If you have an exploding N-sided die that adds another die when it's at max, the average A is:

A = (1+2+...N+A)/N

This is the same as:

A = (N(N+1) + A)/2N

This, in turn, can be simplified to:

2N*A = N(N+1) + A

And that in turn, can be simplified to what Pat Ludwig has so kindly tabulated.

A similar method can be used to calculate averages when the highest roll of a die is substituted by rolling 2 (or more) of the same dice-type.

If you have an exploding N-sided die that adds another die when it's at max, the average A is:

$$ A = \dfrac{1+2+...N+A}{N} $$

This is the same as:

$$ A = \dfrac{N(N+1) + A}{2N} $$

This, in turn, can be simplified to:

$$ 2N \times A = N(N+1) + A $$

And that in turn, can be simplified to what Pat Ludwig has so kindly tabulated.

A similar method can be used to calculate averages when the highest roll of a die is substituted by rolling 2 (or more) of the same dice-type.

If you have an exploding N-sided die that adds another die when it's at max, the average A is:

A = (1+2+...N+A)/N

This is the same as:

A = (N(N+1) + A)/2N

This, in turn, can be simplified to:

2N*A = N(N+1) + A

And that in turn, can be simplified to what Pat Ludwig has so kindly tabulated.

A similar method can be used to calculate averages when the highest roll of a die is substituted by tolling 2 (or more) of the same dice-type.

If you have an exploding N-sided die that adds another die when it's at max, the average A is:

A = (1+2+...N+A)/N

This is the same as:

A = (N(N+1) + A)/2N

This, in turn, can be simplified to:

2N*A = N(N+1) + A

And that in turn, can be simplified to what Pat Ludwig has so kindly tabulated.

A similar method can be used to calculate averages when the highest roll of a die is substituted by rolling 2 (or more) of the same dice-type.