Math question: If I have N N-sided dice (e.g. 100d100) and I roll them repeatedly, removing any that roll maximum (e.g. 100), what is the expected number of rolls until I have none left, in terms of N?
Yes, agreed, sorry - N^2 was from a dumb mistake I made. The results I get currently are ~14 for d6 (which is close to 6/6+6/5+6/4+6/3+6/2+6/1), 28 for d10, 519 for d100, 1176 for d200.
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Yeah, I'm at 517 for d100.
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Which is still O(n^2) growth clearly.
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Ok. So Wolfram Alpha says that series for N = 75 is roughly 367.
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Simulation says 365, 366 -- seems to work! WHY it works I have no idea.
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Go back to your tweet about intuition. It takes on average 1 roll to go from 100 dice to 99. Then it takes on average 100/99 rolls to go down to 99 dice. Then it takes on average 100/98 rolls to go down to 97 dice. Then it takes...
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I guess the possibility of losing multiple dice in one roll just kinda comes out in the wash.
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Yeah, that makes sense, since you have equal probability of undershooting.
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I mean it's no surprise to me that I'm a nerd but I'm enjoying this conversation far too much...
End of conversation
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