316 6 Combinatorics and Probability
Let us find the probability that the first player wins in exactlya+kgames,k=
0 , 1 ,...,b−1. The probability that the first player winsa−1 games out ofa+k−1is
computed using the Bernoulli scheme and is equal to
(a+k− 1
a− 1)
pa−^1 qk, and the probability
of winning the(a+k)th isp. The probability of winning in exactlya+kgames is the
product of the two, namely
(a+k− 1
a− 1)
paqk.
We deduce that the probability of the first player winning the stakes isP=
∑b−^1k= 0(
a+k− 1
a− 1)
paqk,while for the second player this is
Q=qb∑a−^1k= 0(
b− 1 +k
b− 1)
pk.The stakes should be divided in the ratio
P
Q
=
pa∑b−^1k= 0(
a− 1 +k
a− 1)
qkqba∑− 1k= 0(
b− 1 +k
b− 1)
pk.
The first combinatorial identity is equivalent toP+Q=1. For the second combi-
natorial identity, we look for a different way to computeP. Observe that after at most
a+b−1 games have been played, the winner is known. Let us assume that regardless of
the results, the players kept playing all thea+b−1 games. If the first player had won at
leastaof these games, he would have won the stakes as well. HencePis the probability
that the first player wona,a+1,...,a+b−1 of the finala+b−1 games. Each of
these is computed using the Bernoulli scheme, andPis their sum, since the events are
incompatible. We obtain
P=
∑b−^1k= 0(
a+b− 1
a+k)
pa+kqb−^1 −k=
(a+b− 1 )!
a!(b− 1 )!paqb−^1[
1 +
∑b−^1k= 1(b− 1 )···(b−k)
(a+ 1 )···(a+k)(
p
q)k]
.The second identity follows by equating the two formulas that we obtained forP.
This is yet another example of how probability theory can be used to prove identities.
Since “wisdom is the daughter of experience’’ (Leonardo da Vinci), we let you train your
probabilistic skills with the following problems.