Download STATS 50 Solutions to Homework 2 Problem 1 Duel. Two players

STATS 50
Solutions to Homework 2
Problem 1 Duel. Two players start at a distance of 1 unit apart. They walk toward each other, with x
denoting the distance between them. Suppose player I has an accuracy function P1 (x) = 1 − x and II has an
accuracy function P2 (x) = .3 + .7(1 − x)2 . Both functions return the probability of hitting the other player.
At any point, either player may choose to fire. If they hit, they win the duel. If they miss, the players walk
until x = 0 and the non shooting player gets to shoot with certain aim. When should I fire and what is the
value of the game?
Solution. In class we found the answer to this problem was the distance x such that P2 (x) + P1 (x) = 1.
This was because from player I’s perspective, they win if either they shoot and hit, or if player II shoots and
misses. In the beginning, player I shouldn’t shoot because the other player has a greater chance of missing.
In the end, player II has a greater chance of hitting, so I will have the best chance to win at a distance where
his chance to hit equals player II’s chance to miss.
Plugging in the formulas and solving the resulting quadratic equation gives x∗ = .485 or 2.943. Since x needs
to be between 0 and 1, x∗ = .485 and the value of chance of 1 winning will be P1 (x∗) = 1 − .485 = .515.
Problem 2 Value of field position. Let V (x) be the value of the field position, approximated linearly by
V (x) = −2 + 9x/100.
(a) A team punts on fourth down from its own 20 yard line. Suppose that the length of the punt is uniformly
distributed between 30 and 50 yards. What is the expected value of the field position for the receiving
team? How does this compare with a punter who punts 40 yards with probability 1.
Solution. The quick way to do this with virtually no calculation is to exploit linearity (again). The
mean of a uniform random variable lies at the midpoint of its range. Thus, a uniformly distributed
punt with a range of 30 to 50 yards has a mean of 40. By linearity, we immediately get that this
must be the same as a punter that always punts exactly 40 yards. The value of the field position is
V (100 − 60) = −2 + 9(40)/100 = 1.6.
(b) It’s fourth down and five yards to go on the opponent’s 30 yard line. A team has a .5 chance of making
a field goal and a .4 chance of getting to the opponent’s 25 yard line, a first down. (If either play fails,
the opponent takes over at its own 30 yard line.) Based on V (x), is it a better strategy to go for it or
kick a field goal?
Solution. This was the problem that the most people in the class had difficulty with. There is a subtlety
to consider: If the field goal is made, then the opponent will receive the kickoff (and the associated
field position). We must incorporate this into the value of a made field goal. We will assume that
the opponent always returns the kickoff to the 20 yard line (see below for a related remark). For the
field-goal attempt we get
1
1
E V = (3 − V (20)) + (−V (30)) = 1.25 .
2
2
If we go for it on fourth down, then
E V = 0.4V (75) + 0.6(−V (30)) = 1.48 .
Based on this analysis, we should go for it on fourth down (the expected number of points is greater!).
Now, suppose that we don’t like the additional assumption that we had to make regarding what happens
after the kickoff. So, where would the kickoff need to be returned to in order to make both scenarios
have equal value? We need E V (x) = 1.48 in the field-goal scenario. So, we have
E V (x) =
1
1
(3 − V (x)) + (−V (30)) = 1.48 ⇒ x ≈ 14.9 ,
2
2
i.e., the team kicking the field goal would then have to expect to pin the opponent in at the 14.9 yard line
(or less) in order for the field goal to be worth considering. Furthermore, we could relax the assumption
Leo Pekelis
1
22 Oct 14
STATS 50
Solutions to Homework 2
in the second scenario that we always either turn the ball over at the 30 or get exactly to the 25 yard
line. Let x be the number of yards we get beyond the 25 if we succeed and let y be the number of yards
past the 30 that we progress to if we fail to get to the 25 (so, y ≤ 5). Then the expected value of going
for it is
E V (x, y) = 0.4V (75 + x) − 0.6V (100 − (70 + y)) = 0.4V (75 + x) − 0.6V (30 − y) ≥ 1.48 ,
provided that even if we fail to get the first down, we don’t lose yardage. That is, relaxing the assumptions in the second scenario can only increase the expected number of points in the scenario (provided
we ignore penalties or negative-yardage plays), making it look (even) more desirable compared to a field
goal. In some sense, the problem statement in the homework is a “worst case” analysis (in terms of
expected points) for going for it on fourth down.
Problem 3 Batting average. Suppose A has a higher batting average than B in the first season and also
in the second season. Is it true that A has a higher overall batting average than B? Prove or provide a
counterexample.
Solution. This is a classic example of Simpson’s Paradox. The idea is that the total number of at bats have
to be different between the two players in a season where they both perform comparatively poorly. One of
the simplest examples is the following: 2 seasons, 2 players, 17 total at bats

 P1
P2
S1
2/3
1/3
S2
1/10
0/1

TTL
3/13 = .231 
1/4 = .250
Season 2 is a hard year to get hits, which is lucky for player 2 since he managed to sit out most of it due to
an injury, so even though P2 had a terrible year (0 ba) his overall ba wasn’t affected as much as P1.
Simpson’s paradox does come up in the real world. For example (see the wikipedia page), in passing the Civil
Rights Act, and gender bias for graduate programs.
A few more interesting, and worked out examples including the hot hand theory for basketball, and death
penalty cases in Florida can be found here http://www.morris.umn.edu/~sungurea/introstat/public/
instruction/simpson/simpson.html.
Leo Pekelis
2
22 Oct 14