# A Game Theory / Strategy Question

An acquaintance is the claimant in a case which is going to arbitration before the American Arbitration Association. The first step in the process is to select an arbitrator.

The way the process works is that the Association gives each side (claimant and respondent) a list of 10 possible arbitrators. Each side gets to strike 2 arbitrators and then rank the remaining 8 in order of preference. The Association takes the list, removes all arbitrators which have been stricken by either side, and then adds the rankings for each arbitrator. The arbitrator with the lowest total is selected. (Ties are resolved by choosing the arbitrator who is least busy.)

Obviously the claimant wants to end up with the most liberal arbitrator possible; the respondent wants to end up with the most conservative arbitrator possible. Both sides have a rough idea of which arbitrators are more liberal and which are more conservative but those ideas won’t be exactly the same.

Let’s assume that the respondent will strike the two most liberal arbitrators and then rank the remaining arbitrators from most conservative to most liberal.

The question is whether the claimant’s best strategy is to strike the two most conservative arbitrators and then rank the remaining arbitrators in true preference order? Intuitively, it seems a waste for the claimant to give a number one rank to the most liberal arbitrator, since that arbitrator is very likely to be stricken by the other side. Might not the claimant do better by choosing the 3rd or 4th most liberal arbitrator as his or her number one choice?

Do the remaining arbitrators move up the rankings after the four have been removed? If so, just list them in your natural order of preference, Whichever ones your opponent removes, you will be left with your favourite remaining arbitrators in the correct order.

If not, then it depends on whether the assumption that the respondent will strike the two most liberal arbitrators is correct. If it is, then obviously it is better to put the third most liberal arbitrator at the top of your list. But if he reasons as you do, then it resembles a game of Scissors, Paper, Stone and there is no best strategy. You just have to guess what the other guy is going to do.

Sure seems that way.

To make sure I understand, let’s suppose there are ten arbitrators, Anderson through Jones, with the most liberal being Anderson and the most conservative being Jones. If each side has perfect knowledge, and there are no other criteria, and acually ranks the arbitrators in their preferred order, then the lists would look like so:

CLAIMANT

1. Anderson
2. Baker
3. Carter
4. Dennison
5. Evans
6. Foster
7. Galloway
8. Hunter
[del]–Iggy Stooge[/del]
[del]–Jones[/del]

RESPONDANT
[del]–Anderson[/del]
[del]–Baker[/del]
8. Carter
7. Dennison
6. Evans
5. Foster
4. Galloway
3. Hunter
2. Iggy Stooge

1. Jones

In which case Carter through Hunter all wind up scoring 11, and it’s a toss-up who gets the case.

You’re suggesting the claimant rank Carter on top and drop Anderson and Baker, which would ensure getting the best possible arbitrator who hasn’t been stricken, rather than relying on some luck. Whether that’s a winning strategy depends on what the other side does, and just how complex the real decision process is compared to the simplified liberal-conservative model here.

But if I, as claimant, rank Carter as 1, Dennison as 2, Evans as 3, Foster as 4, and so on - they’ll still all tie, just with a score of 9.

Let’s see. If I rank Carter with 1, Anderson and Baker with 2 and 3, then continue on in series, then Carter is going to get a score of 9, and every other eligible candidate has 11! That’s the winning strategy, as far as I can tell.
Assuming our estimates of the other guy’s choices is correct.

This reminds me just a bit of the game ‘Quinto’

Thanks for the replies. It seems that my intuition is correct that one should not waste one’s top choices on arbitrators who are sure to be stricken by the other side. What was confusing me is that assigning a low number to an arbitrator also has the effect of increasing the numbers for the remaining arbitrator.

So, as chrisk points out, it would be a mistake to rank A and B at the bottom. Given that the process is somewhat inexact, I would say the strategy should be to (1) order all 10 arbitrators in terms of how liberal they are; (2) strike the two most conservative arbitrators; and (3) take the 2 or 3 names and the top and move them below the following 2 names.

Yeah, there’s a bit of a paradox in that just giving more weight to all your choices doesn’t really increase the relative weight of your first choice - so if you know what the other guy is doing, there’s some value in sabotaging some of the less-good alternatives.

Now, if both players are aware of this kind of strategizing, then you’d want to adopt some sort of randomizing strategy. Unfortunately, with 10 choices and 2 vetoes, there’s around 1.8 million possible choices for each player, (10! / 2), and thus 3.3*10^12 possible outcomes on the game theory grid.

I may try mapping out some simpler variants, (3 possible lawyers, 1 veto per side and so on) to see if there are patterns that will generalize up.

I’ve worked out the payoff matrices for the three-lawyer version, both with and without one veto per player, and if I’ve analyzed it correctly the strategic equilibrium is the same, and slightly interesting.

Both players get best results by giving the least precedence to the opposing team’s best lawyer, (vetoing them if they have that option,) but it doesn’t matter if they pick their own best or the comprimise candidate first. In the veto version, the compromise lawyer in the middle will always be picked if both players choose any variation of this strategy, for the obvious reasons. In the no-veto version, then if both players choose their own best lawyer first, then all 3 are in a draw, which is still considered a neutral outcome.

The only way the vetoes matter in the 3-lawyer version is if player 1(who prefers A,) picks A, then C, and veto B, and player 2 picks B, then C, and veto A. With the vetoes, C is picked, but if that spot was just ‘lowest vote’ then all three would be tied.

Also ran the 2-lawyer variant, with no surprises - neither player has any incentive to not give their guy precedence, therefore draw.

4-lawyer would be about 16 times more complicated than 3-lawyer, but I may see if I can write a computer program to analyze some of these. (But Nanowrimo starts tomorrow, so I won’t have a lot of free time. Hmm…)

I would think it pretty likely that the respondant would see the same thing and by ranking Hunter 1 and Galloway 2 would achieve a toss of the coin for who was least busy amongst Carter, Dennison, Hunter and Galloway.

If however you rank Dennison 1 Evans 2 and Carter 3 you ensure that you get one of “your” guys regardless. The fact that Carter becomes last choice is meaningless. Your opponent can only prevent you getting your choice by mirroring your rankings or not vetoing your top 2.

I’m not sure I believe any statement that you can get such-and-such a result ‘regardless’ in a game like this without a much more rigorous demonstration. The opportunity to get ‘outsmarted’ is tremendous.

I think I’ve worked out the idea strategy for the four-lawyers, no-veto version. This involves a randomizing strategy, involving four-possible plays. Assuming that I want to get A preferentially, and that the other possibilities are evenly spread out to D who I want least, and the other guy wants them in exactly the same order and spread, my four plays are:

Naive: A, B, C, D
Top switch: B, A, C, D
Bottom switch: A, B, D, C
Double switch: B, A, C, D

I throw an eight-sided (octahedral) die in secret to determine my strategy. If the result is 5 or above, I pick the top switch. For 1, I pick the naive move, for 2, the bottom switch, for 3 the double switch, and if it’s a 4 I keep on rolling hoping for a different result, (until I lose patience and default to the top switch.)

With that probability distribution, no move or combination of moves that the other guy can settle on will tend to give him a better chance than fifty-fifty of getting one of his choices overall. If he makes his strategy from those 4 moves, then we’ll tend to even out, while if he tries doing anything else, then I’ll have a better chance.

More explanations later.

Continuing…

If I pick the top switch and he picks the naive strategy, I get my second-choice guy, B. Same thing if I do the double switch and he does the top switch.

If I play the naive strategy and he does the bottom switch, then I get my first-choice guy A. Same thing if O do the bottom switch and he does the double switch.

If I play the naive strategy and he plays the double switch, then A and C are tied, so I might get my first-choice guy or his second-choice guy.

(All of this applies to him just as well with the names reversed appropriately.)

If we both do the top switch, then it’s the middling guys, B and C, who are tied. If we both do the bottom switch, then it’s A and D. In all other cases, I believe, all four names are tied.

This kind of thing is fun, but it does take time to work out.

One more post before I give up for the day.

In four-lawyers with one veto, the bottom switch (veto C and rank A, B, D) seems to drop out as an ultimately self-defeating move. The other three strategies assume equal weights, essentially a game of rock-paper-scissors.

Naive strategy is veto D, rank A B C

this is beaten by the top switch: veto D and rank B A C (The player pulling the top switch will get his second best choice.)

which is, in turn, beaten by the double switch: veto C and rank B A D (Again, getting second best.)

Which is vanquished by the naive, with a tie for best and third best.

I’m not quite sure when I’ll have the fortitude to tackle five of these guys.

Emphasis mine. Did you mean to type something else for the Double switch? It is identical to the Top switch.

Adds the rankings, or adds the points-as-rank? If I insincerely vote my preferred candidate down and give my third choice an 8, is it the 8 that gets added if he is not struck, or does the 8 become a 7 or 6 depending on my opponent’s strikes?

Yes, sorry. The double switch is B, A, D, C

Incidentally, this is really like three strikes. A candidate cannot win if either side votes “1” for that candidate (the best that can be hoped for is a tie; actually, a 6-way tie when both parties’ preferences are exactly opposite). I don’t know what happens to break a tie here, either. (This supposes preferences are adjusted after striking.)

I’ve tried to reduce the 10 player game to a four-player game but it is impossible. You would think the dominant strategy would be to strike/vote 1 for your three least-favorite candidates, but it turns out that unless you also split this strategy based on probabilities, your opponent can locate a strategy in which one of her more favored candidates wins the election by voting insincerely. (chrisk’s analysis of the 4-candidate race.)

Suppose you say, “I will always strike the least-favored and give the third-least favored a ‘1’”, and preferences aren’t reordered (my previous question) then this third least-favored candidate can win. (Anyone who can get a ‘2’ can win; if preferences aren’t adjusted, it may be that someone insincerely rated their most-preferred candidate a ‘1’, which subsequently was dropped.)

If preferences are reordered, and both sides vote sincerely, and preferences are exactly opposite, then this strategy leads to a tie. The only way to avoid a tie is to vote insincerely.

Of course, if both sides vote sincerely and agree to some small extent, this method will pick that individual, which after all is probably the point.

Incidentally, this is a Borda Count election with 8 points for 10 candidates (which is, Borda count with forced truncated voting and mandatory ranking). It is susceptible to compromise (insincerely raising a preferred candidate) and burying (insincerely lowering a less-preferred candidate). It is also very vulnerable to clones, but this has nothing to do with voter strategy (it does mean candidate selection could be subject to bias, though.)

Fixed.

An update:

The acquaintance ended up with her bona fide #3 choice out of 10 arbitrators. She had ranked this person #2 in her submission to AAA. So the strategy we came up with seems to have worked pretty well.

Not sure how the arbitrator himself will work out, but still it seems like a pretty good start.