D&D 5E Rolled character stats higher than point buy?

[Arial Black]

Okay, @Arial Black, here's my fundamental point:

Do you think that, in the long run, you will get consistently and without fail characters that are indiscernibly different using 4d6k3 than using "4d6k3 reroll 1s ignore results with net mod less than 1/highest stat of 12 or lower"? That because you can generate the same characters with either one, there is no meaningful difference between one method and the other?

We seem to be talking past each other, so permit me to clarify my fundamental point.

Imagine that we know are using 4d6k3 six times to generate a set of six ability scores, but we don't know whether or not we can re-roll sets that don't meet a certain criteria. In fact, one of us is allowed to re-roll, but the other is not, but we don't know this! In fact, it will be decided by a flip of the coin which of us would be allowed to re-roll, and this coin will be flipped after we have each generated our set of six.

We roll our stats as instructed. We expect the average distribution to be about 16/14/13/12/10/9, because that is what is expected for 4d6k3. As it turns out, neither of our sets would qualify to be re-rolled anyway.

At what point is one of us expected to have a higher set than the other, based on a re-roll that was never used?

Explain to me how the one that was allowed to re-roll (but never did) has an expectation of a higher stat distribution than is expected for 4d6k3!

See, this is the reality of rolling for scores. Sets of six are not generated from some average that is observed after the effect. Time only moves one way, and so does causality. Your possible decision to discard a set doesn't make any rolled set higher than it was before! If you discard one unsuitable set after another, five, ten, fifty times until you finally roll a set that doesn't qualify to be discarded and re-rolled, that final set has exactly the same expectation as every other: 16/14/13/12/10/9 in our case.

Re-rolling does not increase any set you roll, discarded or not. This is the Gambler's Fallacy; the idea that if the last seven spins of the roulette wheel have spun a red number, that the next spin is more likely to be black, on the grounds that eight reds in a row only has a 1 in 256 chance. This is bogus! If you're interested, the chances of red/black remain equal, because the sequence '7 reds followed by 1 black' also has only a 1 in 256 chance.

With (possible) discarding, the sets you discard have absolutely no effect on the next set you generate. Discarding doesn't give a mathematical expectation of generating higher scores.

The fact that you can observe averages of already generated sets and change the average of those by discarding low sets doesn't actually increase any rolled set at all! You increased the average, not by increasing any scores, but by taking low results away! This manipulation of the observed average has no impact on character creation beyond taking low scores away! It doesn't increase any kept set at all!

 

[FormerlyHemlock]

See, this is the reality of rolling for scores. Sets of six are not generated from some average that is observed after the effect. Time only moves one way, and so does causality. Your possible decision to discard a set doesn't make any rolled set higher than it was before! If you discard one unsuitable set after another, five, ten, fifty times until you finally roll a set that doesn't qualify to be discarded and re-rolled, that final set has exactly the same expectation as every other: 16/14/13/12/10/9 in our case.

Are you using some other definition of "expectation" than "probability-weighted average of possible values"? Because it's not meaningful to talk about the expected value of a particular sample (a particular roll), only of a distribution, and the two distributions (methods of rolling) have different expected values.

 

[Arial Black]

Are you using some other definition of "expectation" than "probability-weighted average of possible values"? Because it's not meaningful to talk about the expected value of a particular sample (a particular roll), only of a distribution, and the two distributions (methods of rolling) have different expected values.

We might very well be using the same word to mean different things, and this may be the cause of our continuing disagreement.

Although different methods of generating sets of scores may well be expected to produce different results, choosing to discard some of those already generated results does not change the way that method produces sets of scores in future.

Honestly, you and I sit side by side and roll 4d6k3 six times. I am playing by the 'discard' rules, you are not. Is there any reason to expect that my first set is more likely to be higher than yours? If so, why?

If my first rolled set does not qualify for a re-roll, then our two methods are identical!

If my first set does qualify, and I re-roll, my second set is rolled on the exact same 4d6k3 that we both had before! Is there any reason to expect that this roll is more likely to be higher than the others? If so, why?

 

[delericho]

Honestly, you and I sit side by side and roll 4d6k3 six times. I am playing by the 'discard' rules, you are not. Is there any reason to expect that my first set is more likely to be higher than yours? If so, why?

Yes.

If my first rolled set does not qualify for a re-roll, then our two methods are identical!

If my first set does qualify, and I re-roll, my second set is rolled on the exact same 4d6k3 that we both had before! Is there any reason to expect that this roll is more likely to be higher than the others? If so, why?

If you both rolled a set that would qualify for a discard, but you get to use that rule and he doesn't, then of course your resultant set would be higher than his.

To consider a trivial example, consider the average of 1d6.

With no rerolls, the average is a simple 3.5 - there are six results, each equally likely, so it's a simple calculation.

If you reroll 1s and 2s (and keep rerolling as needed), the average is then 4.5 - there are four valid results, each equally likely, and (3+4+5+6)/4 is 4.5. (There's a tiny, but not quite zero, chance that you'll have to keep rerolling for the rest of your life.)

If you reroll 1s and 2s once, the average is 4.167 (with a rounding error) - there are six valid results, but some are more likely than others. The calculation is more complex than the simple cases, but still easy enough.

And if you don't reroll but instead count 1s or 2s as 3 instead, the average is then 4 - that is (3+3+3+4+5+6)/4.

 

[FormerlyHemlock]

We might very well be using the same word to mean different things, and this may be the cause of our continuing disagreement.

Although different methods of generating sets of scores may well be expected to produce different results, choosing to discard some of those already generated results does not change the way that method produces sets of scores in future.

Honestly, you and I sit side by side and roll 4d6k3 six times. I am playing by the 'discard' rules, you are not. Is there any reason to expect that my first set is more likely to be higher than yours? If so, why?

If my first rolled set does not qualify for a re-roll, then our two methods are identical!

If my first set does qualify, and I re-roll, my second set is rolled on the exact same 4d6k3 that we both had before! Is there any reason to expect that this roll is more likely to be higher than the others? If so, why?

In the post you quoted, you said "final set," not "first set." The distribution of our first sets rolled is identical; the distribution of our final sets (first set "that doesn't qualify to be discarded and re-rolled, that final set has exactly the same expectation as every other") are not identical.

 

[Arial Black]

In the post you quoted, you said "final set," not "first set." The distribution of our first sets rolled is identical; the distribution of our final sets (first set "that doesn't qualify to be discarded and re-rolled, that final set has exactly the same expectation as every other") are not identical.

That's the Gambler's Fallacy right there!

The second (and any subsequent sets) are generated using the same 4d6k3 as the first. These subsequent sets have no way to generate a higher average.

 

[Arial Black]

Yes.



If you both rolled a set that would qualify for a discard, but you get to use that rule and he doesn't, then of course your resultant set would be higher than his.

To consider a trivial example, consider the average of 1d6.

With no rerolls, the average is a simple 3.5 - there are six results, each equally likely, so it's a simple calculation.

If you reroll 1s and 2s (and keep rerolling as needed), the average is then 4.5 - there are four valid results, each equally likely, and (3+4+5+6)/4 is 4.5. (There's a tiny, but not quite zero, chance that you'll have to keep rerolling for the rest of your life.)

If you reroll 1s and 2s once, the average is 4.167 (with a rounding error) - there are six valid results, but some are more likely than others. The calculation is more complex than the simple cases, but still easy enough.

And if you don't reroll but instead count 1s or 2s as 3 instead, the average is then 4 - that is (3+3+3+4+5+6)/4.

No, the observed average of previously generated rolls that are not discarded would be higher, but the roll itself will be the same average!

1d6 generates an average of 3.5.

1d6, re-roll 5 or less; what's the average?

Well, which average? The average result? Exactly 6. The average number generated? That stays at 3.5!

If you have 1d6 and discard any result of 2 or less, the average number generated by that 1d6 remains 3.5. It still has a 1 in 6 chance to generate each of the six results. If you roll a 1 or 2 and re-roll, the average number generated is still 3.5! If you roll a 3, 4, 5 or 6, you don't re-roll at all, and whether or not this is your first roll or your 50th, the average result of the die you rolled will still be 3.5. You don't have a higher chance of rolling a 3, 4, 5, or 6, just because you choose to roll again if you roll 1 or 2!

And that is the reality of generating a set of six stats! No re-roll changes the expected number generated by that method, and it is the set you generate which becomes your used set. That set has no expectation of generating a higher set than was expected previously.

Any average you observe after the fact has absolutely no impact on the average you are expected to generate.

In fact, the actual rolls of a group may be higher or lower than the expected average, but this doesn't mean that the rules you used to generate those sets were expected to be higher using that method; it just turned out that way. You wouldn't make the mistake of looking at a high set of results and say that the method used must therefore have expected this average, would you? And yet this is what you're doing with discards; seeing the results of the average sets after discarding, and erroneously concluding that the method of generation must have changed the expected average generated by that method!

 

[FormerlyHemlock]

That's the Gambler's Fallacy right there!

The second (and any subsequent sets) are generated using the same 4d6k3 as the first. These subsequent sets have no way to generate a higher average.

That's a false positive on the gambler's fallacy there.

If I asserted that the distribution of { r } changes because previous samples out of { r } have been low, that's the gambler's fallacy.

If I asserted that the distribution of { r | r exceeds certain stat minima } is not the same as the distribution of { r }, that is not the gambler's fallacy. That's true. You're trying to assert that they are the same, which might be a form of the Special Pleading fallacy. In any case, it's wrong. { r | r exceeds stat minima } has a higher expected value than { r }. This is mathematically demonstrable and seems to have been demonstrated several times in this thread already by various posters.

I just wanted to check and see if I was misunderstanding how you are using "expected value." It appears that I wasn't misunderstanding you. That's all I need to know.

 

[delericho]

And that is the reality of generating a set of six stats! No re-roll changes the expected number generated by that method, and it is the set you generate which becomes your used set. That set has no expectation of generating a higher set than was expected previously.

This is simply wrong. I would explain qhy, but I'll leave it to someone better qualified:

1d6, re-roll 5 or less; what's the average?

Well, which average? The average result? Exactly 6.

 

[Mercule]

We might very well be using the same word to mean different things, and this may be the cause of our continuing disagreement.

Although different methods of generating sets of scores may well be expected to produce different results, choosing to discard some of those already generated results does not change the way that method produces sets of scores in future.

Honestly, you and I sit side by side and roll 4d6k3 six times. I am playing by the 'discard' rules, you are not. Is there any reason to expect that my first set is more likely to be higher than yours? If so, why?

If my first rolled set does not qualify for a re-roll, then our two methods are identical!

If my first set does qualify, and I re-roll, my second set is rolled on the exact same 4d6k3 that we both had before! Is there any reason to expect that this roll is more likely to be higher than the others? If so, why?

There seems to be a basic misunderstanding of statistics, here. We're discussing the final results. If you discard some rolls, that's part of the method. Any sets that are discarded are explicitly outside the sample, which means they don't factor into the average. It's true that there is no difference in the odds of a particular result being rolled. There is a definite difference in the odds of a given result being used, which changes the sample mean.

To determine what the average scores under a given method are, you must include only those scores that are accepted as part of that method. For ease of math, let's talk about the 1d6, reroll 1s method.

Using a d6, the probability of rolling a "1" is 16.6%. The probability of having a result of 1 is 0% -- because 1s are explicitly excluded from the sample. The probability of rolling a 6 is 16.6%, but the probability of having a result of a 6 is 20%. You've just turned the d6 into a functional d5+1 for actual results and you'd probably be better served by using a d10 with results of 6-10 counted as 1-5. The meaningful results are identical.

Because we really only care about results, trying to separate the mid-point rolls is pointless pedantry. In fact, most folks don't give a rip about the rolls, as such, and are just as happy to use the term "roll" to refer to the final results because that's the only meaningful point of conversation.

If you exclude all sets (a.k.a. rolls) that have any scores lower than 8 or have two scores below 10 or have an average of under 12, then those sets are, well, excluded from the sample. That means the mean is going to be different (higher) than a table that has different rules for discards.

Yes, it's true that statistics tend to suck with a small sample size. Any given table could have quirky luck, up or down. With sets of dice, however, we aren't limited to a small sample of a theoretic infinite population. We have quantum, identifiable values that can be projected out over that theoretic infinite population. The more times we use a method the more it will tend to have cumulative results that fit the infinite curve.

You raise an excellent point about the coin flip for whether a given roll is discarded, though. This, IME, is one of the biggest issues with rolling for stats: There's almost always an arbitrary threshold for rerolls. By "arbitrary", I mean that it generally isn't a hard-and-fast, explicit rule. The DM looks at the set and rules. That makes the actual impact pretty much impossible to calculate because it can't be measured.

I don't mind some randomness in my character generation -- in fact, I think I'd prefer it. I just think that's its really important to be able to know three things: 1) the mean, 2) the distribution range, 3) the standard deviation.

I can live with an average result of 10.5 or 12 or 16, but each is going to give me slightly different expectations. Also, how much could I suck compared to the "average" (or best) character in the group? I don't really want to be relegated to sidekick status, and definitely not to page for the sidekick. Finally, what are my odds of having an unusually good or bad character? I'm not comfortable with having a high likelihood or playing that lackey, but it might be fun to get stuck with it once.

Ideally, I think I'd like to see a mean around 12, maybe slightly higher. I wouldn't want the range to be too broad; maybe +/-4 (8-16), for most characters. The standard deviation is key, though; having mathematic sameness isn't ideal, but it's a lot better than archmages and grogs.

Just as a gut reaction, using 4 FATE dice might give the right spread. It might also be too tight to use for all stats.