Supose the chance of winning each point is p. What is the chance of winning the match?

It is very difficult to tell. Such things have to take into account tiebreaks, deuces, and the last set, each of which could technically remain unresolved forever.

Neil M

This is quite a popular recurring theme. Try reading "Game, Set and Math" by Ian Stewart, which contains enough to show the probability of winning one set, and the methods should allow you to tackle the final set too.

Of course, this is assuming that during a match, the probability of winning a point is constant. This is clearly not realistic anyway, since it doesn't take levels of pressure and exhaustion into account.

-Dave

Also surely the chance of winning a point would be dependent on whether you're serving or not.

But suppose you meant that all points are won with probability p independently.

This looks like it is going to be a very messy final answer. One thing is relatively clear. If p> 1/2 then probability of winning the match > probability of winning any individual game > p (unless p = 1 obviously). If p< 1/2 then probability of winning match < probability of winning any game < p.

This is because a match is played over several points, so this increases the chance of winning for the better player. Therefore a small difference in the chance of winning each point will cause an exaggerated chance of winning the match.

I will do the probability of winning a game once it gets to deuce first. Then later I'll see if we can find the probability of winning a game.

If it's at Deuce or 30-30, let the chance of winning the game be x.

x = chance of winning in next two points + chance of it being a deuce after two points and winning from there.

So x = p² + 2p(1-p)x

x[1-2p+2p² ] = p²

x = p² /[1-2p+2p² ]

That is therefore the chance of winning from deuce.

Now the chance of winning from 40-30 is obviously:

p + (1-p)x

because you can either win the next point, or lose the next point and win via deuce.

The chance of winning from 40-15 is:

(1-p)² x + 2p(1-p)

because the two ways of doing it are:

- losing the next two points then winning from deuce
- winning one of the next two points

These are mutually exclusive so we can add them.

The chance of winning from 40-0 is:

[1-(1-p)³ ] + (1-p)³ x

Chance of winning from 30-30 = x

Chance of winning from 30-15 is:

p * R + (1-p)x

where R = chance of winning from 40-15 from above.

Chance of winning from 15-15 is p * last result + (1-p)*[last result with ps replaced by 1-ps]

And we could then do the same for 15-0, and hence 0-0. We then have all the probabilities of winning from a-b where a³ b. Then we could switch p for 1-p to find the chance of winning from b-a.

This looks absolutely horrible. Maybe we should try a numerical example. e.g. p = 0.51. What would be the chance of winning the game? My guess would be about 55%. The match - maybe 60%-65%. The reason this is so high is because the match is dependent on SO many points, so the 1% above 50% should possibly be very significant.

Anybody got a more efficient approach?

Yours,

Michael

The relevant thing is that you can never know if the player is playing on their favourite surface. A lot of things have to be taken in account which we would find hard to predict:
e.g weather
level of fitness
motivating factors(family)
The truth is that this could be very difficult to predict.

Dan

So are you suggesting we can work out an absolute probablility of winning a tennis match? Including all these factors? I think we'd be using game theory to get a decent answer for a match.

Neil M

In order to make any sort of problem solvable that isn't just pure mathematics, we need to assume that those conditions are already included in p.

Well I think we have to assume the Anonymous sender wanted to assume that none of these factors above matter, and just to assume all points are independent at probability p.

Then the interest in the problem is not so much predicting the score of any individual match, but rather seeing whom the scoring system in tennis favours (and by how much)? For example, suppose tennis matches had the more obvious (fairer but less exciting) system, e.g. first player to 250 points wins. Would this system be better for the better player than the conventional system? Obviously both systems will give the better player a > 1/2 chance of winning, but by how much in each case? That is the sort of question we may be able to estimate by working out the probability of winning a match assuming all points are independent.

So if the chance of winning a point is 51%, I've worked out the chance of winning from deuce is 52% (not as much difference as I thought). I had a go at the others numerically (using a calculator):

OK, I've made a spreadsheet to work out the result for any particular p. (A spreadsheet seemed prefarable to a program in this instance.) Again we're assuming all points are won with probability p (whether serving or not). Here are the results:

to 6 decimal places.

Obviously the big fault with this is that we're pretending it doesn't matter who serves, when in fact the game is almost entirely dominated by serve. I should try it out for table tennis (a game I play much more than tennis) as serving is less important here.

Yours,

Michael

But if the chance of winning a point is 70% then surely this is scaled up to the whole match, then the chance of winning the match is 70% not 100%

Michael - How did you go about getting the result from deuce? did you individually consider all the conditions of this. I mean, if no-one wins two points in a row, it can go on indefinitely (until one of the players falls asleep)

Neil M

Anonymous - it is not true to say that the probability of winning a point "scales up" to the probability of winning the whole match. The longer a match is, the more time the better player has to assert his superiority. A general principle in statistics called the law of large numbers says that the greater the sample (i.e. number of points played) the more likely it is that the number of occurences of the event = number of trials * probability to any degree of accuracy. So if the game was a million sets, then even if the players were really close each point (50.1-49.9), the chance of the worse player sustaining that and winning a majority of the million sets is vanishingly small.

Neil (you forgot to type in your password). Let x = probability of winning from deuce.

If you win from deuce there are two possibilities

- you win the next two points
- you win one of the next two points, lose the other one but then win the game from deuce.

The probability of the first possibility = p²
The probability of the second possibility = P(winning exactly one of the next two points)*P(winning it from deuce) = 2p(1-p)x, as x = probability of winning from deuce.

So the sum of these is equal to x:

x = p² + 2p(1-p)x

x[1-2p+2p² ] = p²

So x = p² /[1-2p+2p² ].

So that is the chance of winning it from deuce.

Michael