What's the effective bonus of a double d20 roll take the highest?

[Pielorinho]

Okay, finished. Here are the shortcuts I used to create it:
-A1 gets a 1.
-Fill A2:A20 with a series.
-Copy the series 5 times.
-Copy that 5x series 4 times. You have the A1:A400 filled.
-B1 gets a 1.
-Copy that through b20.
-B21 gets "=sum(B1+1).
-Copy that cell through B400. You have B1:B400 filled.
-C1 gets "=max(a1:b1)"
-Copy that through C400.
-C401 gets "=average(c1:c400)"

Your result=13.825, which is equal to +3.325.

Daniel

 

[Pielorinho]

Nifft, what kind of graph would you run on this for illustrative purposes? I'm not seeing how that'd be helpful.

Daniel

 

[Achan hiArusa]

To set up the spreadsheet in Excel (just another way of doing it), just do a matrix with rows labeled as 1 to 20 (A2 to A21) and columns as 1 to 20 (B1 to U1), then do =MAX(B$1,$A2) and copy and paste this formula throughout the whole table. Then label a new column for results as 1 to 20 (W2 to W21) and then do =COUNTIF($A$2:$U$21,W2) in Row X and in X22 do =SUM(X2:X21) which should come out 400, then in row Y do =X2/X$22 then in cell Z2 do =Y2*W2 and copy and paste. And then in Z22 do =SUM(Z2:Z21) and you will get the previously mention 13.825, so the bonus is about +3.325.

 

[Nifft]

Nifft, what kind of graph would you run on this for illustrative purposes? I'm not seeing how that'd be helpful.

How likely you are to get each specific number. That is: number of occurences of each result divided by total number of results.

For a linear function like a single d20, it's a "uniform" distribution, and thus just a flat line. For a "normal" distribution like 3d6, it's a bell curve.

Cheers, -- N

 

[DM-Rocco]

To set up the spreadsheet in Excel (just another way of doing it), just do a matrix with rows labeled as 1 to 20 (A2 to A21) and columns as 1 to 20 (B1 to U1), then do =MAX(B$1,) and copy and paste this formula throughout the whole table. Then label a new column for results as 1 to 20 (W2 to W21) and then do =COUNTIF($2:$21,W2) in Row X and in X22 do =SUM(X2:X21) which should come out 400, then in row Y do =X2/X$22 then in cell Z2 do =Y2*W2 and copy and paste. And then in Z22 do =SUM(Z2:Z21) and you will get the previously mention 13.825, so the bonus is about +3.325.

Which is fine, but there are two things all the bonuses in the world can't correct that an actual second roll could, a result of 1 on the first roll or a critical roll or 20 on the second which is, IMO, why they give a second roll and not just a bonus. I don't think you can calculate a way to take this into account.

 

[Steveyd]

[FONT=&quot]The success of the d20 roll is binary; a roll of a certain number or above is a success and a lower than that is failure. There's not just a simple effective bonus because the bonus will change depending on the target number.

Let's say you need a 12 or better to hit the monsters Armor Class. The probability of a success is .45 on the first dice, and probability of failure is .55. Now of those failures, the second dice has a probability of bailing you out of .45. Thus the total probability of a failure on the first and a success on the second is .45*.55 = .2475, close to an effective +5 bonus, total possibility to succeed .6975.

That changes though as AC increases, because the second dice also has a lower probability to roll high enough. Say you need a 16 or better to hit, then P(Success) = .25, P(Failure)=.75, and P(Failure then Success) = .1875, an effective bonus of 3.75. Total Success would be .4375.

if you were really looking to give a bonus to eliminate dice rolling (though why is rolling 2 dice at once any harder?) you could look over the range of rolls typically needed and just go with something in there you feel appropriate. +4 over the range of 14-17 on the dice required to hit the target number might be appropriate, though you are actually giving them a little bit better bonus against harder things and slightly shortchanging them against easier things. If the BBEG that they need a 19-20 to actually hit shows up, then you are giving them a huge effective bonus to hit them if you stay with the keep 4. I suggest just having them roll both dice at once.

[/FONT][FONT=&quot] Of course, this is only to hit at all. Factoring in criticals on 20 and automiss on 1 skews to having many more criticals and practically 0 critical misses in a session.


[/FONT] Here's the table:
View attachment 2d20 Probabilty.xls

 

[Pielorinho]

You're absolutely right, DM_Rocco. And Nifft, graphing it returns a *very* interesting result.

On creating the spreadsheet describe above, I added a couple new columns. In d1, I entered "=a1", and copied down to d20. This showed the instance of each die roll. In e1, I entered "=Countif(c1:c400,d1)", giving a count of all dice rolls in which the highest of two d20 rolls (the results in cells c1:c400) equaled d1 (i.e., 1). I copied this down to e20.

The results? You'll roll a 1 as the highest of two d20 rolls 1 in 400 times. A 2 will be your highest roll 3 in 400 times. Each number will occur a number of times equal to the next highest odd number--e.g., a 3 will be your highest roll 5 times, a 4 will be your highest roll 7 times, etc. You'll get a 20 as your highest roll 39 in 400 times, almost a 1 in 10 chance. I definitely wasn't expecting that!

The graphing was a great idea.


Daniel

 

[jaldaen]

Thanks everyone for your answers... they were quite enlightening as I weigh whether to go with double rolls or single rolls with bonuses. I think that getting rid of bad rolls is definately a selling point for the double roll. What about straigth rerolls (i.e. I don't like this roll I'll take the next)? Would they have a higher effective bonus or lower one? Just curious...

 

[DM-Rocco]

You're absolutely right, DM_Rocco. And Nifft, graphing it returns a *very* interesting result.

On creating the spreadsheet describe above, I added a couple new columns. In d1, I entered "=a1", and copied down to d20. This showed the instance of each die roll. In e1, I entered "=Countif(c1:c400,d1)", giving a count of all dice rolls in which the highest of two d20 rolls (the results in cells c1:c400) equaled d1 (i.e., 1). I copied this down to e20.

The results? You'll roll a 1 as the highest of two d20 rolls 1 in 400 times. A 2 will be your highest roll 3 in 400 times. Each number will occur a number of times equal to the next highest odd number--e.g., a 3 will be your highest roll 5 times, a 4 will be your highest roll 7 times, etc. You'll get a 20 as your highest roll 39 in 400 times, almost a 1 in 10 chance. I definitely wasn't expecting that!

The graphing was a great idea.


Daniel

Thanks, in spite of what my players tell me, I am right from time to time

I can and do appreciate that you guys could figure out the exact math on such a thing though. Even though I am not the best at math, I do find it interesting to read you guys going off. Makes me wish I didn't spend all day in math class drawing dungeons on graph paper

Well, almost

 

[DM-Rocco]

oops double post