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Correct me if my calculation is wrong.It is easier to calculate the probability that there is no two consecutive balls, and the probability isnchoosek(44, 6)/nchoosek(49, 6) = 0.5048.However this does not agree with alexandreC's simulation.The basic idea is that we can think of choosing 6 non-consecutive balls in the following pattern...../*/..../_/*/...../_/*/...../_/*/...../_/*/...../_/*/.....where /*/ stands for the ball we select. "...." stands for the balls between the selections, it can be zero, or any other possible values. For the latter 5 balls, imagine that we combine it with the ball right before it. It ensures a) no two balls selected are consecutive, b) every selection can be mapped into one and only one pattern specified above. The # of possible pattern is nchoosek(49-5, 6) = nchoosek(45, 6).

That is what I got ... the answer would be then 1-Binomial[44,6]/Binomial[49,6] (or 0.495198). The general result matched the direct counting results for a lower number of balls.