The following information applies to questions 1 through 5. Let’s play two yard

Question

The following information applies to questions 1 through 5. Let’s play two yard football. The field has 2 goal lines and the one yard line. The home team starts with the ball on the 1 yard line. The transition probabilities are given in the table

Outcome                                               Probability

Lose 1 yard-lose the game 0.4

No gain-turn the ball over 0.2

Gain 1 yard (touchdown)-win the game 0.4

The first team to score wins. So, for example, if the home team gains 1 yard, it wins; if it loses 1 yard, it loses. If it gains zero yards, the ball is turned over to the other team at the 1 yard line. The game is symmetric; so if the probability that the home team wins the game is ?, then if the visitor stops the home team for no gain, then it has the ball on the 1 yard line and it also has a probability of winning the game equal to ?.

Two yard football is played for prize money. The winning team takes home $10,000. The losing team gets nothing.

Drawing the tree for two-yard football will help answer the following questions.

1. This game is an example of a

Question 1 options:

a. Sequential Game

b. Recursive Dynamic Program

c. Non-Recursive Dynamic Program

d. Simultaneous game in mixed strategies

2. Beginning from the state where the home team has the ball on the 1 yard line, what is the probability that the home team wins the game?

Remember (from Q1),

Outcome Probability

Lose 1 yard-lose the game 0.4

No gain-turn the ball over 0.2

Gain 1 yard (touchdown)-win the game 0.4

Question 2 options:

a.0.75

b.0.5

c.0.25

d. 0.1

3. Beginning from the state where the visiting team has the ball on the 1 yard line, what is the probability that the home team wins the game?

Question 3 options:

a.1

b.0.75

c. 0.5

d.0.25

4. What is the value of the game for the home team in the state at the start of the game where the home team has the ball?

Remember: Prize for winning is $10,000

Question 4 options:

a. $7500

b. $5000

c. $2500

d.$1000

5. On the first play the home team is stopped for no gain and turns the ball over to the visitors. Now, what is the value of the game for the home team?

Question 5 options:

a. $10000

b.$7500

c.$5000

d. $2500

a. Sequential Game

b. Recursive Dynamic Program

c. Non-Recursive Dynamic Program

d. Simultaneous game in mixed strategies

2. Beginning from the state where the home team has the ball on the 1 yard line, what is the probability that the home team wins the game?

Remember (from Q1),

Outcome Probability

Lose 1 yard-lose the game 0.4

No gain-turn the ball over 0.2

Gain 1 yard (touchdown)-win the game 0.4

Question 2 options:

a.0.75

b.0.5

c.0.25

d. 0.1

3. Beginning from the state where the visiting team has the ball on the 1 yard line, what is the probability that the home team wins the game?

Question 3 options:

a.1

b.0.75

c. 0.5

d.0.25

4. What is the value of the game for the home team in the state at the start of the game where the home team has the ball?

Remember: Prize for winning is $10,000

Question 4 options:

a. $7500

b. $5000

c. $2500

d.$1000

5. On the first play the home team is stopped for no gain and turns the ball over to the visitors. Now, what is the value of the game for the home team?

Question 5 options:

a. $10000

b.$7500

c.$5000

d. $2500

Explanation / Answer

a) The given problem is a sequential problem as the teams are not simulataneously taking their chances but are following a sequential order to have their chance. Thus answer is (A) Sequential Game.

b) For the home team to win, when the home have the ball in the begining state, there must be either a win of 1 yard by visitors or they have to be no gain by visitors followed by a lose by the visitors team. This is possible in the following ways

Let Lv, Lh demote loss by visitors and home teams, N denotes no gain, Wv, Wh denotes win by visitors and home teams.

if visitors lose 1 yard= NLv+NNNLv+NNNNNLv+NNNNNNNLv....=NLv/(1-N2)

if home wins 1 yard = Wh+NNWh+NNNNWh+....=Wh/(1-N2)

Therefore the total ways that home team wins = NLv/(1-N2)+Wh/(1-N2) = (NLv+Wh)/(1-N2)=((0.4*0.2)+0.4)/(1-0.22) =0.5

Thus the answer is b) 0.5

3)

For the home team to win, when the visitors have the ball in the begining state, there must be either a lose of 1 yard by visitors or they have to be no gain by visitors followed by a win by the home team. This is possible in the following ways

Let Lv, Lh demote loss by visitors and home teams, N denotes no gain, Wv, Wh denotes win by visitors and home teams.

if visitors lose 1 yard=  Lv+NNLv+NNNNLv+NNNNNNLv....=Lv/(1-N2)

if home wins 1 yard = NWh+NNNWh+NNNNNWh+....=NWh/(1-N2)

Therefore the total ways that home team wins = Lv/(1-N2)+NWh/(1-N2) = (Lv+NWh)/(1-N2)=(0.4+(0.2*0.4))/(1-0.22) =0.5

Thus the answer is c) 0.5

4) When the home team starts the value of the game for them is nothing but the expected amount they can win, which is nothing but the product of the prize amount and the probability of win

thus the value of the game = 0.5*10000= $5000

Thus the answer is b) $5000

5) In this case the probability of win will be equal to that of the case when the visitors had to start, ie., probability equals 0.5. {Since the expectation will not depend on the happened events, rather depend on the present occuring event}

therefore the value of the game will now be = 0.5*10000 =$5000

Thus the answer is c) $5000

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