jdoerrie

3 ms·

1. 1.▲
   by jdoerrie 3y ago

   Here's a link to that talk and mentioning of the 50.3%: https://youtu.be/ncHmEUmJZf4?t=2610
2. 2.▲
   by jdoerrie 11y ago

   Given that we stop after the first failure technically this should be a negative binomial distribution ( http://en.wikipedia.org/wiki/Negative_binomial_distribution ) with p = 0.979 and r = 1. The mean is p / (1-p)
3. 3.▲
   by jdoerrie 12y ago

   Not true. Assuming both players move completely random (both choose either 1 or 0 with 50% probability each) and independently (past actions don't influence the current event) player 1 has a 2:1 advantage, i.e. he will win with probabi