What is the Average Result of 2d20, Drop the Lowest.

[Dirigible]

13.825

I'm not buying that answer until Hong says it.

 

[orsal]

Anybody have the shape of the curve?

Linear, increasing. The probability of k (on the better of two dn) is (2k-1)/n, which is an increasing linear function.

 

[hong]

I'm not buying that answer until Hong says it.

It.

 

[hong]

Just a question... is it a Troll to ask a Statistics question on a D&D board?


The grade for this thread is B+. Extra Credit available for actually building the formula in MiniTab and posting it

Good heavens, do people still use MiniTab?

 

[QuaziquestGM]

Given real world circumstances, I'll go with an average result of 11.

As a quazi mathematical justification: the long term average roll of a D20 is 10. The result of any two die rolls, even if made at the same time with separate dice, are statistically independent, so the average overall roll of each die is 10. As you are selecting the better result of 2 independent simultaneous events with identical long term averages, and the result can only be a whole number, the new "average result" should be the next whole number better than the average of one set alone.

In short, I'm going with Murphy's Law.

And as Murphy sets a pretty step curve, I'm arbitrarily setting the accepted average roll of "20 d20s keep the best" at 15.

 

[Nifft]

the long term average roll of a D20 is 10.

This statment is not good for your math cred...

- - -

What's really awesome about max(2d20) is what it does for critical threat generation. The chance to threaten on a 20 goes from 5% to 9.75%; if you threaten on a 19, the chance goes from 10% to 19%; if you threaten on an 18, the chance goes from 15% to 27.75%.

Cheers, -- N

 

[orsal]

As a quazi mathematical justification: the long term average roll of a D20 is 10.

10.5, actually.

As you are selecting the better result of 2 independent simultaneous events with identical long term averages, and the result can only be a whole number,

An individual result can only be a whole number; the same is not true for the average.

the new "average result" should be the next whole number better than the average of one set alone.

Why the next? Even if the average had to be an integer (which it clearly doesn't), why smallest one greater than the simple average of a single die?

Look at it this way: You roll a die, then get to increase your result by rolling again, but with no loss if your reroll is less than your original roll. Suppose instead you were doing this with d1000s. Do you really think the reroll, which has approximately 50% chance of increasing your result, but if it does could increase it by up to 999, would on average increase it by less than 1?

In short, I'm going with Murphy's Law.

Captain Murphy would have the average as significantly less than 10. For example, any time you make an important roll that needs 5 or higher to succeed, Murphy would guarantee you a result no higher than 4. Unless you only make rolls with relatively low probabilties of success, there's no way Murphy's Law can be consistent with such results.[/QUOTE]

 

[Roman]

for "higher of two dn":

The possible results are 1,2,3,...,n

The probability of getting 1 is 1/n^2
The probability of getting 2 is 3/n^2
The probability of getting 3 is 5/n^2
The probability of getting k is (2k-1)/n^2
The probability of getting n is (2n-1)/n^2

Now the expected value (mathematical jargon for mean average in this context) is
the sum from k=1 to k=n of k(2k-1)/n^2
Distribute the 1/n^2 and focus on the sum of k(2k-1). This is the sum of 2k^2-k.

The sum of k, k=1 to k=n, is n(n+1)/2
The sum of k^2, k=1 to k=n, is n(n+1)(2n+1)/6
So the sum of 2k^2-k is
2n(n+1)(2n+1)/6 - n(n+1)/2
= [n(n+1)/2][(4n+2)/3 - 1]
= [n(n+1)/2][(8n-2)/6]
= n(n+1)(8n-2)/12

Now remember that 1/n^2 we put aside a little earlier. Then we have

(n+1)(8n-2)/12n

This is the formula. Checking it for n=20, to see if it agrees with the people who did that case by itself, we get
(21)(158)/240=13.825

which agrees with what two people got from calculating this specific case. So I believe my algebra is correct.

Thanks! This is indeed useful.

Just to expand on the question: How did you obtain the (2k-1) part of the formula?

Also, is can this be generalized for multiple (more than two) rolls keep the highest, multiple rolls drop the lowest or even multiple rolls drop X rolls?

 

[pawsplay]

If you're going to take that approach, why use random numbers? It's no sweat to have the program (or spreadsheet -- it's not hard to implement) systematically generate all 400 possibilities. That's both more efficient than 10,000 random cases, and guaranteed to give you the correct answer instead of merely a good approximation.

Because it's easier.

 

[drothgery]

Because it's easier.

Easier here meaning that you don't really have to think about how the problem works to do a Monte Carlo approach to solving it.