Silvanus said:
...Also, according to Bloomberg, this event (the $6000 one) happened during a special event in which the chances are doubled. 3% is the regular, non-event drop-rate, so it should have been 6%.
If we take that number, then here is the actual chance of failing a 6% change 2200 consecutive times (assuming a fair uniform distribution):
7.6081299e-60
This expressed as a percentage is 7.6e-58%. Before I expand it, here is something fun I wanted to throw in as well - the chance of winning the lottery 6/49 is 1 in 13,983,816 which expressed as a percentage is 7.2e-6%, while 6/99 is 1 in 1,120,529,256 or 8.9e-8%. Now, with that in mind, let's just show all of these expanded
0.00000000000000000000000000000000000000000000000000000000076% - chance of failing at 6% 2200 times
0.00000000000000000000000000079% - chance of failing at 3% 2200 times
0.000000089% - chance of winning a 6/99 lottery
0.0000072% - chance of winning a 6/49 lottery
I can't be bothered right now, but it'd be fun to try and find how many lottery tickets do $6000 buy you, find what's the lottery around where the guy was and if it would have been a better bet to just go for the lottery instead. OK, the answer would be "no", but at least looking at the statistics would have been interesting.