Instead of Vanilla 2d6

My work back here rolling "lower of two dice" made me wonder if I could emulate a 2d6 curve that way. " 7 + (lower of 2d8) -- (lower of 2d8) " turns out to be almost indistinguishable from a 2d6 curve! Were I actually using this die mechanic for something, I'd be tempted to have results of 0 and 14 "explode" with further "lower of 2d8" rolls.

Tabletop play seems to have a tension between providing interesting results (usually means increasing numbers of dice) vs. keeping record keeping reasonable (means keeping pip totals small). World of Darkness and Shadowrun solve this by rolling lots of dice and counting each die as a pip; Betrayal at House on the Hill is similar, in that its dice are numbered from 0 to 2. ...I guess my personal holy grail here would be a simple, manual die mechanic that'll produce a smooth 1/x curve.

2d6 is really easy to use on the tabletop. It does have some problems--cannot roll higher than 12 or lower than 2, "lucky" streaks are pretty common, and "unlikely" results may be more or less common than players think appropriate.

My first thought is that I can break "lucky" streaks by rolling lots of times and discounting all results until a given value has been discounted a certain number of times. (The quota for each value would be proportionate to the chance of rolling that value.) This would be terrible to execute manually but could work with a computerized die roller.

My second  thought is to replace 2d6 with 4d6 / 2. This makes middling results more likely and extreme results much less likely.

I'm not sure I'd want to use any of these as a straight replacement for 2d6 rolls in (say) BattleTech, but I think it might be interesting to designate individual pilots as cautious (4d6/2) or swingy (quotas).
(I'd like to eventually do my own computerization of BattleTech, and I have some thoughts on how I'd modify specific BattleTech mechanics, but that's not a project I can start soon.)

Math Puzzle

I spent way too much time today trying to figure out this math puzzle. The results he wants are equivalent to rolling 1d4 twice and keeping the lowest result:

He wants to use d2s exclusively, and you can reproduce a d4 with two 1d2s easily enough. If you're actually sticking to coins, for instance, you could say "flip a shiny penny, a shiny nickel, a dull penny and a dull nickel. Face-up coins get their full value, tails-up coins get zero value, take the lower value of shiny+shiny and dull+dull." That would give you 6 cents 1/16th of the time, 5 cents 3/16ths of the time, 1 cent 5/16ths of the time, and 0 cents 7/16ths of the time.

The trick is that he wants all d2s to have uniform face values, and because d2s only have two faces, all of those other tricks he mentions are going to come out the same as just adding the dice together. At least, everything else I've tried so far has come out that way, and I'm pretty sure you can't get the 1/3/5/7 ratio just by straight summing the roll.

Okay, [(3d2 - 2)*1d2]/2, round up, seems to work. It's not the friendliest looking thing, but I don't think the solutions can get much better.

Anyways, I was looking at that chart I drew of the 1/3/5/7 ratio, and I thought, hey, that kind of looks like my bullseyes. But, instead of summing two dice like the 2d6 rolls I graphed before, it chooses between two independent rolls! That's more flexible, mechanically; and it's nice to see that the stripes don't thin towards the rim. I guess I'm now hoping to find a 1dX roll that thins evenly and infinitely towards the center. So I can roll two of them for every roll, and allocate them between disconnected aspects of the action.

Weird thing about the chance one roll has to beat another:

A while ago, EastwoodDC posted a spreadsheet for calculating the chance of rolling any given result on a number of whatever sized dice. I started looking at the chances one roll had to beat another and I noticed something weird.

When rolling 1d6 against 2d6, the chance that they'll tie is equal to the chance of rolling exactly 7 on 3d6, and the chance that 1d6 will beat 2d6 is equal to the chance of rolling 6 or less on 3d6.

When rolling 2d6 against 3d6, the chance that they'll tie is equal to the chance of rolling exactly 14 on 5d6, and the chance that 2d6 will beat 3d6 is equal to the chance of rolling 13 or less on 5d6.

When rolling 3d6 against 4d6, the chance that they'll tie is equal to the chance of rolling exactly 21 on 7d6, and the chance that 3d6 will beat 4d6 is equal to the chance of rolling 20 or less on 7d6.

So I'm pretty sure that, anytime you roll Nd6 against (N+1)d6, the chance they'll tie is equal to the chance of rolling N*7 on (N+N+1)d6 and the chance that Nd6 will win is equal to the chance of rolling (N*7)-1 or less on (N+N+1)d6.

Messed around a little more, and it looks like 1d6 ties 3d6 as often as 4d6 will roll exactly 7; 1d6 beats 3d6 as often as 4d6 rolls 6 or less; 2d6 ties 4d6 as often as 6d6 will roll exactly 14; and 2d6 beats 4d6 as often as 6d6 rolls 13 or less. So I expect that, when rolling Nd6 against (N+2)d6, they'll tie as often as (N+N+2)d6 rolls exactly N*7, and that Nd6 will beat (N+2) as often as (N+N+2) rolls (N*7)-1 or less.

Wait. Those look the same. Can I just follow this pattern for any number of d6?

Nd6 ties (N+M)d6 as often as (N+N+M)d6 rolls exactly (7*N).
Nd6 beats (N+M)d6 as often as (N+N+M)d6 rolls (7*N)-1 or less.

And then, since the "7" there is a result of the average outcome of a single d6, can I extend it to other sizes of dice?

NdD ties (N+M)dD as often as (N+N+M)dD rolls exactly ((D+1)*N).
NdD beats (N+M)dD as often as (N+N+M)dD rolls ((D+1)*N)-1 or less.

Seriously just guessing at this point (could probably figure it out by hand with more spreadsheets, or by pure math from Pascal's Triangle) but I think this is going to work.


EDIT: VanVelding has tackled the math for this here and here.
EDIT ²: and here.