I simulated it and it's definitely N^2, which is a very neat result. I haven't yet figured out *why* that's true.
-
-
Replying to @avibryant
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.
1 reply 0 retweets 0 likes -
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.
1 reply 0 retweets 0 likes -
Replying to @avibryant
Urgh am I going to have to simulate this myself....
1 reply 0 retweets 0 likes -
Replying to @DjangoWexler
My Scala simulation code (fixed to work, I think), fwiwhttps://gist.github.com/avi-stripe/6eb925629e1ae8352552242ef7cf3ede …
2 replies 0 retweets 0 likes -
Replying to @avibryant
Unless my sim is very wrong it's not close to N^2 actually. Results are in the 400-700 range.
1 reply 0 retweets 0 likes -
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.
3 replies 0 retweets 0 likes -
Replying to @avibryant @DjangoWexler
All of which match this.pic.twitter.com/dyyEZ6LIQW
2 replies 0 retweets 2 likes -
Replying to @avibryant
Ok. So Wolfram Alpha says that series for N = 75 is roughly 367.
2 replies 0 retweets 1 like -
Replying to @DjangoWexler
That matches my simulation, yep. Was the goal to figure out how many dice you'd need so that it took about a year to run out?
2 replies 0 retweets 0 likes
Rolling once a day I mean.
Loading seems to be taking a while.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.