Which is the correct way to determine the mean result of 1d12 reroll 1s & 2s?

Question

I know the mean is simply the sum of the members divided by the number of them, so the average value given by rolling one $12$ sided die once is $(1+2+3...11+12)/12$ for $6.5$. Straightforward enough.

In this case, however, if that one roll lands on a $1$ or a $2$, you roll one more time and take the second value instead (even if that is a $1$ or $2$).

Now there are far more possible eventualities; if you roll a $1$ or a $2$, that’s just the first element of $12$ different possible results each, but $3-12$ are their own possibilities. That gives $2$ ways of getting a $1$ $(1, 1; 2, 1)$ or $2$, and $3$ ways of getting a $3$ $(1, 3; 2, 3; 3)$ through $12: 34$ possible eventualities.

Treating them as individual possibilities like that gives $(1+1+2+2+3+3+3+4+4+4+5...11+11+11+12+12+12)/34$, which is $6.794117647058824$.

Alternatively it seems to me that you could simply add the average values yielded by an initial $1$ or $2$ instead of going through all the actual results, for $(6.5+6.5+3+4+5...11+12)/12$. This yields $7.3333...$ though.

Which is right, and why is the other wrong?

Answer 1

The second one is correct. The problem with your first approach is that although there are three ways of getting 3, they aren't all equally likely, since getting 1 then 3 has probability $\frac1{12}\times\frac1{12}$, but getting 3 direct has probability $\frac1{12}$. (One way to correct for this is to pretend there are actually $14$ ways of getting 3, by considering what you would have rolled on the next die if you hadn't kept with your 3 - this makes all the ways equally likely.)