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Mathematically Modelling Basketball Shots

Extracts from this document...

Introduction

Mathematically Modelling Basketball Shots

Situation

The manager of a professional basketball team is having a tough decision in choosing which of his two top scorers this season are better at free-throw shots. The final decision will go towards picking the team for Saturday’s Cup Final match.

        On a training session one week before the match the coach decides to “go all out” and bring some mathematical genii in to model a situation where Lee Grimes and Dominic Aspbury, the goalscorers, will shoot at the basketball net.

        The mathematical genii are students from Cambridge and are benefiting from this opportunity in that they will be able to show evidence of coursework for their final exam. Their coursework will be using their abilities to collect data and “test the appropriateness of a probability model” on a real situation whilst the coach’s aim will be to pick the better of the two players for the “big game.”

If the random variables X and Y count the number of independent trials before the event, having a probability p, occurs then X and Y have geometric distributions:

P ( X = r ) = q r – 1 p        where  r = 1,2,3,……

X~G ( p )           and         Y~G ( p )

I will define X         as being the number of shots required before Lee shoots a basket. Therefore, Y is defined as the number of shots required before Dom shoots a basket. I will be attempting to see if X and Y have geometric distributions by taking samples of X and Y.

The populations are the infinite range of shots capable from the two throwers taken in a discrete time period under varied conditions at the same level of skill.

...read more.

Middle

Therefore P(no score) = 1 – 0.231884058

                            = 0.768115942

Therefore for Dom: P (Y = 2) = 0.768115942 x  0.231884058

                                 = 0.1781138416

                                 P (Y = 3) = 0.768115942(3-1)  x  0.231884058

                                                 = 0.1368120813

Expected Frequencies will be: (Y = 1) = 0.231884058 x 80

                                          = 18.55072464

                                             (Y = 2) = 0.1781138416 x 80

                                          = 14.24910733

Chi Squared Distribution

The chi-squared distribution can be applied to measure the ‘goodness of fit’ for the geometric models. It will examine the ‘goodness’ of the model by considering the number of possible outcomes of the events and will analyse the validity of the assumptions.

        Theimage10.pngvalue will be expected to be small to suggest that the model fits the real distribution. A large value would suggest that the model is unlikely to be correct so I will use a 10% critical region to test it.

  • If theimage10.pngvalue lies within the critical region then, assuming the model is correct, it would mean that there is less then 10% chance of a result as high as this occurring. We reject the model as a consequence and conclude insufficient sampling etc.
  • Alternatively, if the image10.pngvalue lies outside the critical region, the result is valid and there is a larger possibility of the value being what it is. The model is assumed to be correct and the model is accepted. Conclusion would be to state that the statistical model is appropriate to the situation and the assumptions are correct.

In the tables, the expected and observed frequencies were calculated but how close together are the values? The closer the observed value to the expected value the more accurate the geometric model will be.

The goodness of fit statistic is:

image09.pngimage09.pngimage02.png         where O = Observed Frequency

                                                                                  E = Expected Frequency

To find the best measure of goodness of fit, add up all values for each statistic and compare with the image03.png2 probabilitydistribution tables.

image04.png

...read more.

Conclusion

A problem with this is time, as it would take a year to go through just one season, therefore it is impractical and illogical. The physical form of the player should also alter throughout the season so a random sample of more than one season would have to be made.

        A much better way is to watch all training sessions and take a general overview of who supplies the most points in miniature matches from free throws. This gives more of a view of consistency than “on the day” performance but during game situations the performer will be thinking more logically. A sample of eighty straight baskets is tedious and will affect performance.

Modifications

  • Use a longer time period. The performers were rushed to collect their sample size within two hours as a result of school timetabling and so one of them had to rush his last twenty shots.
  • Use the same time period i.e. one performer did it one day and the other completed it the next day. Conditions may have been different and morale, energy etc may be variated for both Dom and Lee
  • Use foot-mats on the floor so that it indicates an exact position for the feet to stand instead of just using the line. This may be an insignificant difference but to improve the coursework it is better than no difference at all.
  • Using the same basketball. Half way through the sample collection the basketball was lost leaving us the trouble of having to use another basketball – maybe of different weight, age etc and possibly affecting the results

Improvements

  • I would like to calculate confidence intervals for both expected values (X and Y) to determine my degree of confidence in Lee being a better freethrower.
  • I would also like to be able to see if my result E[X] = E[Y] was statistically significant

...read more.

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