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.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.

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.

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).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?

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

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

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

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

*EDIT*

^{2}: and here.