13
Jun
12

The Value of a Rerolled Die

I’ve seen a couple of blogposts about the advantage/disadvantage mechanic in WotC’s playtest materials that suggested some interest in the underlying math.  I actually find these functions useful to have around in other situations, too, so I’m reposting this bit I wrote up for our campaign wiki.

Average Value Re-rolling a N-Sided Die, Taking the Highest Result

Average Value Re-rolling a N-Sided Die, Taking the Lowest Result

So, for a D20…

The highest of two dice averages 13.825, the lowest of two dice averages 7.175.


10 Responses to “The Value of a Rerolled Die”


  1. June 13, 2012 at 4:27 pm

    So basically, rolling 2 d20’s and taking the higher/lower result is pretty much a +3/-3 bonus or penalty? I assume that the average roll of 1 d20 is 10.5.

  2. 2 Charlatan
    June 13, 2012 at 4:49 pm

    That’s the average effect, yeah. It’s swingier- I missed this earlier, but Roger over at Roles & Rules covered it: http://rolesrules.blogspot.co.uk/2012/05/d-next-disadvantage.html

  3. June 13, 2012 at 8:22 pm

    That is a much more interesting look at it. I like the fact that it gives better chances of increasing even-odds success, but not much more at long-shot success. This is just the sort of breakdown of the mechanic I wanted to see.

  4. June 13, 2012 at 11:25 pm

    “Roll 2dx and take the lower” has to be my favorite use of dice. It creates a nice Gaussian distribution with common results at one end and rare at the other. You can also tweak such a chart for mundanity or weirdness by rolling min(3dx), or a straight dx. The average is only the beginning of the fun …

  5. 5 Adam
    June 14, 2012 at 1:04 am

    Yeah, I think the key here is that you’re frequently not actually interested in the effect on the average, but rather on the effect on a true-false success test–if you’re using advantage for an attack or a skill check, the question is really “how much more often do you hit/succeed?” not “what’s your average roll?” That’s not entirely true, and would be almost completely untrue for something like advantage applied to damage rolls or hit point rolls or whatever, but…

  6. 6 Charlatan
    June 14, 2012 at 5:05 am

    In fact, you hit on exactly the reason (damage rolls) I worked the formulas out to begin with! But since everyone’s talking about the advantage/disadvantage mechanic…

  7. June 14, 2012 at 4:52 pm

    It’s a lot swingier than +/- 3%. Here’s some brute-force numbers:
    http://pastebin.com/QPmdtG65

  8. 8 Charlatan
    June 14, 2012 at 5:19 pm

    GDorn: For the reasons Roger and Adam pointed out, the average isn’t exactly the concern for advantage/disadvantage checks, but insofar as it is, the +/- 3 is in relation to the d20 roll ie +/- 16.625%.

    You should be able to work out exact probabilities at different DCs without figuring out the series or brute-forcing anything: With the advantage the probability of failure is (DC – 1)^2 / 400. I think that matches Roger’s original graph: If you need a 10+ to succeed, there’s a 45% chance of failure on an unmodified roll, but only a 20.25% chance of failure on the advantage, which is effectively a +5. With the disadvantage, your chance of success is (DC + 1)^2 / 400, so if you need a 10+ your chance of success is 55% unmodified but only 30.25% with the disadvantage, effectively a -5.

  9. 9 Charlatan
    June 14, 2012 at 6:02 pm

    Good gravy, this is what I get for multitasking: With the disadvantage, your chance of success is (21-DC)^2/400.

  10. 10 Drew
    June 16, 2012 at 2:35 am

    This mechanic is hardly new. First place I ran across it was with the Tween in the original Fiend Folio.


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