Concrete mathematics : a foundation for computer science

392 DISCRETE PROBABILITY
But the coin-flipping problem can be solved in a much simpler way,
without the complexities of the general finite-state approach. Instead of six
equations in six unknowns SO, S, , . , . , Ss, we can characterize S with only
two equations in two unknowns. The trick is to consider the auxiliary sum
N = SO + S1 + SJ + S3 + Sq of all flip sequences that don’t contain any occurrences
of the given pattern THTTH:
We have
N = 1 + H + T + HH + . . . + THTHT + THTTT + .
l+N(H+T) = N+S, (8.67)
because every term on the left either ends with THTTH (and belongs to S) or
doesn’t (and belongs to N); conversely, every term on the right is either empty
or belongs to N H or N T. And we also have the important additional equation
NTHTTH = S+STTH, (8.68)
because every term on the left completes a term of S after either the first H
or the second H, and because every term on the right belongs to the left.
The solution to these two simultaneous equations is easily obtained: We
have N = (1 - S)( 1 - H - T) ’ from (&X67), hence
(1 -S)(l -T-H) ‘THTTH = S(1 +TTH).
As before, we get the probability generating function G(Z) for the number of
flips if we replace H by pz and T by qz. A bit of simplification occurs since
p+q=l,andwefind
(1 - G(z))p2q3t5
1-Z
hence the solution is
G(z) =
= G(z)(l +pqV);
p2q’;z5
p2q325 + (1 +pqV)(l - 2)
(8.69)
Notice that G( 1) = 1, if pq # 0; we do eventually encounter the pattern
THTTH, with probability 1, unless the coin is rigged so that it always comes
up heads or always tails.
To get the mean and variance of the distribution (8.6g), we invert G(z)
as we did in the previous problem, writing G(z) = z5/F(z) where F is a polynomial:
F(z) =
p2q3z5+ (1 +pq2z3)(1 - 2)
p2q3
(8.70)