D&D 5E Bladesinger - a criticism of its design

[Ovinomancer]

The average array on 4d6 drop lowest as I understand it is 16, 14, 13, 12, 10, 9. I did not Monte Carlo that.

I checked the frequency of the above average array for BS, which is 16, 15, 13, 12, 10, 9 or better. About a third of characters.

I'll take a look at the "average" array and revert.

Sorry, but can you confirm that you get at least 16, 15, 13, 12, 10, 9 about 1/3 of the time on 4d6k3?

Okay, running 25k trials of 4d6k3 sets (as in 25k sets of 6), I have the frequency of 16, 15, 13, 12, 10, 9 or better as 17.5% with 95% error of +/- 1.6%. So maybe 19% likely?

For 16, 14, 13, 12, 10, 9 it's 23.4% with 95% error at 1.8%, or maybe 1:4 for that or better.

If I run all sets that get at least 1 16 and at least 1 15, my numbers look like yours, or slightly more than 1/3 of sets. Again, at 25k runs.

 

[Ovinomancer]

Yes, that's right.

From the Monte Carlo in Excel about a third of characters generated by the standard method have that array or better.

Crossposted: my number disagree with yours for the set, but agree on having at least 1 16+ with at least 1 15+. You're not testing what you're claiming.

 

[clearstream]

Crossposted: my number disagree with yours for the set, but agree on having at least 1 16+ with at least 1 15+. You're not testing what you're claiming.

I'm happy to refine this as I started with a quick look. We don't really care about the whole array. We want to see what proportion of characters can have the salient +8 to their AC at level 4? It's a third.

The other attribute to look at is the 12 Con. So I'll update my sheet to look for one 16+, one 15+ and one 12+. The Con isn't nearly so important, as the Dexterity and Intelligence, but worth checking.

 

[Ovinomancer]

Fun fun and for [MENTION=996]Tony Vargas[/MENTION], I took the average over 10K trial sets for the highest, second highest... lowest:

15.7
14.1
12.9
11.8
10.4
8.5

The Students T test 95% confidence on this is +/- 0.1. That surprised me, so I double checked that twice -- I believe it correct. This array is the mean array with high confidence.

 

[Ovinomancer]

I'm happy to refine this as I started with a quick look. We don't really care about the whole array. We want to see what proportion of characters can have the salient +8 to their AC at level 4? It's a third.

The other attribute to look at is the 12 Con. So I'll update my sheet to look for one 16+, one 15+ and one 12+. The Con isn't nearly so important, as the Dexterity and Intelligence, but worth checking.

Pretty sure that players in real games will care about the whole array. Having a 16 and a 15 and no other score over 9 would play very differently as a bladesinger than the average array you're claiming, for instance.

 

[clearstream]

Crossposted: my number disagree with yours for the set, but agree on having at least 1 16+ with at least 1 15+. You're not testing what you're claiming.

Also looking for 12+ Con, I found (over >100,000 arrays) the fewest per hundred = 26 and the most per hundred = 31. Does that agree with your estimate for same?

That is an array of 16, 15, 12... or better. That allows the BS to have +8 AC by level 4, and +1 HP per level and HD.

 

[clearstream]

Pretty sure that players in real games will care about the whole array. Having a 16 and a 15 and no other score over 9 would play very differently as a bladesinger than the average array you're claiming, for instance.

That is nitpicking. We can equally focus on the fact that this is for the desired array or better.

The chance of the other scores being horribly low is the same as that of them being problematically high.

 

[Ovinomancer]

The average array on 4d6 drop lowest as I understand it is 16, 14, 13, 12, 10, 9. I did not Monte Carlo that.

I checked the frequency of the above average array for BS, which is 16, 15, 13, 12, 10, 9 or better. About a third of characters.

I'll take a look at the "average" array and revert.

Also looking for 12+ Con, I found (over >100,000 arrays) the fewest per hundred = 26 and the most per hundred = 31. Does that agree with your estimate for same?

That is an array of 16, 15, 12... or better. That allows the BS to have +8 AC by level 4, and +1 HP per level and HD.

No, it does not agree. I get a mean of 34.2% with a 95% confidence margin of +/- 8.3%. So, slightly more than you're getting. How are you sorting your numbers?

I generate 1000 sets of 6 4d6 rolls and then compute the k3 scores for each group of 4 in the 1k sets. I then sort largest to smallest. I have a set of reference cells for the array I'm looking for and then count the number of the largest scores equal to or greater than the first reference stat that also have the second highest stat equal to or greater than the second reference, etc.

Generating 4d6 rolls: randbetween(1,6)
Finding 4d6k3: sum(range)-min(range)
Sorting for largest to smallest: large(range,place) ex: large(range,1) for highest, large(range,4) for 4th highest
Counting number of sets that qualify: countifs(range of highest, ">="&reference, range of second highest, ">+"&reference, ...)

This returns the number of the 1000 rolls that meet or exceed the required array. If I'm looking for a partial array, I set the unwanted numbers to 0, which will count all sets with the partial numbers.

I run this test 25 times, generating 25k runs, and average the values using Average(range). Standard deviation is found with stdev.s(range). Confidence margins are found using Confidence(margin, SD reference, number of trials). Since confidence is altered by taking averages, I run it per 1000 and take the highest value given across 25 trials. Doesn't underestimate and means i don't have to have 25k fields in my spreadsheet.

 

[Ovinomancer]

The average array on 4d6 drop lowest as I understand it is 16, 14, 13, 12, 10, 9. I did not Monte Carlo that.

I checked the frequency of the above average array for BS, which is 16, 15, 13, 12, 10, 9 or better. About a third of characters.

I'll take a look at the "average" array and revert.

That is nitpicking. We can equally focus on the fact that this is for the desired array or better.

The chance of the other scores being horribly low is the same as that of them being problematically high.

When we're talking about the effectiveness of the tradition and your claims rest solely on a high stat array rolled, no, it's not nitpicking to point out that just a 16 and a 15, while sufficient to power the narrow combo you're promoting, will not provide a character that is overall as effective as one with the assumed array.

Having a -1 CON definitely changes play over having a +1 CON, for instance.

 

[clearstream]

No, it does not agree. I get a mean of 34.2% with a 95% confidence margin of +/- 8.3%. So, slightly more than you're getting. How are you sorting your numbers?

Hmm, so I'm finding it quite a bit less frequent than you. Perhaps I'm doing something wrong.

I generate 1000 sets of 6 4d6 rolls and then compute the k3 scores for each group of 4 in the 1k sets. I then sort largest to smallest. I have a set of reference cells for the array I'm looking for and then count the number of the largest scores equal to or greater than the first reference stat that also have the second highest stat equal to or greater than the second reference, etc.

Similar. I generate 2000 and count arrays with the desired three values or better.

Generating 4d6 rolls: randbetween(1,6)
Finding 4d6k3: sum(range)-min(range)

Snap!

Sorting for largest to smallest: large(range,place) ex: large(range,1) for highest, large(range,4) for 4th highest
Counting number of sets that qualify: countifs(range of highest, ">="&reference, range of second highest, ">+"&reference, ...)

I'm using countif() here in a similar, but not identical fashion. Why do you sort first? I check every array.

Confidence margins are found using Confidence(margin, SD reference, number of trials). Since confidence is altered by taking averages, I run it per 1000 and take the highest value given across 25 trials. Doesn't underestimate and means i don't have to have 25k fields in my spreadsheet.

Similar. I run 50 trials (of 2000 per trial). What do you find one SD to be?