I have a closed form solution and its hideous
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goddamn robots can't you _think_
Thanks. Twitter will use this to make your timeline better. UndoUndo
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The probability of it going to one on the first roll would be 1/N then (with n being the new number of sides) it would be 1/n on the second roll, and so on.
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Also, I think the lowest number of the average would have to be 2 and it would have no maximum.
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was about to go ahead and write a sim on this but came across this tweet
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the hero we Need
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This Tweet is unavailable.
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r(1) = 0 r(2) = \sum_1^\infty x/2^x = 2 r(N) =pic.twitter.com/xlrNJGjM5o
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Too lazy to closed form. I put it in python with lru_cache to make the recursion not violently bad and it turns out to grow sloooowwwwpic.twitter.com/l5Ja1mGbyj
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