Ah, this would mean that my table is wrong againgreatkingrat wrote:The station missing was Limehouse. Actually this is a problem with Matt's R15 generator as it does not seem to be in the system at all.

If there is no redraw, N = 141 and m = 23 (as Peter had correctly used before):
P( k = 0 ) = 7.21923%
P( k = 1 ) = 22.139%
P( k = 2 ) = 29.8667%
P( k = 3 ) = 23.4468%
P( k = 4 ) = 11.9405%
P( k = 5 ) = 4.16274%
P( k = 6 ) = 1.02176%
P( k = 7 ) = 0.178841%
P( k = 8 ) = 0.0223552%
P( k = 9 ) = 0.00197833%
P( k = 10 ) = 0.000121477%
P( k = 11 ) = 4.99351e-06%
P( k = 12 ) = 1.29142e-07%
P( k = 13 ) = 1.86794e-09%
P( k = 14 ) = 1.13071e-11%
If there is a redraw, N = 142 and m = 24:
P( k = 0 ) = 6.50748%
P( k = 1 ) = 20.8239%
P( k = 2 ) = 29.3696%
P( k = 3 ) = 24.1544%
P( k = 4 ) = 12.9159%
P( k = 5 ) = 4.73978%
P( k = 6 ) = 1.22803%
P( k = 7 ) = 0.227589%
P( k = 8 ) = 0.0302267%
P( k = 9 ) = 0.00285326%
P( k = 10 ) = 0.000187714%
P( k = 11 ) = 8.30989e-06%
P( k = 12 ) = 2.3282e-07%
P( k = 13 ) = 3.67369e-09%
P( k = 14 ) = 2.44616e-11%
So, from a statistical point of view, redrawing the stations to include Limehouse would lower the probability of drawing no DLR stations.
It's funny you should mention that - I was thinking about how it compared with winning the National Lottery. To put it into perspective, there is more chance of drawing 12 DLR stations than winning the jackpot in the Euro Millions.Mitchell&BrownLook wrote:I'll also add that 0.0000000000003% is equivalent to a 1 in one quadrillion chance. Winning Euro Millions is 1 in 76,275,360.