• I can tell youï»¿ one thing without having to watch this video. I'll only have to wait, at most, 10^10^10^10^10^1.1 years to see this again.

• I trust numberphile's math of coarse, butï»¿ I was bored so I made a program that "flipped a coin" (produced one of two values) until it got 14 wins in a row and kept track of how many tries it took. I had it run 10,000,000 times and find the average tries, it usually comes out relativity close to 32,776 :D yay math

#### Video Responses

This video is a response to 13,983,816 and the Lottery - Numberphile

• Still. That's the max I'llï»¿ have to wait.

• Not quite. That was for the very large universe. It will actuallyï»¿ happen in

10^10^10^10^2.08

• Well, afterï»¿ second look I realized that the phrase "the probability of winning under 14 occasions in a row in 45 games is equal to 1" is not correct. But I think, you've got the idea. Basically, if each time the team had approx. 87.3% chance to win, then this event would not be unlikely at all. (The future will tell :o) )

• What's about considering the following issue (asymmetric coin?): assuming the probability of winning under 14 occasions in a row in 45 games is equal to 1, find the relation of forces between the two teams, given it is constant with time? In other words, t = 45, but the probability to win (p) is higher than the probability to loose (1-p). The equation would turn: 45 = 46*(1-p) + 47*p*(1-p) + 48*(p^2)*(1-p) +...+ 59*(p^13)*(1-p) + 14*p^14. It can be rewritten asï»¿ a 14 order polynomial, p=0.87269..

• 10^10^10^10^2.8ï»¿ for our universe...

• Wrong.ï»¿

• I'm confused by the mathematics used by James to find the average time needed to get 3 heads inï»¿ a row. Can someone point me to a video or something? Thanks.

• This is just for one side. HHH would be winningï»¿ for whoever called heads, losing for whoever got tails, and vice versa for TTT.

• My mistake, I was wrong!

The answer is 2 + 4 + ... +ï»¿ 2 to the power of n