It is the fifth game in a best of 7 series of the national volleyball championship series. The home team is 3 games down to 2 so it must win the next two games in order to win the championship. The coach has to decide which players he will use in the game, especially which opposite hitter. His best opposite hitter has played a lot and could use some rest. The next best opposite hitter is fit and could play the next two games. The coach believes that his best opposite hitter, although not fully rested, is still better than his next best opposite hitter. Letting him play the 6th game would mean that he can’t play the last game. The opponent’s coach has chosen his next best opposite hitter, maybe saving his best player for the last game? The home team coach assesses that the opponent’s opposite hitters have about the same ability as his own opposite hitters. What should the coach decide, let his best opposite hitter play or his next best opposite hitter? Can Operations Research help in this decision?
Traditionally Operations Research has been used in sports for timetabling, like the construction of a round robin tournament schedule. Sports leagues and teams need schedules that satisfy different types of constraints. For example the three soccer teams of Rotterdam cannot play at home at the same time. Other examples are the restrictions due to the international games played by the teams, like the Champions league. Timetable construction in sports competitions is a difficult problem to which several O.R. techniques have been applied. Most of the time a Mixed Integer Problem (MIP) needs to be solved. But Operations Research can be used in other areas of sports as well.
Let’s go back to the decision the coach of the volleyball team has to make. In what way could we assist him in this decision? Should he use his best player in the 6th game and use the next best in the final game, or the other way around? Most people I talked to on this decision say to put up the best player in game 6 and use the next best player in game 7. For obvious reasons they say, because if you don’t win game 6 there won’t be a game 7. Is this the best strategy?
The decision the coach has to make is similar to the examples used by Kahneman and Tversky. They demonstrate that human intuition is notoriously bad at processing even the simplest probability problems. Here is one of their examples:
Linda is 31 years old, single, outspoken and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in antinuclear demonstrations. Which of the following statements is more probable?
A: Linda is a bank teller.
B: Linda is a bank teller and active in the feminist movement.
C: Linda is a bank teller and takes yoga classes.
Most respondents choose B or C. Few choose A. Yet, by the laws of probability Pr(A) ≥ Pr(A ∩ B)
In a similar way we can prove that saving the best player for the last game is the best strategy to win the series and therefore the championship. To see this, let P1 be the probability that the best player wins game 6. Then the probability of winning game 7 for the best player will be P2 = P1 + ε where ε > 0. ε is positive since the player has additional rest and the opposing players are judged to be the same. Let the probability that the next best opposite hitter wins game 6 be P3 = P1 – δ, where δ > 0. Note that this definition takes into account the assessment of the coach that his best opposite hitter is better than his next best even though he isn’t fully fit. The probability of winning game 7 (P4) with the next best player is the same as winning the 6th game, so P3=P4.
The probability of winning the championship starting with the best opposite hitter in game 6 is C1 = P1*P4. Letting the best opposite hitter play in the final game the probability of winning the championship is C2= P3*P2. Now note that C2 = P3*P2 = P3*(P1 + ε) = P3*P1 + P3*ε =P4*P1+ P4*ε. Hence C2 = C1 + P4*ε. Since P4*ε >0, the best strategy therefore would be to save the best opposite hitter for the last game. As you can see, sports can benefit from the application of Operations Research in various ways. Can you think of others?
Traditionally Operations Research has been used in sports for timetabling, like the construction of a round robin tournament schedule. Sports leagues and teams need schedules that satisfy different types of constraints. For example the three soccer teams of Rotterdam cannot play at home at the same time. Other examples are the restrictions due to the international games played by the teams, like the Champions league. Timetable construction in sports competitions is a difficult problem to which several O.R. techniques have been applied. Most of the time a Mixed Integer Problem (MIP) needs to be solved. But Operations Research can be used in other areas of sports as well.
Let’s go back to the decision the coach of the volleyball team has to make. In what way could we assist him in this decision? Should he use his best player in the 6th game and use the next best in the final game, or the other way around? Most people I talked to on this decision say to put up the best player in game 6 and use the next best player in game 7. For obvious reasons they say, because if you don’t win game 6 there won’t be a game 7. Is this the best strategy?
The decision the coach has to make is similar to the examples used by Kahneman and Tversky. They demonstrate that human intuition is notoriously bad at processing even the simplest probability problems. Here is one of their examples:
Linda is 31 years old, single, outspoken and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in antinuclear demonstrations. Which of the following statements is more probable?
A: Linda is a bank teller.
B: Linda is a bank teller and active in the feminist movement.
C: Linda is a bank teller and takes yoga classes.
Most respondents choose B or C. Few choose A. Yet, by the laws of probability Pr(A) ≥ Pr(A ∩ B)
In a similar way we can prove that saving the best player for the last game is the best strategy to win the series and therefore the championship. To see this, let P1 be the probability that the best player wins game 6. Then the probability of winning game 7 for the best player will be P2 = P1 + ε where ε > 0. ε is positive since the player has additional rest and the opposing players are judged to be the same. Let the probability that the next best opposite hitter wins game 6 be P3 = P1 – δ, where δ > 0. Note that this definition takes into account the assessment of the coach that his best opposite hitter is better than his next best even though he isn’t fully fit. The probability of winning game 7 (P4) with the next best player is the same as winning the 6th game, so P3=P4.
The probability of winning the championship starting with the best opposite hitter in game 6 is C1 = P1*P4. Letting the best opposite hitter play in the final game the probability of winning the championship is C2= P3*P2. Now note that C2 = P3*P2 = P3*(P1 + ε) = P3*P1 + P3*ε =P4*P1+ P4*ε. Hence C2 = C1 + P4*ε. Since P4*ε >0, the best strategy therefore would be to save the best opposite hitter for the last game. As you can see, sports can benefit from the application of Operations Research in various ways. Can you think of others?
