Author Topic: Estimating the Best of Two Rolls  (Read 353 times)

Offline Garryl

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Estimating the Best of Two Rolls
« on: August 30, 2018, 12:54:21 PM »
So, for the longest time, I had it in my head that when rolling 2d20 and taking the better result (2d20b1), the average result was 15. I don't know where I got this number from (probably from extrapolating from the far more trivial case of 2d2b1, a pair of coins), but suffice it to say, it's wrong. Today I learned of my mistake and a better estimator.

As it turns out, a good estimator is that for rolling a d20 twice and taking the better result, the average result is actually 13.825. A good estimation for die-rolling (and the actual exact average result if working with a random number generator not restricted to whole numbers only) is that the average = (min + 2*max) / 3, or that average = min + (max - min)*2/3.

Now, I haven't double-checked my math, but I think that the generalized case of XdNb1 (rolling X N-sided dice and taking the best result) has an average of approximately (1 + X*N) / (X+1). Again, dice only give integer results, so this is only an estimate; with a random number generator that could give any result from A to B, not just whole numbers, the average would be exactly (A + X*B) / (X+1), assuming my napkin math is right. If anybody feels like double-checking me, I'd appreciate it.

Offline SorO_Lost

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Re: Estimating the Best of Two Rolls
« Reply #1 on: August 30, 2018, 07:35:04 PM »
I haven't checked it but I did want to plug AnyDice in here. It's a really old site but it calculates odds for you.

For example,
Code: [Select]
output [highest 1 of 2d20]
Gives you the odds of rolling any certain face along with a two point decimal value of its average which is 13.82.

If you want the formulas, you can just read the source code.
« Last Edit: August 30, 2018, 07:37:06 PM by SorO_Lost »
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Offline Keldar

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Re: Estimating the Best of Two Rolls
« Reply #2 on: September 03, 2018, 02:27:30 AM »
The 4E Avenger rolled best of  2d20 for attack rolls, it was generally thought of as analogous to a +4 in guides.  I didn't pay attention to the math then, but it does further support your arithmetic.   :cheers

Offline Garryl

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Re: Estimating the Best of Two Rolls
« Reply #3 on: September 04, 2018, 01:13:57 AM »
I haven't checked it but I did want to plug AnyDice in here. It's a really old site but it calculates odds for you.

For example,
Code: [Select]
output [highest 1 of 2d20]
Gives you the odds of rolling any certain face along with a two point decimal value of its average which is 13.82.

If you want the formulas, you can just read the source code.

Yeah, AnyDice is an awesome tool. I've used it for some surprisingly complex calculations in the past.

The 4E Avenger rolled best of  2d20 for attack rolls, it was generally thought of as analogous to a +4 in guides.  I didn't pay attention to the math then, but it does further support your arithmetic.   :cheers

If you need anywhere from a 7 to a 15 to succeed on your attack roll, 2d20b1 is actually slightly better than a +4 bonus (topping out at being equal to a +5 bonus if you need exactly 11 to hit) in terms of your odds of success. It's slightly worse than +4 outside of that range, although 7-15 should cover most situations. However, rolling twice also gets you better odds of rolling a natural 20 or whatever you need for a crit, so that's an extra benefit to account for.

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Offline awaken_D_M_golem

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Re: Estimating the Best of Two Rolls
« Reply #4 on: September 04, 2018, 06:13:01 PM »
ninja'd except this detail , iirc the +5 is the actual value for Passive checks, perhaps PHB p175 or there'bouts.


The 4E Avenger rolled best of  2d20 for attack rolls, it was generally thought of as analogous to a +4 in guides. I didn't pay attention to the math then, but it does further support your arithmetic.   :cheers

bolded ... but that sorta didn't matter, you either played without having the right math, or somebody blew money on the paid-for-errata and was auto-part of a build.
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