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

[Ovinomancer]

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


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


Snap!


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

It allows me to do other things with the data.

I'm not following how you're counting. If you're checking with individual count functions on each array, how are you avoiding doublecounting values?

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

IIRC around 8. I'm mobile right now, will double check when I get back to the spreadsheet.

 

[clearstream]

I'm not following how you're counting. If you're checking with individual count functions on each array, how are you avoiding doublecounting values?

=IF((COUNTIF(E1:E6,">11")-COUNTIF(E1:E6,">14"))>=1,1,0)+IF((COUNTIF(E1:E6,">14")-COUNTIF(E1:E6,">15"))>=1,1,0)+COUNTIF(E1:E6,">15")

Clunky and inelegant.

 

[Ovinomancer]

=IF((COUNTIF(E1:E6,">11")-COUNTIF(E1:E6,">14"))>=1,1,0)+IF((COUNTIF(E1:E6,">14")-COUNTIF(E1:E6,">15"))>=1,1,0)+COUNTIF(E1:E6,">15")

Clunky and inelegant.

Yikes! XP awarded for even getting that to work!

If you want a 1 cell formula, go with a countifs with large.

Countifs((large(range,1)),">=16",(large(range,2)),">=15",(large(range,3)),">=12")

That will return if the highest score is 16+, the second highest is 15+, and the third highest 12+.

 

[Ovinomancer]

=IF((COUNTIF(E1:E6,">11")-COUNTIF(E1:E6,">14"))>=1,1,0)+IF((COUNTIF(E1:E6,">14")-COUNTIF(E1:E6,">15"))>=1,1,0)+COUNTIF(E1:E6,">15")

Clunky and inelegant.

Figured your error. Run through what happens with an all 16 set.

You're only counting arrays that have a 16+, and then a 15, and then a 12-14. That's why you're undercounting, you're excluding better sets.

 

[clearstream]

Yikes! XP awarded for even getting that to work!

If you want a 1 cell formula, go with a countifs with large.

Countifs((large(range,1)),">=16",(large(range,2)),">=15",(large(range,3)),">=12")

That will return if the highest score is 16+, the second highest is 15+, and the third highest 12+.

countifs needs the first argument to be a range... I'm not sure that large(range,k) works there...

well, at least I can't get that function to work as you have it here...

any thoughts?

 

[clearstream]

Yikes! XP awarded for even getting that to work!

If you want a 1 cell formula, go with a countifs with large.

Countifs((large(range,1)),">=16",(large(range,2)),">=15",(large(range,3)),">=12")

That will return if the highest score is 16+, the second highest is 15+, and the third highest 12+.

This seems to work

=IF(LARGE(E1:E6,1)>=16,1,0)+IF(LARGE(E1:E6,2)>=15,1,0)+IF(LARGE(E1:E6,3)>=12,1,0)

Then only count arrays with that = 3

 

[Ovinomancer]

countifs needs the first argument to be a range... I'm not sure that large(range,k) works there...

well, at least I can't get that function to work as you have it here...

any thoughts?

The range there would be E1:E6. The error you're getting may be because the ">=X" needs to be in quotation marks for countifs.

This seems to work

=IF(LARGE(E1:E6,1)>=16,1,0)+IF(LARGE(E1:E6,2)>=15,1,0)+IF(LARGE(E1:E6,3)>=12,1,0)

Then only count arrays with that = 3

Yes, this works. Still a little clunky. The If command has more arguments, and you still need another step.

 

[clearstream]

Yes, this works. Still a little clunky. The If command has more arguments, and you still need another step.

With that, I'm finding arrays containing 16+, 15+ and 12+... at a rate of 29%. It occurs to me that 17+, 14+, 12+... can also give us a +8 AC at level 4, so add some %age more for that. (Any quick estimate how much?) A party of 4 is therefore more than 75% likely to contain one on their first set of characters. Which explains what I've seen at the table.

Our TotYP group (4 characters) included one character qualifying with 17, 14, 12, 12, 11, 11 another with 18, 15, 13, 12, 10, 8 and a third with 17, 17, 15, 13, 13, 9. It was that third character that took BS: they'll have +9 AC at level 4.

Our OOTA group (6 characters + 5 replacements) included six qualifying characters (I have their rolled arrays, if of interest). The worst qualifying array from that group is perhaps this one 17, 16, 14, 9, 9, 6 which can be set up for +9 AC at level 4, but has some low scores to work around. Could go 6, 19, 14, 17, 9, 9 which gives it a couple of low saves partly compensated for by the Wizard class. So far, I have found that the way 5th edition is structured makes having a few low scores not particularly punitive.

My experience has been that an array enabling a +8 or +9 AC at level 4 has been the norm not the exception on a per party basis: usually more than one. What sort of arrays has your group seen? Any qualifying?

 

[clearstream]

With that, I'm finding arrays containing 16+, 15+ and 12+... at a rate of 29%. It occurs to me that 17+, 14+, 12+... can also give us a +8 AC at level 4, so add some %age more for that.

@Ovinomancer looking for 16+, 15+, 12+... OR 17+, 14+, 12+ (and avoiding double counting) I found in 100 runs of 100 arrays a rate of 28% to 52%. (Using only 100 arrays gives a wider SD, but I was after a quick estimate.)

Perhaps about 40%? So the BS we are looking for appears about 87% of the time in a party of four (on their first set of characters).

For me, this scotches earlier objections that high AC (+8 or better at level 4) would be rare.



[Edited to correct mistakenly entered values. Conclusion is unaffected.]

 

[Ovinomancer]

@Ovinomancer looking for 16+, 15+, 12+... OR 17+, 14+, 12+ (and avoiding double counting) I found in 100 runs of 100 arrays a rate of 28% to 52%. (Using only 100 arrays gives a wider SD, but I was after a quick estimate.)

Perhaps about 40%? So the BS we are looking for appears about 87% of the time in a party of four (on their first set of characters).

For me, this scotches earlier objections that high AC (+8 or better at level 4) would be rare.



[Edited to correct mistakenly entered values. Conclusion is unaffected.]

I am slightly below you numbers, but I'm putting that on running 25k trails vs 10k and having a tighter estimate of the probability. I'm centered on 86.5% +/- 0.45%.

For interest, the odds of matching the criteria (17,14,12 OR 16,15,12) in four sets:
No times: 13.5%
One time: 35.1%
Twice: 34.2%
Thrice: 14.8%
All: 2.4%

This is a function of how many requirements you're setting, though, and isn't entirely representative of the actual sets produced. Even setting a limit that the lowest score rolled is at least an 8 reduces the overall probability of getting your set by 12%, but, most notably, doubles the chance you get no qualifying sets. And that's just avoiding the outside 2 SD on 4d6k3 boundary.

Not saying you're wrong, just cautioning that you might be overcertain about your outcomes.