CS代考 Tutorial questions: making group decisions – cscodehelp代写

Tutorial questions: making group decisions
Solutions version 2, corrected on 6 January 2022 Correction to the answer to Question 1 made in red below
1. Consider the following voter profile representing the truthful preferences of Alice, Bob, and Charlie:
Alice: B≻A≻C≻D Bob: C≻A≻B≻D Charlie: B≻A≻C≻D
Alice, Bob, and Charlie decide to hold a vote which will be decided according to the Borda count protocol, with ties decided randomly.
Is it in anyone’s interest to vote strategically and falsely report their preference (assum- ing they know the truthful preferences of the other two voters)? If your answer is yes, then specify the person as well as the ordering they should report instead of their true preference.
Hint. A voter v has an incentive to vote strategically if and only if it is the case that:
• the winner if they submit their true preferences is ωx
• the winner if they submit some false preferences is ωy, and • v prefers ωy to ωx
Answer. If all parties report their true preferences the winner is: B.
A = (3×0)+(2×3)+(1×0)=6
B = (3×2)+(2×0)+(1×1)=7
C = (3×1)+(2×0)+(1×2)=5(not4,asitsaidinpreviousversion) D = (3×0)+(2×0)+(1×0)=0
Alice and Charlie want this outcome and so don’t have an incentive to vote strategically. Bob would prefer C or A to B. He could falsely report his preferences as A ≻ C ≻ D ≻ B,
which would give the following, meaning A would win:
A = (3×1)+(2×2)+(1×0)=7 B = (3×2)+(2×0)+(1×0)=6 C = (3×0)+(2×1)+(1×2)=4 D = (3×0)+(2×0)+(1×1)=1
It is in Bob’s interest to falsely report: A ≻ C ≻ D ≻ B.
2. The plurality voting protocol is an example of a positional scoring rule voting protocol.
What is the scoring vector that captures the plurality procedure?
Answer. The plurality voting protocol is a positional scoring rule voting protocol with the scoring vector (1, 0, …, 0). That is, a candidate only receives a point from a particular voter if that voter has listed the candidate in first place.
1

3. Consider the following voter profile:
3voters: A≻B≻C 6voters: B≻C≻A 4voters: C≻A≻B 3voters: C≻B≻A
Suppose that (s1, s2, s3) is an arbitrary scoring vector, where s1 ≥ s2 ≥ s3 ≥ 0 and s1 > s3. Show that the corresponding positional scoring rule does not select the Condorcet winner.
Hint. This is essentially an algebraic manipulation question. You can represent the scores received by each of the three candidates in terms of s1, s2 and s3. Let score(ωi) be the score received by candidate ωi. If you can show that score(ωi) − score(ωj ) < 0 then you know that score(ωi) < score(ωj). If you can show that score(ωi) − score(ωj) ≤ 0 then you know that score(ωi) ≤ score(ωj). Don’t forget that you know s1 ≥ s2 ≥ s3 ≥ 0 and s1 > s3.
Answer. To determine the Condorcet winner, we have to consider pairwise majority elections.
• A vs B. A gets 7. B gets 9. B wins. • AvsC. Agets3. C gets13. C wins. • B vs C. B gets 9. C gets 7. B wins.
B is the Condorcet winner.
Under the positional scoring rule protocol, we get:
score(A) = 3s1 + 4s2 + 9s3
score(B) = 6s1 + 6s2 + 4s3
score(C) = 7s1 + 6s2 + 3s3
Now let’s manipulate this algebraically to determine the winner under the positional scoring rule protocol.
score(A) − score(B) = −3s1 − 2s2 + 5s3
= 3(s3 −s1)+2(s3 −s2)
Sinceweknowthats1 >s3,weknowthats3−s1 <0,hence3(s3−s1)<0. Sinceweknowthats1 ≥s2 ≥s3 ≥0,weknowthats3−s2 ≤0,hence2(s3−s2)0. It follows that 3(s3 − s1) + 2(s3 − s2) < 0. So score(A) − score(B) < 0, therefore score(A) < score(B). score(C) − score(B) = s1 − s3 Sinceweknowthats1 >s3,weknowthats1−s3 >0.
So score(C) − score(B) > 0, therefore score(C) > score(B).
Hence score(C) > score(B) > score(A) and C wins under the positional scoring rule protocol with any arbitrary scoring vector (s1, s2, s3) where s1 ≥ s2 ≥ s3 ≥ 0 and s1 > s3.
2

As the Condorcet winner for this example is B, we have shown that the positional scoring rule protocol violates the Condorcet principle with any scoring vector (s1,s2,s3) where s1 ≥s2 ≥s3 ≥0ands1 >s3.
There are other ways you might have approached this. For example, instead of determin- ing score(C) − score(B), you might have done:
score(A) − score(C) = −4s1 − 2s2 + 6s3
= 4(s3 −s1)+2(s3 −s2)
Sinceweknowthats1 >s3,weknowthats3−s1 <0,hence4(s3−s1)<0. Sinceweknowthats1 ≥s2 ≥s3 ≥0,weknowthats3−s2 ≤0,hence2(s3−s2)0. It follows that 4(s3 − s1) + 2(s3 − s2) < 0 So score(A) − score(C) < 0, therefore score(A) < score(C). 3