The discovering mathematics module has made me curious about things in everyday life that involve maths. Squash is a fast game which requires a good level of physical fitness and agility but also a great deal of maths is involved in playing the game.

In a game of squash you can only win a rally on your own serve and the serve only passes from one player to another when the serving player loses a rally. The winner is usually the player first to reach 9 points. However if it is 8 all the game can be played to 9 or 10. The person who is receiving or in other words lost the point decides this. So what should you decide? This is where maths comes into the game?

The answer all comes down to probability which of course is a fundamental mathematics principle.

According to Beardon (2001)

‘Assume players are called A and B. Assume that the probability of A winning a rally is p, whether he serves or not, and whatever the score is. The probability for B is q, where p+q=1

There are four outcomes to be considered and need to find the four probabilities:

Pr(A wins the next point given that A serves first)  =θ,

Pr(A wins the next point given that B serves first)  =φ,

Pr(B wins the next point given that A serves first)  =λ,

Pr(B wins the next point given that B serves first)  =μ.

Let us find θ. One possibility is that A wins the first rally, and  the first point; he does this with probability p. If not, loses with probability q and B then serves.

For A to win the next point he must win the next rally (with probability p) and then the situation returns to that in which A serves and wins the next point with probability θ. Therefore  θ=p+qpθ, and hence Pr(A wins the next point given that A serves first)=θ=p1−pq.

Next, we find φ. As B serves first, A must win the next rally (with probability p). The position now is that A is serving and must win the next point (which he does with probability θ). Thus φ=pθ, and hence Pr(A wins the next point given that B serves first)=φ=p21−pq.

By interchanging A with B, and p with q, we see that Pr(B wins the next point given that A serves first)=λ=q21−pq, andPr(B wins the next point given that B serves first)=μ=q1−pq.

Note that

Pr(A or B wins the next point given that B serves first)=φ+μ,

and

φ+μ=p21−pq+q1−pq=p(1−q)+q1−pq=p+q−pq1−pq=1−pq1−pq=1.

It follows that if B serves first, then the probability that A or B eventually wins a point is one; hence the probability that the game goes on for ever is zero! You may like to draw a tree diagram to illustrate this, and you will find that a succession of rallies in which nobody scores a

point is represented by a long ‘zig-zag’ in your tree diagram. The start of the tree diagram is

.

The rules of squash say that if the position is reached when the score is (8,8) (we give A’s score first), and B is to serve, then A must choose between the game ending when the first player reaches 9 or when the first player reaches 10. We want to decide what is the best choice for A to make (when p and q are known).

If A chooses 9, then he wins the game with probability φ. Suppose now that A chooses 10; then the sequence of possible scores and their associated probabilities are as follows:

Sequence of scores	Probability	Probability in terms of p,q
(8,8)→(9,8)→(10,8)	φ×θ	p3/(1−pq)2
(8,8)→(8,9)→(9,9)→(10,9)	μ×φ×θ	p3q/(1−pq)3
(8,8)→(9,8)→(9,9)→(10,9)	φ×λ×φ	p4q2/(1−pq)3

It follows that the choice of 9 by A is best for A if φ>φθ+μφθ+φ2λ, or, equivalently, 1>θ+μθ+φλ. In terms of p and q this inequality is

1>p1−pq+pq(1−pq)2+p2q2(1−pq)2, and this is equivalent to (1−pq)2>p(1−pq)+pq+p2q2.

After simplification ( p+q=1), this is equivalent to p2−3p+1>0.

This means that A should choose 9 if p<0.38 and should choose 10 if p>0.38.’

References

https://nrich.maths.org/1390