On Feb 11, 8:52 am, Archie <kohmover...@yahoo.com> wrote: > Say you can choose six numbers out of 59. What is the probability that > three (and only three) will be consecutive? This is complicated by the > fact that numbers appear in order from least to greatest. What if the > six numbers contain 2 sets of 3 consecutive numbers? > > I'm not finding any easy formula for this. > > Thanks!
I'll assume that you want the probabilities after the six numbers are sorted in ascending order. In other words, the order in which the six are selected is irrelevant.
Let n be the number of numbers from which the six are chosen (in your case n = 59). The total number of ways to choose the six, disregarding order, is C(n, 6) ("n choose 6"), where C(x, y) is a binomial coefficient calculated as C(x, y) = x!/(y!*(x - y)!).
Number of ways to choose six such that there are two separated runs of three in consecutive order =
n - 6 ways if the lower run is 1, 2, 3 + n - 7 ways if the lower run is 2, 3, 4 + n - 8 ways if the lower run is 3, 4, 5 + ... + 1 way if the lower run is n - 6, n - 5, n - 4
= (n - 6)*(n - 5)/2
So probability = (n - 6)*(n - 5)/(2*C(n, 6))
Number of ways to choose six such that there is exactly one run of three and no longer run =
C(n - 4, 3) if the run is 1, 2, 3 (remaining three can be any three from n - 4) + C(n - 5, 3) if the run is 2, 3, 4 (remaining three can be any three from n - 5) + C(n - 5, 3) if the run is 3, 4, 5 ... + C(n - 5, 3) if the run is n - 3, n - 2, n - 1 + C(n - 4, 3) if the run is n - 2, n - 1, n
= 2*C(n - 4, 3) + (n - 4)*C(n - 5, 3)
BUT... every selection containing two separated runs of three will be included twice in this count, so we need to subtract twice the answer to the first question to get the required