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# Modelling Probabilities in Games of Tennis

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Introduction

Math HL Portfolio Type II:

Modelling Probabilities in Games of Tennis May 2009

In this portfolio we will look at the probability involved in playing tennis.  Our calculations will be based on the estimated probability a player has of scoring a point.  We will develop models for different kinds of tennis games and use Excel to explore up to what extent we can exploit the two probabilities with which we start.  Furthermore, we will differentiate between probability and odds, comparing them and analyzing how they can affect the way we look at the same numbers.  In my conclusion I will mention the possibility of involving other kinds of distribution in this portfolio, such as Poisson.

Part 1: Club Practice.

1. Games to 10 points.

a) Since we know that Adam wins about twice as many points as Ben does, we can say that the probability of Adam winning a point is , and the probability of Ben winning a point is . So, given that P(A) is the probability of Adam winning a point and P(B) is the probability of Ben winning a point, we have that:

P(A)= P(B)= Clearly, this is a binomial distribution.  Hence, we will use the formula , where n is the total number of trials, x the number of successes, p the probability of success and q the probability of failure.  Because we want the variable x to represent the number of points won my Adam, we will substitute P(A) for p and P(B) for q.  Regardless of what is considered a success or a failure, n will be 10.  Following these guidelines our model will be the following: Where

Middle .  In other words, the total number of points played now does not need to be less than 7.  However, because 4 is still the minimum amount of points required to win a match, the cases of still apply as the “non-deuce cases.”  With this in mind we can already separate three probabilities which will be unaffected by whether the game goes to deuce or not:
 Y, number of points played: P(A), Probability that Adam wins: P(B), Probability that Ben wins: Y=4   Y=5  Y=6  Sum: Sum: From this point on, if the game goes past 6 points it means that a deuce has been called.  However, the game cannot be settled at 7 points because a superiority of 2 points is required.  This means that if deuce is called at 6 points, and both Adam and Ben win one point again, deuce will be called once more.  If A is a point won by Adam and B a point won by Ben, it would look like this:

AAABBB (+ all other possible combinations for the first deuce) AB

AAABBB (+ all other possible combinations for the first deuce) BA

For the game to end at 8 points the same player must score two points consecutively:

AAABBB (+ all other possible combinations for the first deuce), AA

AAABBB (+ all other possible combinations for the first deuce), BB

This basically states that once the first deuce is called the games can only end on an even number of points.  Either way, the probability of the game going to the first deuce should be calculated if we want to know what the probability of the consecutive deuces is.  The probability of going to the first deuce,

Conclusion

An interesting field of further investigation for this portfolio would be to compare how different distributions can apply to the tennis games.  For example, Poisson distribution.  In question 1 we found that the expected value for X was about .  If we use this as λ we can try to find the probability that Adam scores X number of points in a match.  We can use Excel to construct another simple spreadsheet, this time showing the Poisson distribution:

 X, number of points won by Adam P(X=x) 1 0.008483985 2 0.028280091 3 0.062844961 4 0.104742126 5 0.139656867 6 0.155175072 7 0.147786522 8 0.123156051 9 0.09122716 10 0.060818411

The following spreadsheet is the same, only that it shows the Binomial distribution:

 X, number of points Adam wins P(x=X), probability that Adam wins X points 1 0.000339 2 0.003048 3 0.016258 4 0.056902 5 0.136565 6 0.227608 7 0.260123 8 0.195092 9 0.086708 10 0.017342

The following is a comparison between both probabilities:

 P(X=x) for Poisson P(x=X) for Binomial 0.008483985 0.000339 0.028280091 0.003048 0.062844961 0.016258 0.104742126 0.056902 0.139656867 0.136565 0.155175072 0.227608 0.147786522 0.260123 0.123156051 0.195092 0.09122716 0.086708 0.060818411 0.017342

It is interesting to observe that when using different distributions the probability values can vary.  Regardless, the correct method is binomial, as a Poisson distribution could technically still work for x=11.  What I mean by this is that since the binomial distribution is restricted by the presence of n, it will most likely be more accurate as its values will add up to one.  The Poisson distribution, on the other hand, will not always add up to exactly one for the value of n we are using. This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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