Re: [EM] Chicken problem (was: SODA and the Condorcet criterion)
On 6.8.2011, at 19.40, Jameson Quinn wrote: > More thoughts on the "chicken problem". > > Again, in Forest's version, that's a scenario like: > > 48 A > 27 C>B > 25 B>C > > C is the pairwise champion, but B is motivated to truncate, and C to > retaliate defensively, until A ends up winning. > > In my opinion, scenarios like this make the single most intractable practical > strategy problem in voting theory: > Approval, Range, and median-based systems all suffer directly. > Most winning-vote-like Condorcet systems fall prey, including otherwise-great > systems like Schulze. > Margins systems have no truncation incentive - but as a direct consequence, > they give extremely difficult-to-justify results if the B block truncates; in > fact, they allow a strategic C block to fool the system into thinking it's > seeing this scenarion when actually B and C are mortal enemies. > IRV does relatively well with this scenario - but in return, pays no > attention at all to the second choice of the A voters, which should be > decisive if it exists. > At the other extreme, some systems resolve this problem by forcing strict > rankings from the A voters - but if they really don't have a preference, that > ends up being just statistical noise, and doesn't even necessarily remove the > game-of-chicken incentives if things are balanced right. Moreover, forcing B > and C voters into strict rankings only makes them escalate their truncations > into burials. > > Most of us, when we want to "test" our voting systems with a difficult case, > use a strict-ranking Condorcet cycle of three; the old, standard ABC BCA CAB > scenario. That's nice and simple, but not very realistic. To me, the "game of > chicken" scenario; the resulting Condorcet cycle if B truncates; and related > scenarios that could strategically be made to masquerade as these; are better > practical tests for a voting system. In fact, I'd go so far as to guess that > a real-life Condorcet cycle would be more likely to be the result of playing > chicken than of honest preferences. > > As Forest already explained, SODA, as currently formulated, resolves the game > of chicken — if all votes are delegated. It can do that because games among > finite candidates are much more tractable than those among oceans of voters. > SODA's "sequential trick" would be ridiculous with voters; imagine "Your turn > to vote is on Sunday at 2:35:58 PM." > > In my previous message in this thread (Re: SODA and the Condorcet criterion), > I pointed out that there's still a problem if voters explicitly truncate by > refusing to delegate. But I've been considering this issue, and eventually I > found a solution that I think is simple enough to include in SODA: > > Make all candidate's predeclared rankings into strict rankings by breaking > declared ties in order of the current approval totals when it's their turn to > use their delegated votes. > > So if B voters truncated, candidate A would see that B was headed for a win, > and would have the option to delegate to C. All the truncation would have > accomplished would be to make A into a kingmaker between B and C. Since A > could have had this kingmaker power, if she had wanted it, from the start, > that's not a problem. The only difference between this end-game kingmaker > power of A's, and if she had simply declared a preference from the start, is > that the end-game power could in theory arise no matter which of B or C has > more approvals, whereas an initial preference would only confer kingmaker > power if the preferred candidate ended up with fewer approvals. > > Is this version of SODA really the only system to have a fully-satisfactory > resolution to the chicken problem? Even if it is, is it worth adding this > additional complexity to SODA? Can anyone make a chicken-like scenario which > still stumps this SODA version? (If your scenario has more than 4 candidates, > please use DAC instead of approval to find the SODA order of play.) Or do you > know of a different system which creatively resolves the chicken problem? Remember trees :-). In a tree where B and C form one branch they and their voters are bound to support each others. Juho > > JQ > > 2011/8/5 Jameson Quinn > > > 2011/8/5 > > Jameson, > > as you say, it seems that SODA will always elect a candidate that beats every > other candidate majority > pairwise. If rankings are complete, then all pairwise wins will be by > majority. So at least to the degree > that rankings are complete, SODA satisfies the Condorcet Criterion. > > Also, as I mentioned briefly in my last message under this subject heading, > SODA seems to completely > demolish the "chicken" problem. > > Well almost. See below. > > > To review for other readers, we're talking about the scenario > > 48 A > 27 C>B > 25 B>C > > Candidates B and C form a clone set that pairwise beats A, and in fact C is > the Condorcet Winner, but >
Re: [EM] Chicken problem (was: SODA and the Condorcet criterion)
>>> To review for other readers, we're talking about the scenario >>> >>> 48 A >>> 27 C>B >>> 25 B>C >>> >>> Candidates B and C form a clone set that pairwise beats A, and in fact C >>> is the Condorcet Winner, but >>> under many Condorcet methods, as well as for Range and Approval, there is >>> a large temptation for the >>> 25 B faction to threaten to truncate C, and thereby steal the election >>> from C. Of course C can counter >>> the threat to truncate B, but then A wins. So it is a classical game of >>> "chicken." >>> >>> Some methods like IRV cop out by giving the win to A right off the bat, >>> so there is no game of chicken. Wait a minute! IRV elects C in this scenario, if that is how the voters actually vote, and those are the sincere preferences (A voters have no preference between B and C). Much as I hate to say it, IRV works OK in that scenario. On the other hand, if the A voters prefer B over C, (as in the 2009 Burlington, VT mayoral election, http://scorevoting.net/Burlington.html) IRV ignores the preference and still elects C, which seems to be the wrong choice. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Chicken problem (was: SODA and the Condorcet criterion)
More thoughts on the "chicken problem". Again, in Forest's version, that's a scenario like: 48 A 27 C>B 25 B>C C is the pairwise champion, but B is motivated to truncate, and C to retaliate defensively, until A ends up winning. In my opinion, scenarios like this make the single most intractable practical strategy problem in voting theory: - Approval, Range, and median-based systems all suffer directly. - Most winning-vote-like Condorcet systems fall prey, including otherwise-great systems like Schulze. - Margins systems have no truncation incentive - but as a direct consequence, they give extremely difficult-to-justify results if the B block truncates; in fact, they allow a strategic C block to fool the system into thinking it's seeing this scenarion when actually B and C are mortal enemies. - IRV does relatively well with this scenario - but in return, pays no attention at all to the second choice of the A voters, which should be decisive if it exists. - At the other extreme, some systems resolve this problem by forcing strict rankings from the A voters - but if they really don't have a preference, that ends up being just statistical noise, and doesn't even necessarily remove the game-of-chicken incentives if things are balanced right. Moreover, forcing B and C voters into strict rankings only makes them escalate their truncations into burials. Most of us, when we want to "test" our voting systems with a difficult case, use a strict-ranking Condorcet cycle of three; the old, standard ABC BCA CAB scenario. That's nice and simple, but not very realistic. To me, the "game of chicken" scenario; the resulting Condorcet cycle if B truncates; and related scenarios that could strategically be made to masquerade as these; are better practical tests for a voting system. In fact, I'd go so far as to guess that *a real-life Condorcet cycle would be more likely to be the result of playing chicken than of honest preferences.* As Forest already explained, SODA, as currently formulated, resolves the game of chicken — if all votes are delegated. It can do that because games among finite candidates are much more tractable than those among oceans of voters. SODA's "sequential trick" would be ridiculous with voters; imagine "Your turn to vote is on Sunday at 2:35:58 PM." In my previous message in this thread (Re: SODA and the Condorcet criterion), I pointed out that there's still a problem if voters explicitly truncate by refusing to delegate. But I've been considering this issue, and eventually I found a solution that I think is simple enough to include in SODA: *Make all candidate's predeclared rankings into strict rankings by breaking declared ties in order of the current approval totals when it's their turn to use their delegated votes.* So if B voters truncated, candidate A would see that B was headed for a win, and would have the option to delegate to C. All the truncation would have accomplished would be to make A into a kingmaker between B and C. Since A could have had this kingmaker power, if she had wanted it, from the start, that's not a problem. The only difference between this end-game kingmaker power of A's, and if she had simply declared a preference from the start, is that the end-game power could in theory arise no matter which of B or C has more approvals, whereas an initial preference would only confer kingmaker power if the preferred candidate ended up with fewer approvals. Is this version of SODA really the only system to have a fully-satisfactory resolution to the chicken problem? Even if it is, is it worth adding this additional complexity to SODA? Can anyone make a chicken-like scenario which still stumps this SODA version? (If your scenario has more than 4 candidates, please use DAC instead of approval to find the SODA order of play.) Or do you know of a different system which creatively resolves the chicken problem? JQ 2011/8/5 Jameson Quinn > > > 2011/8/5 > > Jameson, >> >> as you say, it seems that SODA will always elect a candidate that beats >> every other candidate majority >> pairwise. If rankings are complete, then all pairwise wins will be by >> majority. So at least to the degree >> that rankings are complete, SODA satisfies the Condorcet Criterion. >> >> Also, as I mentioned briefly in my last message under this subject >> heading, SODA seems to completely >> demolish the "chicken" problem. >> > > Well almost. See below. > > >> >> To review for other readers, we're talking about the scenario >> >> 48 A >> 27 C>B >> 25 B>C >> >> Candidates B and C form a clone set that pairwise beats A, and in fact C >> is the Condorcet Winner, but >> under many Condorcet methods, as well as for Range and Approval, there is >> a large temptation for the >> 25 B faction to threaten to truncate C, and thereby steal the election >> from C. Of course C can counter >> the threat to truncate B, but then A wins. So it is a classical game
