Concrete mathematics : a foundation for computer science

8.4 FLIPPING COINS 395
But there’s an interesting interplay between the patterns when both are
considered simultaneously. Let SA be the sum of Alice’s winning configurations,
and let Ss be the sum of Bill’s:
SA = HHT + HHHT + THHT + HHHHT + HTHHT + THHHT + . . . ;
Ss = HTT + THTT + HTHTT + TTHTT + THTHTT + TTTHTT + . . . .
Also- taking our cue from the trick that worked when only one pattern was
involved-let us denote by N the sum of all sequences in which neither player
has won so far:
N = 1 +H+T+HH+HT+TH+TT+HHH+HTH+THH+... . (8.77)
Then we can easily verify the following set of equations:
l+N(H+T) = NfS~f.5.s;
NHHT = SA ; (8.78)
NHTT = SATTS~.
If we now set H = T = i, the resulting value of SA becomes the probability
that Alice wins, and Ss becomes the probability that Bill wins. The three
equations reduce to
1 +N = N +sA +Ss; ;N = s,; ;N = $A +sg;
and we find SA = f , Ss = f . Alice will win about twice as often as Bill!
In a generalization of this game, Alice and Bill choose patterns A and B
of heads and tails, and they flip coins until either A or B appears. The
two patterns need not have the same length, but we assume that A doesn’t
occur within B, nor does B occur within A. (Otherwise the game would be
degenerate. For example, if A = HT and B = THTH, poor Bill could never win;
and if A = HTH and B = TH, both players might claim victory simultaneously.)
Then we can write three equations analogous to (8.73) and (8.78):
1 +N(H+T) = N+SA+S~;
min(l,m)
NA = SA i A(lPkj [A (k’ =A(kj] + sp, x A(lmk) [Bck) =Aiki];
k=l k=l
min(l,m)
NB = SA x B lrnpk’ [Atk’ = B(k)]
k=l
+ Ss 5 BCmPk) [Bck) = B,,,] .
k=l
(8.79)