Calcualting Odds - 2d6 take the lowest (or highest)

edited August 2011 in Story Games
I'm working on a Fudge/Fake Hack and would like to use 7 levels of competency. Although it goes against my very essence, my players seem to need numbers associated with Terrible through Superb (thus Terrible=1 and Superb=7) with die rolls to typically move "up the ladder" rather than in both directions with the 2d6-7 or 1d6-1d6. Before anyone suggests it, they do NOT get the simplicity of traditional Fudge Dice, (sort of like my kids not eating pineapple because they think it couldn't possibly taste as good as Skittles).

The Mechanic is based on rolling 2d6 and keeping either the highest or lowest die rolled (with some modifiers on either end). So my question is simply this: How do I calculate what the odds are for rolling 2d6 and keeping only the highest or the lowest? The mechanic looks like this:

Terrible 2d6 take the lowest and subtract 2
Poor 2d6 take the lowest and subtract 1
Mediocre 2d6 take the lowest
Fair 1d6
Good 2d6 take the highest
Great 2d6 take the highest and add 1
Superb 2d6 take the highest and add 2

Each character has a set of Fudge Points broken down into Luck, Stamina, and Will which they can use to raise a result when making a roll to ensure success. These points replenish throughout the game, but are low enough to encourage mindful resource management and "saving one's 'oomph' for when one really needs that extra little bit of luck or effort to succeed."

Thank you for any pointers you can offer a mathematically impaired librarian...


  • Hi Todd, there's a whole game design sub-forum here, you might have better luck asking there.
  • The basic die rolling probability looks like this:

    To roll a pair of six-sided dice, and have both of the face be the same (rolling 1 and 1, for example), is a 1/6 probability for each die.

    So, for the pairs (1-1,2-2,3-3,4-4,5-5, and 6-6) the probability for each set is: .0277777 (1/6 * 1/6).

    To roll two different numbers (say, a 3 and a 6) is slightly more trickier. The first die can be 2 numbers (since you don't care if the first die is a 3 or a 6), so that prob is 2/6. The second number would be remaining unrolled number, so that prob would be 1/6.

    So for each set of unmatching numbers (3-6, 1-2, 4-5, etc.), the probability is: .05555555 (2/6 * 1/6).

    Here's a google spreadsheet:

    Or is there something else you were looking for?
  • @Jason - Thank you... I forgot about the Praxis Discussion. I'll use it in the future.

    @Doho123 - That was exactly what I couldn't wrap my mind around. Thank you, the spreadsheet is precisely what makes sense to those of us who have difficulty with odds. I'd tried using but could not come up with what you have presented. Again, thank you!
  • AnyDice functions for this are as follows:

    output [highest 1 of 2d6]
    output [lowest 1 of 2d6]
  • Thank Eero, This really makes it understandable...
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