Talk:Smith//IRV

The double-slash notation needs to be explained somewhere. Maybe one centralized place for all such articles? — Psephomancy (talk) 19:41, 27 February 2020 (UTC)

Agreed, but no idea where to place it. We could explain the notation somewhere in the article for Condorcet methods, and maybe create a "Artificial Condorcet methods" category to handle all the Condorcet// and Smith// methods. BetterVotingAdvocacy (talk) 20:20, 27 February 2020 (UTC)

As a side-note, I think it'd generally be a good policy to try to write up who would win in various Smith-efficient methods when the Smith set has 3 candidates (unless it's obvious, like with Smith//Score), since that's the most common case when there's no CW. BetterVotingAdvocacy (talk) 20:39, 27 February 2020 (UTC)

Smith//IRV almost good enough

Smith//IRV does a much better job than IRV, and STAR. It fails to win me over when it falls back on IRV and the problems it has. For discussion, I compare the results of 4 voting systems: IRV, Smith//IRV, STAR (points 5-4-3) and my own suggested voting result (SV) in the following scenario.

Suppose there have been millions of votes cast, and it’s a three way tie. It all comes down to these last nine votes, one of them is yours.

4 votes for A and possibly a second preference.

3 votes for B and possibly a second preference.

2 votes for C and possibly a second preference.

I sincerely doubt you can be sure which voting block you are in. In a large election, with a secret ballot, it is unlikely you and yours can figure out what the other guys are going to do. I suggest you all vote honestly and let the voting system do what it does for you.

There are 27 possible combinations of honest opinions. In 18 of these combinations IRV, Smith//IRV, STAR (5-4-3) and SV all agree. They elect A 12 times and B 6 times. There are 3 variations that are not unanimous and not cyclical. IRV gives you a chance to compromise or betray your honest opinion in all of these.

4 A>C,

3 B>C,

2 C

Or (the same results when the block of 2 supports A)

4 A>C,

3 B>C,

2 C>A

The winner is: A (IRV), C (Smith//IRV), C (STAR), C (SV). IN IRV, if you think you are in the block of 3, you should change your vote to C>B.

4 A>C,

3 B>C,

2 C>B

The winner is: B (IRV), C (Smith//IRV), C (STAR), C (SV). IN IRV, if you think you are in the block of 4, you should change your vote to C>A.

There are 6 cyclical combinations left. IRV and SMITH/IRV invite compromise in all six. STAR invites compromise in two.

4 A,

3 B>C,

2 C

Or (the same results when the block of 2 supports A)

4 A,

3 B>C,

2 C>A

The winner is: A (IRV), A (Smith//IRV), C (STAR), tie (SV). IN IRV and Smith/IRV, if you think you are in the 3, you should change to C>B.

4 A>B,

3 B>C,

2 C

The winner is: A (IRV), A (Smith//IRV), B (STAR), tie (SV). In IRV and Smith/IRV, if you think you are in the 3, you should change to C>B.

4 A>B,

3 B>C,

2 C>A

The winner is: A (IRV), A (Smith//IRV), A (STAR), tie (SV). In STAR, if you think you are in the 3, you should change B>C to just C. In IRV and Smith/IRV, if you think you are in the 3, you should change B>C to C>B.

4 A>C,

3 B,

2 C>B

The winner is: B (IRV), B (Smith//IRV), C (STAR), tie (SV). In IRV and Smith/IRV, If you think you are in the 4, you should change A>C to C>A.

4 A>C,

3 B>A,

2 C>B

The winner is: B (IRV), B (Smith//IRV), A (STAR), tie (SV). In STAR, if you think you are in the 2, you should change C>B to B>C. In IRV and Smith/IRV, If you think you are in the 4, you should change A>C to C>A.

Conclusion: SV does not invite any voter to compromise in these 4-3-2 scenarios. To avoid the temptation of compromise, and the resulting confusion, an honest opinion that ends in a cyclical tie is exactly that. It’s a tie. It needs to be broken randomly (SV’s step 11).

RalphInOttawa (talk) 09:22, 11 December 2023 (UTC)