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?
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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
New conversation -
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I'm still standing by my answer: X= [log(1/n)]/Log((n-1)/n)], but apparently I'm answering " "when is it more likely than not that I have no dice left" not "what is the expected number of rolls until I have none left" which according to my slack is a WILDLY DIFFERENT question.
Thanks. Twitter will use this to make your timeline better. UndoUndo
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