NARETTA SOLUTIONS

Problem Set #2

Another popular sport on SASN is One-out, Two-base baseball. In this sport there are, obviously, two bases: home and 2nd. The batter hits the ball and runs to 2nd. If he or she (it’s a co-ed sport) gets a hit, then the batter is on second. If it’s an out, then the inning is over and the other team gets to bat. Suppose that it’s a hit. Then the next batter bats with a runner on 2nd base. If that batter makes an out, the inning is over. If that batter gets a hit, the run scores and the game and his or her team wins. The probability of getting a hit is 0.6.

Draw the tree.
The tree is drawn below. Note that Pr(out) symbolizes the probability of getting an out [similar for Pr(hit) with hit]. The payouts are at the end of the tree. In this case, we have not assigned dollar values to either winning the game or the inning being over, so I have just written out the outcomes. On a side note, I tend to prefer writing my game trees vertically (so it looks like roots of a tree), but doing it horizontal (as in the lecture slides) is just as valid of an approach.

Write an equation for the value of being the team at bat with no one on base.. (Hint: let V be the value of being at bat with no one on base. Then –V is the value of your opponent being at bat with no one on base.)
In this situation, the probability of winning in a given inning is 0.36 (the probability of getting two hits = 0.6*0.6). This implies that probability of getting an out before getting two hits is 0.64 (and at this point your opponent would be batting with no one on base). Since your value of batting with no one on base is V and your value of your opponent batting with no one on base is -V, we get the following equation (which I am also going to solve for V to get the solution for part c):

V=0.36+(0.64)(-V)=0.36-0.64V
1.64V=0.36
V=0.2195

c. Solve the equation in part b. Value = 0.2195

What is the probability that the team at bat with no one on base will win the game?
Probability = 0.6098 – This solution is kind of complicated to get (and Professor Martin doesn’t show you the process in the lecture), but here is the process you would go through to calculate it [if you plan on taking additional upper-level or graduate-level economics or statistics courses, you should really learn this]: First, you need to write down all the possible ways the team at bat with no one on base will win and their respective probabilities.
i) Get two hits = probability is 0.36 (as discussed in part b)
ii) Get an out (0.64 probability of happening in a given inning), then the opposing team also get an out (again 0.64 probability), and then get two hits (0.36 probability) = 0.64 * 0.64 * 0.36
iii) Go through two full innings with neither team getting two consecutive hits [(0.64)4 probability], and then get two hits (0.36). So, (0.64)4 * 0.36
iv) Go three full innings with neither team getting two consecutive hits [(0.64)6 probability], and then get two hits (0.36). So, (0.64)6 * 0.36
v) At this point you might notice a pattern to the probabilities and that this loop continues infinitely.

So, our next step is to sum up all the probabilities of winning and we get the following equation:
Pr?(Team at Bat Winning)= 0.36+(0.64)^2*0.36+(0.64)^4*0.36+?
Factoring out 0.36:
Pr?(Team at Bat Winning)= 0.36[1+(0.64)^2+(0.64)^4+?]

Here, the ellipsis (i.e. the “…”) indicates that the series goes on for infinity. Luckily, math has a way of dealing with such an infinite series. This type of summation – where it is the same coefficient (that is less than 1) repeatedly added with an increase exponent – is called a geometric series. In the general sense, a geometric series is equal to 1+r+r^2+r^3+? (where0<|r|<1) and can be rewritten as
1/(1-r)
In this case, r=(0.64)^2. So, we can rewrite this as:
1/(1-(0.64)^2 )
We can substitute this in for the infinite summation in the equation above to get:
Pr(Team at Bat Winning)=0.36[1/(1-(0.64)^2 )]=0.36[1/(0.5904)]
This is roughly equal to:
Pr(Team at Bat Winning)=0.36*1.694=0.6098
Hence, the probability of the team at bat winning the game is equal to 0.6098.

More on One-out, Two-base baseball. The key strategic variable for the team in the field is where to position the rover fielder. The key strategic variable for the batting team is where to hit the ball. The following table gives the strategies and payoffs for the first play. The number in the table is the probability of a hit. The team at bat wants a large probability, the team in the field wants a small one. Experience has taught the rovers to wait at the mound and then move suddenly to the left, right or center field as the ball is hit. Thus, you can think of this as a simultaneous move situation.

Team in the field

right center left pitcher’s
field field field mound

Bunt .300 .400 .350 .250

Hit to right .100 .250 .300 .200

Team at bat

Hit to left .350 .250 .100 .220

Hit to center .250 .100 .250 .240

Does the team at bat have a dominant strategy? If so what is it?
No, there is no dominant strategy. Bunting is close (dominates with three of the four possible fielder positions), but if the fielder is in right, it is better to hit to left.

b. Does the team at bat have a dominated strategy? If so what is it?
Yes, hitting to center is dominated. This is because regardless of where the fielder is located, the team at bat can improve their batting average by hitting to some other location besides center field. Additionally, hitting to right is also a dominated strategy.

c. Write out the best responses for both teams.

Team at bat   Team in field
Team in Field   Team at Bat   Team at Bat   Team in Field
Right   Hit to Left   Bunt   Pitcher’s Mound
Center   Bunt   Right   Right
Left   Bunt   Left   Left
Pitcher’s Mound   Bunt   Center   Center

What are the Nash equilibrium strategies?
There is only one Nash Equilibrium (NE) and it is for the Team in Field to position at the Pitcher’s Mound and the Team at Bat to Bunt.

Mike Caruso, the three-time national wrestling champion from Lehigh University had two great take-down moves: a quick single leg and his legendary barrel roll. (By the way, Mike wrestled in 1967 when Michigan State won the NCAA team championship.) His opponent could better defend the barrel roll with a more upright stance. The better option against the single leg was a more crouched stance. Of course, in wrestling everything happens so fast that you can think of it as a simultaneous game. The probability of a take-down for Mike is given in the table
Defender
Stand upright Crouch

Single leg 0.7 0.2
Mike Caruso
Barrel roll 0.1 0.4

Is there a Nash equilibrium in pure strategies? Explain briefly.

There is no pure strategy NE in this game. If Caruso chooses Single leg, Defender will choose Crouch, but if Defender chooses Crouch, Caruso will prefer to choose Barrel Roll, but if Caruso chooses Barrel Roll, Defender will choose Stand Upright, at which point Caruso would choose Single Leg. So, there is no NE.

Determine the optimal mixing probability for Mike.
The optimal mixing strategy for Caruso would be to set the probability of “Single Leg” (which I will refer to as p) such that the Defender would be indifferent between choosing to Stand Upright and to Crouch. Put into equation form (note: 1-p is the probability that Caruso chooses Barrel Roll):
0.7p+0.1(1-p)=0.2p+0.4(1-p)
Now just solve for p:
0.6p+0.1=-0.2p+0.4
0.8p=0.3
p=3/8
This implies that:
1-p=5/8
So, Caruso chooses Single Leg with probability 3/8 and Barrel Roll with probability 5/8.

What is the probability that he makes the take-down.
We just need to plug in the values of p and 1-p into either side of the first equation listed in part b to get the probability that Caruso makes the take-down (I’m doing both to verify that my calculations are correct, you only need to do one):
0.7(3/8)+0.1(5/8)=0.2625+0.0625=0.325
0.2(3/8)+0.4(5/8)= 0.075+0.25=0.325
Based on this, the probability that Caruso makes the take-down is 0.325.

4. Larry and Courtney are settling a grade dispute on the basketball court. It’s one point per basket. They play to 15. Larry has a nice drive and scoop that sometimes works against younger players who don’t know 1950s basketball. He also has a jumper patterned after the jumper of the mayor of Detroit (the current one, not Kwame). Here are his data from similar grade-settling games with students.

Drive Jumper

Score 80 25

Fail 30 20

The Chi-square test for the difference in means gives a p-value of 0.0379. What would you recommend for Larry? (No facetious comments.) Explain.
I would recommend that Larry do more drive and scoops. The chi-square test says that we can reject the null hypothesis that Larry is making more drive and scoops by chance. Since the difference is not just chance, he should utilize this advantage by using his drive and scoop more often.

Again analyzing his past data, Larry does a runs test. He finds the following results.

Expected Number of Runs: 64.9; sd: 5.1064
z-value= 2.96274;

His own data indicate 80 runs. What is he doing wrong? Explain.
Since the 80 runs are above the expected number of runs (and the difference is statistically significant and the 5% level), the data suggests that Larry is switching between using the drive and scoop and using the jumper too frequently. Combining this with the answer to part a, one would suggest that Larry do longer runs of drive and scoops than he is currently doing. That way, he is doing the statistically better maneuver and is lower his number of runs.

5. The diagram below shows the point value of having the ball on each yard line. It is taken from Romer’s “It’s fourth and two.”

a. What is the value of having the ball on your own 30 yard line? About 1 point

b. If you have the ball on your own 30 and gain 18 yards, how many points have you gained in expected value?
A little less than 1 point

c. If you have the ball on the 30 and fumble, and your opponent recovers the ball at the same yard line, how many points have you lost in expected value?

This fumble would put your opponent on your 30 yard line (which is worth about 3 points to your opponent). Additionally, the fumble causes you to lose the value of having the ball on your own 30 yard line (which was worth about 1 point). So, the net loss in expected value from this fumble is about 4 points.

Problem Set #2

1. Another popular sport on SASN is One-out, Two-base baseball. In this sport there are, obviously, two bases: home and 2nd. The batter hits the ball and runs to 2nd. If he or she (it’s a co-ed sport) gets a hit, then the batter is on second. If it’s an out, then the inning is over and the other team gets to bat. Suppose that it’s a hit. Then the next batter bats with a runner on 2nd base. If that batter makes an out, the inning is over. If that batter gets a hit, the run scores and the game and his or her team wins. The probability of getting a hit is 0.6.

a. Draw the tree.

b. Write an equation for the value of being the team at bat with no one on base.. (Hint: let V be the value of being at bat with no one on base. Then –V is the value of your opponent being at bat with no one on base.)
V=0.6*0.6+0.64*(-v)
=0.36-0.64v
=0.36/1.64
=0.2195

c. Solve the equation in part b. Value = 0.2195

d. What is the probability that the team at bat with no one on base will win the game?

Probability =
2. More on One-out, Two-base baseball. The key strategic variable for the team in the field is where to position the rover fielder. The key strategic variable for the batting team is where to hit the ball. The following table gives the strategies and payoffs for the first play. The number in the table is the probability of a hit. The team at bat wants a large probability, the team in the field wants a small one. Experience has taught the rovers to wait at the mound and then move suddenly to the left, right or center field as the ball is hit. Thus, you can think of this as a simultaneous move situation.

Team in the field

right center left pitcher’s
field field field mound

Bunt .300 .400 .350 .250

Hit to right .100 .250 .300 .200

Team at bat

Hit to left .350 .250 .100 .220

Hit to center .250 .100 .250 .240

a. Does the team at bat have a dominant strategy? If so what is it?

b. Does the team at bat have a dominated strategy? If so what is it?

c. Write out the best responses for both teams.

Team at bat Team in field

Team in Team at Team at Team in
field bat bat field

d. What are the Nash equilibrium strategies?
3. Mike Caruso, the three-time national wrestling champion from Lehigh University had two great take-down moves: a quick single leg and his legendary barrel roll. (By the way, Mike wrestled in 1967 when Michigan State won the NCAA team championship.) His opponent could better defend the barrel roll with a more upright stance. The better option against the single leg was a more crouched stance. Of course, in wrestling everything happens so fast that you can think of it as a simultaneous game. The probability of a take-down for Mike is given in the table

Defender
Stand upright Crouch

Single leg 0.7 0.2

Mike Caruso

Barrel roll 0.1 0.4

a. Is there a Nash equilibrium in pure strategies? Explain briefly.

b. Determine the optimal mixing probability for Mike.

c. What is the probability that he makes the take-down.

Drive Jumper

Score 80 25

Fail 30 20

a. The Chi-square test for the difference in means gives a p-value of 0.0379. What would you recommend for Larry? (No facetious comments.) Explain.

b. Again analyzing his past data, Larry does a runs test. He finds the following results.

Expected Number of Runs: 64.9; sd: 5.1064
z-value= 2.96274;

His own data indicate 80 runs. What is he doing wrong? Explain.

5. The diagram below shows the point value of having the ball on each yard line. It is taken from Romer’s “It’s fourth and two.”

a. What is the value of having the ball on your own 30 yard line? _____________

b. If you have the ball on your own 30 and gain 18 yards, how many points have you gained in expected value?
______________

c. If you have the ball on the 30 and fumble, and your opponent recovers the ball at the same yard line, how many points have you lost in expected value?

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