[RyanW]

Assuming my math is right:

The final roll of an "exploding" d6 is effectively a d5 (if it rolls a six, it isn't the last die) which will have an average value of 3.

The number of expected 6s rolled is the infinite sum of 5n/6^(n+1), which is 0.2. If each six adds 5 to the total, that's just 0.2*5 = 1.

The total average is just 3+1 = 4.

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If all 1s are rerolled, the number of expected 6s becomes the infinite sum of 4n/5^(n+1), or 0.25. The last die would average 3.5 (it's now effectively d4+1). So 0.25*5+3.5 = 4.75.

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If only an initial 1 is rerolled, it gets just a little bit trickier, but brute force spreadsheet approximations still work just fine (you just have to have the first line use a different formula from the rest). I get 0.24 rerolls for a total average of 4.6.