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

[Maxperson]

What 'extraordinarily high' roll are you talking about? If you mean 16/14/13/12/10/9, this is the average set produced by rolling 4d6k3 six times.

And the reason for rolling a set has no impact on the expected results of any particular method.

Nobody but you is talking about average.

 

[FrogReaver]

Google definition of independent in probability.

When two events are said to be independent of each other, what this means is that the probability that one event occurs in no way affects the probability of the other event occurring. An example of two independent events is as follows; say you rolled a die and flipped a coin.

 

[Arial Black]

Care to explain how this concession fits with your other claims. If it is wrong in this example why should we have faith that your argument will hold in any other case? What is special about the case I cited that your general statements regarding these probabilities failed to explain it? Could it not be that this is evidence that probabilities of high played stats actually increase due to the discarding and rerolling like we claimed?

Sorry, I was called away before I could explain properly. I'll try again.

I can't use 1d6 as my example because it doesn't produce a bell curve, so I'll use 2d6 because it produces a well-known bell curve which has the easiest maths.

2d6 generates results of between 2 and 12, where both the average roll AND the most common roll equals 7. Specifically, out of 36 rolls, we expect:-

1 result of 2
2 results of 3
3 results of 4
4 results of 5
5 results of 6
6 results of 7
5 results of 8
4 results of 9
3 results of 10
2 results of 11
1 result of 12
36 total results

This is the bell curve.

Now, if you were to be generating a character for an RPG where, instead of having six ability scores generated by rolling 3d6, you instead you have one ability score generated by rolling 2d6....

The average set (of one score in this case) will be 7. Also, as it happens, the most common score for a table full of PCs will also be observed to be 7 (we would expect).

If you were to add a rule to character creation such that any set (of one stat) that was less than, say, 6 had to be re-rolled, what effect would this rule have on the stats of the PCs?

Well, it's totally true that the average observed stats of a group of PCs will be...8.15, if my calculations are correct. Indisputably higher than 7.

Why? If players roll 6+, does the discard rule improve what they roll if that is the result? No, they roll what they roll!

What if they roll 5 or less? Then they roll again, and if that re-roll is 6+ then the fact that this is a re-roll does not change what they roll, and if it's 5 or less then the re-roll again. At no point in this process does any previous re-roll improve any score of 6+, and those are the only rolls that are played.

So the average observed is over (but close to) 8, but does this mean that the most common roll is 8 under a system that includes discards while a no-discard method remains at 7?

No!

If you discard rolls of less than 6, this doesn't change the bell curve at all! Low rolls are just as likely as they ever were, 7 remains the most common result (six times out of every 36 rolls) and results of 6 or 8 remain joint second at 5 rolls out of 36 each. 7 remains the most likely stat because it is not discarded! 2 or 12 remain the joint least likely results, and while 2 cannot be played (because it must be re-rolled), 12 doesn't suddenly become more likely to be rolled than it was before! It's still only a 1 in 36 chance to be a result of 12, still a six in 36 chance to be a 7, and 7 remains exactly six times more likely to be played than a 12, unchanged whether or not there is a discard rule in place.

This is why DMs should not fear that a discard rule skews stats towards higher results, because the proportion of average stats to max stats remains totally identical either way! The only difference is the lack of low sets, not a greater proportion of high sets over medium sets than there would have been without a discard rule!

Now, this breaks down if you discard the average rolls as well as the low rolls, but not because the proportions of high to medium sets have changed but simply because you are no longer allowed to play medium sets!

TL;DR: with or without a discard rule, the proportion of high to medium sets remains totally identical as long as you don't discard medium sets.

I hope that this explains my position more clearly.

 

[Arial Black]

Google definition of independent in probability.

When two events are said to be independent of each other, what this means is that the probability that one event occurs in no way affects the probability of the other event occurring. An example of two independent events is as follows; say you rolled a die and flipped a coin.

If the probability of rolling a 7 on 2d6 equals 1 in 6, then the fact that you discarded a previous roll has no influence on the probability of rolling a 7 on 2d6 the next time you roll; it remains 1 in 6, and the previous discard did not influence that probability.

That makes it an independent event. The reason you might have for rolling 2d6 (again) doesn't affect the performance of that next roll.

 

[EzekielRaiden]

The proportion does not remain the same.

When we exclude potentially roll-able options--using your example, 5 or less--we change nothing about how many possible options produce 7, but we do change the total number of "real" possibilities. (The excluded possibilities thus become "virtual"--yes, real dice will make them, but their existence doesn't matter for the real, "recorded" stats.)

With an unmodified 2d6, there are 36 possible distinct results, which can be seen by pretending that one die is, say, blue and the other is red. With "2d6, reroll if less than 6 until it's not less than 6" we have removed rolls of 2 (one roll), 3 (two rolls), 4 (three rolls), and 5 (four rolls), for a total of ten rolls removed from the distribution. This, then, means that all the other rolls--the ones not removed--are only possibilities out of 26, rather than 36, since those ten excluded rolls no longer hold sway over the "recorded" results. (Again, using "recorded" as I defined it upthread.)

So, we now have our "bell curve," really more a pyramid, as options out of 26. For (say) 1000 actually recorded rolls, this means we should expect 1000*(7/26) = 269 recorded 7s, because the recorded proportion is (7/26) = .2692307692307... approximately. We should expect (6/26) = (3/13) = .230769 repeating as the recorded proportions of both 6 and 8. Etc.

This is caused by exactly the same things that make the mean different. We now have (6*5+7*6+8*5+9*4+10*3+11*2+12)/26 = 8.15. The *only* way you can get that 8.15 figure is by dividing by the reduced number of acceptable possibilities (26), which is what causes the proportions in any sufficiently-large sample size to be different from the proportions you would expect of an unmodified 2d6 roll.

If the probability of rolling a 7 on 2d6 equals 1 in 6, then the fact that you discarded a previous roll has no influence on the probability of rolling a 7 on 2d6 the next time you roll; it remains 1 in 6, and the previous discard did not influence that probability.

That makes it an independent event. The reason you might have for rolling 2d6 (again) doesn't affect the performance of that next roll.

It doesn't affect the symbols that may be shown, or however one might say that better, but it absolutely does affect what the expected result will be on the whole.

To use an even simpler example, let's say we reroll 1s and 2s on a single d6. Then you have (reroll, reroll, 3, 4, 5, 6). Each reroll, itself, is another (reroll, reroll, 3, 4, 5, 6), which makes for an infinite series (as someone else argued previously). When you run that infinite series to its end, the amount of probability that had been present for 1 and 2 becomes proportionally distributed among the other possible results; since all the other results have exactly the same probability, they get the same extra amount, which is (2/6)*(1/4) = (2/24) = 1/12 (the 1/4 is because you're apportioning this into 4 equally-likely acceptable alternatives). You then add that to the original, natural probability of 1/6 for each face: 1/6 + 1/12 = 2/12 + 1/12 = 3/12 = 1/4. So the proportion of recorded 3s will increase from 1/6 to 1/4.

In other words, as I have said twice before, this is a simple and straightforward application of Bayes' Theorem for conditional probability. It has nothing to do with the gambler's fallacy, and asserting that it does is terribly confusing.

 

[FrogReaver]

The probability of the reroll even happening changes depending on whether you roll low or high. That makes the reroll dependent.

no. The reroll is not an independent trial but is dependent upon the the first roll being low.

if you always got a reroll no matter how well you rolled that would make the reroll truly independent. But it is not because the rerolls very existence is tied to you rolling something specific in the first place.

If the probability of rolling a 7 on 2d6 equals 1 in 6, then the fact that you discarded a previous roll has no influence on the probability of rolling a 7 on 2d6 the next time you roll; it remains 1 in 6, and the previous discard did not influence that probability.

That makes it an independent event. The reason you might have for rolling 2d6 (again) doesn't affect the performance of that next roll.

 

[EzekielRaiden]

The probability of the reroll even happening changes depending on whether you roll low or high. That makes the reroll dependent.

Exactly. Rerolling happens 0% of the time that you roll an acceptable value, and 100% of the time that you roll an unacceptable value. Therefore, the probability of rerolling is conditional upon what the result of the original roll was.

Or, putting it very, VERY simply: The probability of flipping a fair coin and getting heads is 0.5. The probability of flipping two coins and getting at least one head is 0.75. This is equivalent to getting exactly ONE re-flip if the first doesn't work. The probability of flipping n coins and getting at least one head is (1-2^n); this is equivalent to getting n re-flips, stopping as soon as you get a head. The probability of getting a head if you flip an infinite number of coins is 1, because lim(n->infinity) 1-2^-n = 1; this is equivalent to flipping coins until you get a head.

This exact same process applies to non-uniform distributions, it's just more complex to determine their probabilities. If the odds of getting a rejectable result on an unmodified roll are, say, 10%, then the infinite series for getting a non-rejectable result is lim(n->infinity) 1-10^-n = 0.999999... = 1. And so on.

 

[FrogReaver]

Arial black,

I don't understand why you think an equal distribution behaves differently in this instance compared to a more bell shaped distribution?

it almost seems you claim it isn't true of the equal/uniform distribution because it's trivially easy to show you are wrong with it and with your chosen distributions it becomes more complicated.

 

[Arial Black]

The proportion does not remain the same.

When we exclude potentially roll-able options--using your example, 5 or less--we change nothing about how many possible options produce 7, but we do change the total number of "real" possibilities. (The excluded possibilities thus become "virtual"--yes, real dice will make them, but their existence doesn't matter for the real, "recorded" stats.)

With an unmodified 2d6, there are 36 possible distinct results, which can be seen by pretending that one die is, say, blue and the other is red. With "2d6, reroll if less than 6 until it's not less than 6" we have removed rolls of 2 (one roll), 3 (two rolls), 4 (three rolls), and 5 (four rolls), for a total of ten rolls removed from the distribution. This, then, means that all the other rolls--the ones not removed--are only possibilities out of 26, rather than 36, since those ten excluded rolls no longer hold sway over the "recorded" results. (Again, using "recorded" as I defined it upthread.)

Okay, lets say that for 2d6 results of 5 or less are defined as 'low', 6 to 9 are defined as 'medium', and 10 to 12 are defined as 'high'.

With the unmodified bell curve, 10 results (out of 36) are low, 20 are medium, and 6 are high.

The proportion of medium to high rolls is 20 to 6, which is 3 and a third.

Now, 2d6 re-roll 'low'. This leaves results of 2 to 5 as re-rolls and so cannot be played, and takes away 10 possibilities, leaving us 26 possible played scores.

Of those 26, 20 are 'medium' and 6 are 'high', therefore the proportion of medium to high rolls remains 3 and a third.

The discard/re-roll rule has not nor ever will change the proportions of the remaining possibilities.

Remember, I never disagreed that the average score goes up. My point was that this doesn't mean that high scores are now proportionately more common than medium scores; it's just that the LOW possibilities have been taken away.

The original assertion which put this particular bee in my bonnet as that, with the discard rule, that this makes more high scores. I wanted to point out that, although not untrue itself, this statement leads to the erroneous impression that the discard/re-roll rule skews the results so that high scores become proportionately more common than medium scores; that the lost possibilities of the low scores were re-distributed more to high scores than medium scores.

In fact, when low scores are lost as possibilities, both medium AND high scores become more common in exactly the same proportion!

Since low scores are rarely played anyway, and since high scores seem to be feared by DMs for some reason, I felt the need to point out that the proportion of high to medium scores remains unchanged by the discard rule, so that DMs need not fear it.

Have I explained myself better now?

 

[FrogReaver]

Actually you claimed that high rolls were probabilisticly the same whether or not rerolls exist. That was false. Your clever change to high and medium rolls remain in the same proportion becomes true. However, high and low rolls are no longer in the same proportion. Same with medium and low rolls.

But yes, overall what you are saying is much better and much more accurate. However, if Dms fear high scores then they will still see more of them even though the proportion of medium to high remains the same.

Okay, lets say that for 2d6 results of 5 or less are defined as 'low', 6 to 9 are defined as 'medium', and 10 to 12 are defined as 'high'.

With the unmodified bell curve, 10 results (out of 36) are low, 20 are medium, and 6 are high.

The proportion of medium to high rolls is 20 to 6, which is 3 and a third.

Now, 2d6 re-roll 'low'. This leaves results of 2 to 5 as re-rolls and so cannot be played, and takes away 10 possibilities, leaving us 26 possible played scores.

Of those 26, 20 are 'medium' and 6 are 'high', therefore the proportion of medium to high rolls remains 3 and a third.

The discard/re-roll rule has not nor ever will change the proportions of the remaining possibilities.

Remember, I never disagreed that the average score goes up. My point was that this doesn't mean that high scores are now proportionately more common than medium scores; it's just that the LOW possibilities have been taken away.

The original assertion which put this particular bee in my bonnet as that, with the discard rule, that this makes more high scores. I wanted to point out that, although not untrue itself, this statement leads to the erroneous impression that the discard/re-roll rule skews the results so that high scores become proportionately more common than medium scores; that the lost possibilities of the low scores were re-distributed more to high scores than medium scores.

In fact, when low scores are lost as possibilities, both medium AND high scores become more common in exactly the same proportion!

Since low scores are rarely played anyway, and since high scores seem to be feared by DMs for some reason, I felt the need to point out that the proportion of high to medium scores remains unchanged by the discard rule, so that DMs need not fear it.

Have I explained myself better now?