Fellow Forum Members I have dusted off the abacus and have 'attempted' to calculate the probability that one team can be given a first round bye twice in a 3 year period in a 17 team competition based on random selection. Here is the result:
The probability that a given team is awarded a first round bye in two of three consecutive years is (48⁄4913), or about 0.980 %.
In any one year only one of the 17 teams receives the first round bye, so the chance that a particular team gets it is (1⁄17).
P=1/17
Across three independent years, the number of first round byes a given team gets follows a binomial distribution;
X Bin(3, \(1⁄17)\).
We want P(X=2)
P(X=2)=(3/2)(1/17)2(1−1/17)
Compute the value.
P(X=2)=3⋅1/289⋅16/17=48/4913≈0.0098
Therefore according to this probability calculation the chances of a team being awarded a first round bye twice out of three consecutive years in a 17 team competition on a random basis is 0.0098, or less than 1%.
As always I accept NO responsibility should this arithmetic be proven wrong.
Only time we have ever "won" the NRL lottery,needless to say it is one that you try to avoid.