shortname=":)" /><img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":)" title="Smile :)" data-smilie="1"data-shortname=":)" /><img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":)" title="Smile :)" data-smilie="1"data-shortname=":)" /><img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":)" title="Smile :)" data-smilie="1"data-shortname=":)" />, low stats.</p><p></p><p>By the way, you can check <a href="http://anydice.com/program/81d5" target="_blank">the equivalence of 4d6k3r1 and 4d5k3+3 in AnyDice</a>, if you like. The fact that a "rejection rule" can be equivalent to literally using a different kind of dice seems pretty conclusive, to me, that rejection rules can in fact be mathematically equivalent to a completely different distribution....which is the same as saying that the results are different (in this case, better; if we were rejecting 6s, it would of course be worse).</p></blockquote><p></p>
[QUOTE="EzekielRaiden, post: 6872465, member: 6790260"] Okay, so you're employing a semantic argument. When I say "roll higher stats," I mean "roll [I]and keep[/I] higher stats." Because rolls you discard don't matter for actual play, by definition. For [I]generating[/I] stats, yes, you're always going to have a percentage chance of getting a stat roll you can't keep--because the generator doesn't care what your manipulations are. But if you generate 10,000 [I]actually acceptable[/I] characters, their average is different, and within that set, there IS a higher probability of having good stats when compared to a set with no rejection rules whatsoever. [I]Teeeeeechnically[/I] speaking that's not accurate. The probabilities--again, EXCLUSIVELY FOR KEPT ROLLS--have to go up, because the sum of the probabilities must be 1. They go up by an amount proportional to the number of options removed. For communication's sake, instead of talking about "rolling" (which seems to be the core of the bugaboo), I'm going to talk about [I]recorded[/I] stats--you never, ever "record" stats that don't match whatever rules you're using, so from the perspective of the "recorded" stats, it doesn't matter how many "rolls" you had to go through to generate a number, it only cares about those sets that play by da rules. Like I said: by mathematical definition, if the average result goes up, EITHER the probabilities are different, or the possible results are different, or both. With your first example case, if the probabilities remained identical, the average result should be 2/6+3/6+4/6+5/6+6/6 = 20/6 = 20/6 = 3.333..., not 4 as you have claimed. The only way to get the 4 you have claimed is if we say that the uniform distribution is now p=0.2, not 0.166...; this gives us 2/5+3/5+4/5+5/5+6/5 = 20/5 = 4, just as you've said. [I]But that means the probabilities are now higher than they were before.[/I] They are, in fact, higher by exactly (probability of number of options removed)/(number of options remaining) = (1/6)/(5) = 1/30. 1/6 + 1/30 = 5/30 + 1/30 = 6/30 = 1/5. For a non-uniform distribution, the changes are harder to calculate, but they still exist. It's true that, for simple rejection of a particular result value, you have to truncate off the rejected part when plotting the "kept" ("recorded") distribution. But the recorded distribution will, as a result, be (overall) [I]taller[/I] than the "rolled" distribution (given the same axes), as you cannot truly "remove" area from a complete histogram--that is, one where the probabilities sum to 1, which will always be the case for any dice-based distribution (since there is only a finite number of possible results, each with positive probability). This is the graphical analogue to the "probability increases from 1/6 to 1/5" statement. Some probabilities may go down, but the net effect will be most probabilities going up; a uniform distribution guarantees all probabilities go up. [URL="http://anydice.com/program/8169"]The data seems to say otherwise.[/URL] The probability of 6 on 4d6k3r1 is [I]much[/I] lower than on just 4d6k3--reduced by an order of magnitude, in fact. Your probability of [I]recording[/I] 17-or-higher has almost doubled. That sounds like a change to the math to me! Of course, if you aren't using a computer to automatically get rid of values you won't accept, you'll spend extra time, because you'll need to reroll all the ones that show up, but that's not the same thing. (Technically, you could just use d10s, dividing even numbers in half and adding 1 to each die--that would produce the same distribution WITHOUT any "rules for removal" parts.) That is: 4d6k3r1 is precisely equivalent to 4d5k3+3--yet with the former I have "rejection rules," while with the latter I do not. Seems hard to argue with that... I haven't been trying to say that it does, and given my repeated statements to that fact, I'm getting a little annoyed here. Rolling 4 dice--whether you keep 3 or not, whether you ignore 1s or not--has a particular distribution. Now, in my opinion, you are being [I]fantastically[/I] pedantic, since as I have repeatedly said, results you [I]could roll[/I] but which you won't (using my new terms) [I]record[/I] don't, and CAN'T, matter. So the difference between "recorded" and "rolled" is, essentially, semantics--characters end up having better stats, on the average, if you remove the possibility of characters having :):):):):):), low stats. By the way, you can check [URL="http://anydice.com/program/81d5"]the equivalence of 4d6k3r1 and 4d5k3+3 in AnyDice[/URL], if you like. The fact that a "rejection rule" can be equivalent to literally using a different kind of dice seems pretty conclusive, to me, that rejection rules can in fact be mathematically equivalent to a completely different distribution....which is the same as saying that the results are different (in this case, better; if we were rejecting 6s, it would of course be worse). [/QUOTE]
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