Decision tree analysis
(a) The total expected payoffs for Tech when going for one point are (0.98)($7.2 million) + (0.02)(-$1.7 million) = $7,014,000 (0.02)(0.20)($7.2 million) + (0.02)(0.80)(-$1.7 million) = -$108,000 Therefore, based on the expected payoffs, Tech should go for one point.
(b) Tech’s probability of winning the game in overtime should be 76.82% to make Tech indifferent to going for either 1 or 2 points.
Decision tree analysis is a graph with different decision options and their potential consequences. Tech can go for two points or one point, and if one point is successful, it will go over time. The decision tree analysis is as follows: The following probabilities and outcomes will be used for decision tree analysis if Tech goes for two points: Probability of winning = 0.33 The payoff for a win is $7.2 million The payoff for a loss is -$1.7 million The probability of losing is 0.67 The payoff for a win is -$1.7 million The payoff for a loss is -$1.7 million If Tech goes for one point, the following probabilities and outcomes will be used: Probability of winning = 0.98 The payoff for a win is $7.2 million The payoff for a loss is -$1.7 million The probability of losing is 0.02 The payoff for a win is -$1.7 million The payoff for a loss is -$1.7 million If it goes for one point and loses, it will have a 20% chance of winning in overtime. Therefore, the probability of going for one point and winning is 0.98, while the probability of going for one point and losing is 0.02. If Tech goes for one point and goes to overtime, the probability of winning in overtime is 0.20. The total expected payoffs for Tech when going for two points are: (0.33)($7.2 million) + (0.67)(-$1.7 million) = $1,386,000 The total expected payoffs for Tech when going for one point are: (0.98)($7.2 million) + (0.02)(-$1.7 million) = $7,014,000 (0.02)(0.20)($7.2 million) + (0.02)(0.80)(-$1.7 million) = -$108,000 Therefore, based on the expected payoffs, Tech should go for one point.
(b) Let P be the probability that Tech will go for one point, and 1 – P be the probability that Tech will go for two points. Then, the overall probability of Tech winning the game is given by: P(Tech wins) = P(Tech wins, goes for one point) + P(Tech wins, goes for two points)Let A denote the event that Tech goes for one point and wins, B denote the event that Tech goes for two points and wins, and C denotes the event that Tech goes for two points and loses. Then we have: P(Tech wins) = P(A) + P(B) = 0.98P + 0.33(1 – P)If Tech goes for one point, there is a 0.98 probability of winning, and if Tech goes for two points, there is a 0.33 probability of winning. Let D denote the event that the game goes into overtime. Then: P(A) = P(Tech wins | A)P(A) = 0.98P(1 – 0.2)P(B) = P(Tech wins | B)P(B) = 0.33P(C) = P(Tech loses | C)P(C) = 0.67(1 – P)Hence: P(Tech wins) = P(A) + P(B) = 0.98P + 0.33(1 – P)If Tech is indifferent between going for one point and going for two points, it follows that the payouts for the Sugar Bowl and the Gator Bowl must be equalized. We have: P(Tech wins) = P(Tech wins, goes for one point) + P(Tech wins, goes for two points) = $7.2 million x P(A) + $7.2 million x P(B) + $1.7 million x P(C)Now we can solve for P as follows:0.98P + 0.33(1 – P) = $7.2 million x P(A) + $7.2 million x P(B) + $1.7 million x P(C)0.98P + 0.33 – 0.33P = $7.2 million x 0.98P + $7.2 million x 0.33P + $1.7 million x 0.67(1 – P)0.65P = $7.2 million x 0.98P + $7.2 million x 0.33P + $1.139 million – $1.139 P0.481P = $7.2 million x 0.98P + $7.2 million x 0.33P + $1.139 millionP = $1.139 million/$1.4815 millionP = 0.7682, or 76.82%Therefore, Tech’s probability of winning the game in overtime should be 76.82% to make Tech indifferent to going for either 1 or 2 points.