COGS 300 — Fall 2013
Assignment 2
Solution
Question 1
Solution
(a) At first, cheat. At the second time, cheat if you didn’t get caught the first time. The expected utility is
81.69.
(b) When the probability of Watched is increased a small amount (e.g., from 0.3 to 0.5), the policy changes
to cheat at the first opportunity and don’t cheat at the second opportunity. When the probability of
increased more (e.g., to 0.8) the policy becomes to not cheat at either time.
(c) We can change P(Grade1|Cheat1) and P(Grade2|Cheat2).
If the probability are changed to P(Grade2 = F|Cheat2 = T) = 0.1 and P(Grade2 = A|Cheat2 =
T) = 0.4 (with the other probabilities for Cheat2 = T remaining at 0.25 each), the policy changes to
not cheat at the second opportunity.
If the probabilities of P(Grade1|Cheat1) are changed to be P(Grade2 = F|Cheat2 = T) = 0.1,
P(Grade2 = C|Cheat2 = T) = 0.25 P(Grade2 = B|Cheat2 = T) = 0.25 and P(Grade2 = A|Cheat2 =
T) = 0.4 the policy changes to not cheat at the first opportunity.
(d) See:
Watched1
Watched2
Caught 1
Punishment
Caught 2
Cheat 1
Utility
Cheat 2
Grade 1
Grade 2
Final Grade
If the probability of watched1 goes up, we can even get to the situation where the policy is to not cheat
at the first opportunity and to cheat at the second opportunity. See:
http://www.cs.ubc.ca/~poole/cogs300/2013/cheat_decision_watched2.xml
for such an example (which didn’t happen by changing the probabilities in the first model).
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(e) At the moment cheating at time 1 only affects the Grade at time 2 by affecting whether the student
will cheat at time 2. We can add an arc from Cheat1 to Grade2. The model can them make it so that
cheating at time 1 reduces grade 2 for the case when the student does not cheat at time 2 (because the
agent doesn’t learn from their mistakes). See:
http://www.cs.ubc.ca/~poole/cogs300/2013/cheat_decision_watched2_grade2.xml
for an example.
(f) You need to say something sensible here.... The probabilities are way off; most students don’t cheat.
(However, it is not clear how we could acquire these probabilities, or why the students should believe
any probabilities the university provides — they are subjective beliefs after all). The utilities for
punishment need to be more extreme. There are more than two opportunities to cheat, but maybe we
could model all subsequent behaviour as a single variable.
(g) Making punishment more severe (suspension or recorded on transcript for first offences) makes cheating less good even if the probability of getting caught is 0.2. See e.g.,:
http://www.cs.ubc.ca/~poole/cogs300/2013/cheat_decision_honour.xml
(h) Write something sensible here. Present your argument for or against.
Question 2
Solution
(a) The decision network is:
Final
Grade
Difficulty
Pass
Midterm
Pass Final
Study
Midterm
Study
Final
Utility
Total Effort
(b) In the following the actual numbers are not significant. However,they have the following characteristics that is important: A is preferred to B which is preferred to C which is much preferred to F for all
efforts. Working little is much better than working a lot. The best outcome is to get an A with little
effort. The worst outcome is to work hard and get an F.
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Grade Total Effort Utility
A
Lot
50
A
Little
100
B
Lot
40
B
Little
90
C
Lot
30
C
Little
80
Lot
0
F
F
Little
20
(c) For this student, getting an A is much better than getting a B, which is a little bit better than a C which
is a little bit better than F. Arguably they will feel better getting a low grade if they worked hard than
if they didn’t (although we won’t take marks off for the other way around). What is important is the
big gap between the A’s and the rest.
Grade Total Effort Utility
A
Lot
95
Little
100
A
B
Lot
40
B
Little
30
C
Lot
20
Little
10
C
F
Lot
10
F
Little
0
Question 3
Solution Chris is right, the fact that a user did not use a particular word is important information that should
not be ignored. For example, if people who want to know about printer problems almost always use the word
“print” (or a derivative), the fact that someone did not use the work “print” is evidence that it is not a printer
problem. This is different from now knowing whether they used the word “print”.
Question 4
Solution
(a) The words are independent of each other given the help page. (But they are dependent on each other
if the help page is not observed.)
(b) The words probably should not be independent of each other, there are some words that are likely to
go together (such as “I” and “am”) and some that are not likely to go together (such as “folder” and
“directory” — someone is likely to use only one of these depending on which system they use).
(c) The topics are independent of each other given no words (but are dependent given words). The words
are independent of each other given the topics (but are dependent of each other given no topics).
(d) If one topic is a superset of another (e.g., shirt and clothes) or are unlikely to go together (e.g., fashion
and science, even they are both associated with “model”) or are likely to go together (e.g., fashion and
shirts). There are many possible answers.
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