3.2. RUIN PROBABILITIES 99
Then add these together,and after some algebra, the total is 1. (Check it
out!)
Forp= 1/2 =q, the proof is simpler, since thenpT 0 = 1−(a−T 0 )/a,
andqT 0 = 1−T 0 /a, andpT 0 +qT 0 = 1 easily.
Corollary 2.The expected gain against the infinitely rich adversary isE[G] =
(1−qT 0 )a−T 0.
Proof. In the game against the infinitely rich adversary, the gambler’s ulti-
mate gain (or loss!) is a Bernoulli (two-valued) random variable,G, where
Gassumes the value−T 0 with probabilityqT 0 , and assumes the valuea−T 0
with probabilitypT 0. Thus the expected value is
E[G] = (a−T 0 )pT 0 + (−T 0 )qT 0
=pT 0 a−T 0
= (1−qT 0 )a−T 0.Now notice that ifq= 1/2 =p, so that we are dealing with a fair game,
thenE[G] = (1−(1−T 0 /a))·a−T 0 = 0. That is, a fair game in the short
run is a fair game in the long run. However, ifp < 1 / 2 < q, so the game is
not fair then our expectation formula says
E[G] =
(
1 −
(q/p)a−(q/p)T^0
(q/p)a− 1)
a−T 0=
(q/p)T^0 − 1
(q/p)a− 1a−T 0=
(
[(q/p)T^0 −1]a
[(q/p)a−1]T 0− 1
)
T 0
The sequence [(q/p)n−1]/nis an increasing sequence, so
(
[(q/p)T^0 −1]a
[(q/p)a−1]T 0− 1
)
< 0.
Remark.An unfair game in the short run is an unfair game in the long run.