Concrete mathematics : a foundation for computer science

396 DISCRETE PROBABILITY
Here 1 is the length of A and m is the length of B. For example, if we have
A = HTTHTHTH and B = THTHTTH, the two pattern-dependent equations are
N HTTHTHTH = SA TTHTHTH + SA + Ss TTHTHTH + Ss THTH ;
N THTHTTH = SA THTTH + SA TTH + Ss THTTH + Ss .
We obtain the victory probabilities by setting H = T = i, if we assume that a
fair coin is being used; this reduces the two crucial equations to
N = S/I x zk ]Alk’ = A.(k)] + Ss x 2k [Bckl = Ackj] ;
k=l
k=l
N =SA 2k [Alk) = B,,,] + Ss x 2k [Bckl = B(k)] .
k=l k=l
(8.80)
We can see what’s going on if we generalize the A:A operation of (8.76) to a
function of two independent strings A and B:
min(l,m)
A:B = x 2kp’ [Alk’ =Bck,] .
k=l
Equations (8.80) now become simply
S*(A:A) + Ss(B:A) = S*(A:B) + Ss(B:B) ;
the odds in Alice’s favor are
SA B:B - B:A
- SB = A:A-A:B
(8.81)
(8.82)
(This beautiful formula was discovered by John Horton Conway [ill].)
For example, if A = HTTHTHTH and B = THTHTTH as above, we have
A:A = (10000001)2 = 129, A:B = (0001010)2 = 10, B:A = (0001001)2 = 9,
and B:B = (1000010)2 = 66; so the ratio SA/SB is (66-9)/(129-10) = 57/l 19.
Alice will win this one only 57 times out of every 176, on the average.
Strange things can happen in Penney’s game. For example, the pattern
HHTH wins over the pattern HTHH with 3/2 odds, and HTHH wins over THHH with
7/5 odds. So HHTH ought to ‘be much better than THHH. Yet THHH actually wins
over HHTH, with 7/5 odds! ‘The relation between patterns is not transitive. In
fact, exercise 57 proves that if Alice chooses any pattern ri ~2 . . ~1 of length
1 3 3, Bill can always ensure better than even chances of winning if he chooses
the pattern ;S2rlr2 . . . ~1~1, where ?2 is the heads/tails opposite of ~2.
Odd, odd.