Generalizing from my analysis for 2 consecutive heads (HH) not occurring in (n) total flips of an unbiased coin, I have developed a general formula for (m) consecutive heads (HH...H(m)) not occurring in (n) total flips.

The solution for (m) = 2 in (n) total flips was the product of p1 x p2 x p3 x ...p(n-(m-1)), where the p's and probabilities for 10 total flips were defined as follows:

p1 = 3/4 = .75 After 2 flips probability no HH = .75
p2 = 5/6 = .83333 After 3 flips probability no HH = .625
p3 = 8/10 = .8 After 4 flips probability no HH = .5
p4 = 13/16 = .8125 After 5 flips probability no HH = .40625
p5 = 21/26 = .80769 After 6 flips probability no HH = .328125
p6 = 34/42 = .80952 After 7 flips probability no HH = .265625
p7 = 55/68 = .80882 After 8 flips probability no HH = .214844
p8 = 89/110 = .80909 After 9 flips probability no HH = .173828
p9 = 144/178 = .80899 After 10 flips probability no HH = .140625

A close look will reveal that the series formed by the differences between the denominators and numerators of each p is the Fibonacci series 1,1,2,3,5,8,13,... (as was noted by Jeff Merrick), with each integer in the series the sum of the previous 2 integers. Therefore, since den(k) = 2 x num(k-1), num(k) = den(k) - Fib(k).

Further analysis shows that the general form of this formula for (m) occurrences of heads (HH...H(m)) not occurring in (n) total flips is as follows: den(k) = 2 x num(k-1), num(k) = den(k) - (m)sFib(k), where (m)sFib is a set of super-Fibonacci series' that each begin like a normal Fibonacci series (1,1,2,...) but continue such that each integer in the series is the sum of the previous (m) integers.

The solution for (m) = 3 in (n) total flips, therefore, is the product of p1 x p2 x p3 x ...p(n-(m-1)), where the p's and probabilites for 10 total flips are as follows:

p1 = 7/8 = .875 After 3 flips probability no HHH = .875
p2 = 13/14 = .92857 After 4 flips probability no HHH = .8125
p3 = 24/26 = .92308 After 5 flips probability no HHH = .75
p4 = 44/48 = .91667 After 6 flips probability no HHH = .6875
p5 = 81/88 = .92045 After 7 flips probability no HHH = .63281
p6 = 149/162 = .91975 After 8 flips probability no HHH = .58203
p7 = 274/298 = .91946 After 9 flips probability no HHH = .53516
p8 = 504/548 = .91971 After 10 flips probability no HHH = .49219

Note, that for (m) = 3 the (3)sFib series is 1,1,2,4,7,13,24,44,...

Note also, that there is a shortcut way to calculate the probability of (m) consecutive occurrences of heads (HH...H(m)) not occurring in (n) total flips without actually multiplying the p's together (p1 x p2 x ...p(n-(m-1)). It is as follows: because the total number of possible ordered outcomes in (n) total flips is 2^n, and because every ordered outcome is equally likely to occur (assuming an unbiased coin), once num(k) for (n) flips has been calculated, the probability of (m) consecutive occurrences of heads not occurring in (n) total flips is given by num(k)/2^n.

Bryce Carlson