How to win a game: the math behind

In an earlier post we discussed the probability of winning a game if you know the probability of winning a point. This is the math behind the formula.

The complicated part of the calculation is that a game of tennis can go over deuce many times, even indefinitely (in theory, at least). We will therefore treat the deuce part and the no-deuce part separately.

Mathematically speaking, the probability of winning a game is the probability of winning without deuce P_{winnodeuce} plus the probability of winning with deuce P_{windeuce} (since these are disjoint events).


Recall that x is the probability of winning a single point. P_{winnodeuce}(x) consists of winning the point after 40-0, 40-15 or 40-30. The question now is how many ways of getting to 40-0, 40-15 or 40-30 are there? The first one (40-0) is easy, there is only one way: 15-0, 30-0, 40-0. The second one is more difficult, as there are already 4 different ways: (1) 15-0, 30-0, 40-0, 40-15, (2) 15-0, 30-0, 30-15, 40-15, (3) 15-0, 15-15, 30-15, 40-15, and (4) 0-15, 15-15, 30-15, 40-15.

Generally, this kind of problem is known as the number of possible selections of k elements amongst a set of n elements, written as {n}\choose{k}. In the context of our problem, n is the number of points played and k is the number of points your opponent scores. Equivalently, k can be seen as the number of points you will score, which is always 3 (you need 3 points to reach 40-?). So there are {n}\choose{3} ways of reaching 40-?. The interpretation is that in a series of n points played, you can select any 3 points among them in order to get to 40-?.

The probability of e.g. winning the game after 40-15 is x times the probability of reaching 40-15. The probability of reaching 40-15 is {{4}\choose{3}}\cdot x^{3}\cdot (1-x)^{1}, as you have to score 3 points (x^{3}) and your opponent scores one point ((1-x)^{1}). So winning after 40-15 is as probable as x\cdot {{4}\choose{3}}\cdot x^{3}\cdot (1-x)^{1}.

Putting it all together for the three different cases (40-0, 40-15, 40-30), we get

P_{winnodeuce}(x)=x\cdot {{3}\choose{3}}\cdot x^{3}+x\cdot {{4}\choose{3}}\cdot x^{3}\cdot (1-x)^{1}+x\cdot {{5}\choose{3}}\cdot x^{3}\cdot (1-x)^{2}. Simplified, this is

P_{winnodeuce}(x)={{3}\choose{3}}\cdot x^{4}+{{4}\choose{3}}\cdot x^{4}\cdot (1-x)+{{5}\choose{3}}\cdot x^{4}\cdot (1-x)^{2} and

P_{winnodeuce}(x)=1\cdot x^{4}+4\cdot x^{4}\cdot (1-x)+10\cdot x^{4}\cdot (1-x)^{2}

We have got part one of P_{wingame}(x). Part two has two main components, the probability of getting to deuce (P_{gettodeuce}(x)) and the probability of winning at deuce (P_{winatdeuce}(x)). In order to get P_{windeuce}(x), these two components can simply be multiplied (P_{windeuce}(x)=P_{gettodeuce}(x)\cdot P_{winatdeuce}(x)).

The probability of getting to deuce is P_{gettodeuce}(x)={{6}\choose{3}}\cdot x^{3}\cdot(1-x)^{3}, as 6 points have to be played, 3 of which either player has to win (this is the number of ways getting to 40-40) times the probability of winning three points each (x^{3},  (1-x)^{3} respectively).

The probability of winning a game at deuce is a function of winning two points in a row, plus winning only one point each (which is deuce again) and winning the game from there. This is the formula: P_{winatdeuce}=x^{2} + 2\cdot x\cdot (1-x)\cdot P_{winatdeuce}. This resolves to P_{winatdeuce}=\frac{x^2}{2\cdot x^2-2\cdot x + 1}.

We get

P_{windeuce}(x)={{6}\choose{3}}\cdot x^{3}\cdot(1-x)^{3}\cdot \frac{x^2}{2\cdot x^2-2\cdot x + 1}=20\cdot x^{3}\cdot(1-x)^{3}\cdot \frac{x^2}{2\cdot x^2-2\cdot x + 1}

Putting it all together, we get the formula we have been looking for:

P_{wingame}(x) = x^{4} + 4\cdot x^{4}\cdot(1-x) + 10\cdot x^{4}\cdot(1-x)^{2} + (20\cdot x^{3}\cdot(1-x)^{3})\cdot(\frac{x^2}{2\cdot x^2-2\cdot x + 1})

For {0}\leq{x}\leq{1}, the function plots like this:



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