That kind of makes intuitive sense since at 100 dice it takes expected 1 trial to lose, and at 1 die it takes 100, which sort of hints to at least an O(N^2) solution.
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Replying to @DjangoWexler
Wait, I screwed my simulation up, but your tweet here's a good way to think about it. It's the sum for i=1 to N of N/i.
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Replying to @avibryant
Urgh am I going to have to simulate this myself....
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Replying to @DjangoWexler
My Scala simulation code (fixed to work, I think), fwiwhttps://gist.github.com/avi-stripe/6eb925629e1ae8352552242ef7cf3ede …
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Replying to @avibryant
Unless my sim is very wrong it's not close to N^2 actually. Results are in the 400-700 range.
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Replying to @DjangoWexler
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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Replying to @avibryant @DjangoWexler
All of which match this.pic.twitter.com/dyyEZ6LIQW
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Replying to @avibryant
Ok. So Wolfram Alpha says that series for N = 75 is roughly 367.
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Replying to @DjangoWexler @avibryant
Simulation says 365, 366 -- seems to work! WHY it works I have no idea.
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Replying to @DjangoWexler
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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Replying to @avibryant
Yeah, that makes sense, since you have equal probability of undershooting.
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Replying to @DjangoWexler @avibryant
I mean it's no surprise to me that I'm a nerd but I'm enjoying this conversation far too much...
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