Concrete mathematics : a foundation for computer science

The relevant derivatives are
F’(1) = 5 - (1 +pq2)/p2q3,
F”(1) = 20 - 6pq2/p2q3 ;
and if X is the number of flips we get
EX = Mean(G) = 5-Mean(F) := pP2qm3 +pmlqP1;
VX = Var(G) = -Var(F)
8.4 FLIPPING COINS 393
(8.71)
= -25+pP2q 3 + 7p ‘q--l +Mean(F)’
= (EX)2 -9~~ 2qP” - 3pP’qm’ , (8.72)
When p = 5, the mean and variance are 36 and 996.
Let’s get general: The problem we have just solved was “random” enough
to show us how to analyze the case that we are waiting for the first appearance
of an arbitrary pattern A of heads and tails. Again we let S be the sum of
all winning sequences of H's and T’s, and we let N be the sum of all sequences
that haven’t encountered the pattern A yet. Equation (8.67) will remain the
same; equation (8.68) will become
NA = s(l + A”) [A(“-‘, =A,,_,,] + A(21 [A’m 2) =A(,- 2,]
+.,.$-Aim "[A~'-Ac,i]), (8.73)
where m is the length of A, and where ACkl and Aiki denote respectively the
last k characters and the first k characters of A. For example, if A is the
pattern THTTH we just studied, we have
Ai” = H, Al21 = TH, Ai31 = TTH, Ai41 = HTTH.
A,,, = T, 42, = TH, A(3) = THT, A,,, = THTT:
Since the only perfect match is Ai21 = A ,l), equation (8.73) reduces to (8.68).
Let A be the result of substituting p-’ for H and qm’ for T in the pattern
A. Then it is not difficult to generalize our derivation of (8.71) and (8.72)
to conclude (exercise 20) that the general mean and variance are
EX = T A/k, [Alk) =A/k,] ; (8.74)
k=l
w = (EX)2 - f (2k- l&k) [ACk’ =A[k)] . (8.75)
k=l