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

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?

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.)