The first method to calculate the average of 2d6 with a single re-roll of one 1:
The average roll on a d6 is (1+2+3+4+5+6)/6 = 20/6 = 3.5
The average roll on a d6 with a single re-roll if the result is 1 is ((2+3+4+5+6) x $\frac16$) + ((1+2+3+4+5+6) x $\frac1{36}$) = $\frac{141}{36}$ = 3.92
To get the average roll of 2d6 with a single re-roll of one 1, I add the two averages of the single d6 rolls together.
3.5 + 3.92 = 7.42
The second method to calculate the average of 2d6 with a single re-roll of one 1:
Calculate the chance of rolling all possible values on 2d6 (ie 2-12) including a single re-roll of one 1.
The formula is (number of 2d6 combinations x $\frac1{36}$) + (number of 3d6 combinations × $\frac1{216}$)
I worked out the combinations manually eg 4 on 2d6 can be made with the following combinations (re-rolled result in brackets[])
(2,2)
(3,1,[1]) (1,[3],1) (1,1,[3]) (2,1,[2]) (1,[2],2)
After identifying all combinations that make up each result 2-12 I have the following:
2 is (0 x $\frac1{36}$) + (1 x $\frac1{216}$) = $\frac1{216}$
3 is (0 x $\frac1{36}$) + (3 x $\frac1{216}$) = $\frac3{216}$
4 is (1 x $\frac1{36}$) + (5 x $\frac1{216}$) = $\frac{11}{216}$
5 is (2 x $\frac1{36}$) + (7 x $\frac1{216}$) = $\frac{19}{216}$
6 is (3 x $\frac1{36}$) + (9 x $\frac1{216}$) = $\frac{27}{216}$
7 is (4 x $\frac1{36}$) + (11 x $\frac1{216}$) = $\frac{35}{216}$
8 is (5 x $\frac1{36}$) + (10 x $\frac1{216}$) = $\frac{40}{216}$
9 is (4 x $\frac1{36}$) + (8 x $\frac1{216}$) = $\frac{32}{216}$
10 is (3 x $\frac1{36}$) + (6 x $\frac1{216}$) = $\frac{24}{216}$
11 is (2 x $\frac1{36}$) + (4 x $\frac1{216}$) = $\frac{16}{216}$
12 is (1 x $\frac1{36}$) + (2 x $\frac1{216}$) = $\frac8{216}$
I add all the possible average outcomes together: $\frac1{216}$ + $\frac3{216}$ + $\frac{11}{216}$ + $\frac{19}{216}$ + $\frac{27}{216}$ + $\frac{35}{216}$ + $\frac{40}{216}$ + $\frac{32}{216}$ + $\frac{24}{216}$ + $\frac{16}{216}$ + $\frac8{216}$ = 7.76
The first method gives me an average of 7.42 and the second gives me an average of 7.76. How do I figure out which one is correct (or are both incorrect?)
Best Answer
The first method is incorrect because it results in $\frac 16$ chance of rerolling instead of $\frac {11}{36}$. It is as if you roll one die, which you reroll if it comes up $1$, then roll a second which you cannot reroll. Th second looks right but I did not work through the details.