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The average roll on a six sided die (d6) is 3.5.

If the result is a 1 you are allowed a single re-roll accepting the second result. What does the average roll increase to? And how to calculate this?

Similarly the average roll on 2d6 is 7. If the result of ONE of the d6s is 1 you are allowed a single re-roll accepting the second result (so if two 1s are rolled you may only re-roll one of them). What does the average roll increase to? And how to calculate this?

My maths skills are poor and I have been unable to find any resources on the internet to explain this to me. Many thanks.

From comment:

I work out the average of 1d6 is (1+2+3+4+5+6)/6= 3.5 The average roll of the 2-6 is (2+3+4+5+6)/5 = 4 The average roll of the 1 will go back to being 3.5 as the re-roll will make it a normal die roll. Where I lost is how to combine the average of the 4 for the 2-6 results with the average of the re-rolled 1 which is 3.5. If I take the average of both 4 + 3.5 = 3.75 but this does not fell right as it is giving equal weight to each result when you have an 5/6 chance of getting 2-6 and only a 1/6 chance of getting 1.

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  • $\begingroup$ Welcome to MSE. Please show your thoughts or efforts towards solving the problem. $\endgroup$
    – Kumar
    Dec 10, 2021 at 11:34
  • $\begingroup$ I work out the average of 1d6 is (1+2+3+4+5+6)/6= 3.5 The average roll of the 2-6 is (2+3+4+5+6)/5 = 4 The average roll of the 1 will go back to being 3.5 as the re-roll will make it a normal die roll. Where I lost is how to combine the average of the 4 for the 2-6 results with the average of the re-rolled 1 which is 3.5. If I take the average of both 4 + 3.5 = 3.75 but this does not fell right as it is giving equal weight to each result when you have an 5/6 chance of getting 2-6 and only a 1/6 chance of getting 1. $\endgroup$
    – ian432
    Dec 10, 2021 at 11:48

2 Answers 2

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There are $11$ possible outcomes : $2,3,4,5,6$; $(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)$.

The first $5$ of these each have probability $\tfrac16$, and the last six each have probability $\tfrac{1}{36}$.

For each outcome, you have a score and a probability.

The scores are $2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6$.
Probabilities are $\tfrac16$, $\tfrac16$, $\tfrac16$, $\tfrac16$, $\tfrac16$, $\tfrac{1}{36}$, $\tfrac{1}{36}$, $\tfrac{1}{36}$, $\tfrac{1}{36}$, $\tfrac{1}{36}$, $\tfrac{1}{36}$.

To get the average, you multiply the scores by their associated probabilities, and add up. So $$ \frac16(2+3+4+5+6) + \frac{1}{36}(1+2+3+4+5+6) = \frac{141}{36} \approx 3.916 $$

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  • $\begingroup$ Thank you that's very helpful explanation. $\endgroup$
    – ian432
    Dec 12, 2021 at 0:24
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Your comment essentially had the answer. You said (edited down)

The average roll of the $2-6$ is $4$. The average roll of the $1$ will go back to being $3.5$ as the re-roll will make it a normal die roll. You have a $5/6$ chance of getting $2-6$ and only a $1/6$ chance of getting $1$.

So the overall mean of the distribution of outcomes is $\frac56 \times 4+\frac16\times 3.5 = \frac{47}{12}\approx 3.9167$

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  • $\begingroup$ thank you for the explanation that makes sense to me now. $\endgroup$
    – ian432
    Dec 12, 2021 at 0:25

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