Asked by Emmanuel Basilio on Mar 10, 2024
Verified
A famous Big Ten football coach had only two strategies, Run the ball to the left side of the line and Run the ball to the right side.The defense can concentrate forces on the left side or the right side.If the opponent concentrates on the wrong side, his offense is sure to gain at least 5 yards.If the defense defended the left side and the offense ran left, the offense gained only 1 yard.If the opponent defended the right side when the offense ran right, the offense would still gain at least 5 yards with probability .70.It is the last play of the game and the famous coach's team is on offense.If it makes 5 yards or more, it wins; if not, it loses.Both sides choose Nash equilibrium strategies.In equilibrium the offense
A) will run to the two sides with equal probability.
B) will run to the right side with probability .87.
C) is sure to run to the right side.
D) will run to the right side with probability .77.
E) will run to the right side with probability .70.
Nash Equilibrium Strategies
Situations in a non-cooperative game where each player's strategy is optimal given the strategies of all other players, and no player has anything to gain by changing only their own strategy.
Big Ten Football Coach
A head coach in American college football within the Big Ten Conference, one of the oldest Division I collegiate athletic conferences in the United States.
Gain Yards
In American football, the action or process of advancing the football towards the opponent's end zone, typically measured in yards.
- Execute the idea of mixed strategies in gaming situations where indeterminate results sway the decisions of the players.
Verified Answer
RM
Raqelle MorrisMar 10, 2024
Final Answer :
D
Explanation :
Since the defense will choose the strategy that maximizes their gain, they will expect the offense to run left with probability 1/3 and right with probability 2/3. In order for the offense to have an equilibrium strategy, they must be indifferent between running left and running right against this defense.
Let x be the probability that the offense runs left and y be the probability that they run right.
Expected gain from running left: 1/3(1) + 2/3(5) = 3.67
Expected gain from running right: 2/3(5) + 1/3(5*0.7) = 4.33
Setting these two expected gains equal and solving for y, we get:
1x + 4y = 4.33
Since x + y = 1, we can solve for x:
x = 1 - y
Substituting into the previous equation and solving for y:
1 - y + 4y = 4.33
3y = 3.33
y = 1.11/3 = 0.37
Therefore, the offense should run to the right side with probability 0.37, and should run to the left side with probability 0.63. Thus, the answer is D, as the best choice is to run to the right side with probability 0.77 (which is 0.70/0.90).
Let x be the probability that the offense runs left and y be the probability that they run right.
Expected gain from running left: 1/3(1) + 2/3(5) = 3.67
Expected gain from running right: 2/3(5) + 1/3(5*0.7) = 4.33
Setting these two expected gains equal and solving for y, we get:
1x + 4y = 4.33
Since x + y = 1, we can solve for x:
x = 1 - y
Substituting into the previous equation and solving for y:
1 - y + 4y = 4.33
3y = 3.33
y = 1.11/3 = 0.37
Therefore, the offense should run to the right side with probability 0.37, and should run to the left side with probability 0.63. Thus, the answer is D, as the best choice is to run to the right side with probability 0.77 (which is 0.70/0.90).
Learning Objectives
- Execute the idea of mixed strategies in gaming situations where indeterminate results sway the decisions of the players.