Two methods for calculating the average roll on 2d6 (including a single re-roll of 1 on one die) give different results - which is correct?

Question

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?)

Answer 1

The first method is definitely not correct as RossMillikan explains.

Your second method is a bit involved, but let me do what I would do, and see if I get the same answer.

OK, I would start with the $36$ possible outcomes for the initial throw of two dice, and then take into account the reroll of those where at least one $1$ appears.

So: for those rolls with no $1$ you get $\frac{1}{36}*(1*4+2*5+3*6+4*7+5*8+4*9+3*10+2*11+1*12)$

For the rolls with a single $1$ you get $\frac{1}{36}*(2*(5.5+6.5+7.5+8.5+9.5)$ (e.g. if you initially throw a $2$ and a $1$, you can expect to get $2+3.5=5.5$)

And getting two $1$'s in the initial throw gives you $\frac{1}{36}*(1+3.5)$

All of this tgether totals: $\frac{1}{36}*(200+75+4.5)=\frac{1}{36}*279.5$

... which I can confirm coincides exactly with your second method's finding, which works out to $\frac{1677}{216}$

Answer 2

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.