# “My opponent is a no-good, rotten cheater out to destroy America…”

Well, the primary campaign season sure has been heating up. As expected, we see that the candidates are really going after each other, making outlandish remarks, the whole shebang. This isn’t anything new: even back in June, former Minnesota governor Tim Pawlenty accused former Massachusetts governor Mitt Romney of being complicit in the formulation of the Democrats’ controversial health care reform plan, calling it “Obamneycare.” Clearly, with the jam-packed field and a necessity to beat the other contenders to get to the main event, there is a strong incentive for the candidates to go after each other’s throats.

Yet there is a major tradeoff in doing so. All of these candidates would very much prefer, even if they personally do not win the Republican nomination, that one of the other Republicans beat Obama in the general election. But by attacking their competitors, they decrease the chance of that happening.

Let’s assume that a candidate’s chance of winning is proportional to the amount of flak that is not directed at him or her. Thus, if each candidate $i$ (of $N$ total) generates $f_{i,j}$ flak towards candidate $j$, then each candidate’s share of the flak is $\frac{\sum_{j=1}^{N}f_{j,i}}{\sum_{k=1}^{N}\sum_{j=1}^{N}f_{j,k}}$

However, by being the victim of more flak, the chances of beating Obama get smaller and smaller: voters are much less likely to vote for a candidate who has a terrible reputation, as bestowed upon him or her by his or her opponents. We can therefore set up a threshold at which the voters will not vote for candidate i: once he or she has taken more collective beatings than threshold $F^{*}$, Obama automatically wins (we can model Obama’s chances of winning as increasing in the amount of trashing his opponent has received; the main idea of the result will remain the same). Scary thought if you are a Republican, no?

Each candidate gets payoff $R$ for being the nominee, as well as an additional payoff $P$ if he or she wins the general election. The other candidates get payoff $0$ if they lose the nomination and the presidency, while they get payoff $V$ if a different Republican candidate wins the general election. Assume that $R>V$ – candidates would rather be the nominee themselves, and get a shot at the presidency, no matter what the other candidates’ chances are.

This being the case, despite the preference to beat Obama, Republican primary candidates can always do better by hacking at each other as much as possible. Think about it – given any fixed amount of flak the other candidates are giving you, your chances of being the nominee go to one as the amount you attack them goes to infinity. This strategy (given finite amount of flak from others) will yield a payoff of approximately $R$, which we stipulated was greater than $V$. Hence the Democrats will automatically win.

Note that this isn’t quite a Nash equilibrium, since the set of possible options is not bounded. Yet some of the normal game theory ideas are still visible here: what will end up happening, based on the incentives, is that the Democratic candidate gets re-elected. Nevertheless, the result depends on several assumptions: that the Republicans can attack each other to an arbitrarily large extent, and that they would always rather be the nominee than let someone else win. Still, it is interesting to observe that the situation as modeled here leads to an automatic Republican loss in November 2012. Pretty ironic, given that one would think that the entire purpose of the campaign is to unseat Obama.

We think of the veto as a very powerful (perhaps even unfairly powerful) bargaining chip, but this is not always the case. Sometimes having a veto is not as good as giving it away.

In this example, you want to select a juror. Candidates arrive randomly — most are acceptable but mediocre for both sides, a few are great for one side and terrible for the other, and a few are pretty good for both.

This is a variation of the Secretary Game. The central question for secretary games is, “Since, once a candidate is rejected, he does not apply again, when should we stop interviewing?”

For simplicity, we assume that there are only three types:

 Type by utility to (x,y) Probability of arrival b, b where {1/2 < b < 1} $1 - 2\epsilon$ 1, 0 $\epsilon$ $1- \epsilon$, $1 - \epsilon$ $\epsilon$

[1]

1. If both players reject the candidate, he is rejected.
2. If both players accept the candidate, he is accepted.
3. If one player accepts the candidate and one player rejects, then the candidate is accepted unless someone uses a veto.

This is a sequential game, so it is a game of perfect information. It is also Markovian, which means that if the candidate is rejected, we return to the same state in which we started.

Suppose neither side has any vetoes. Then, X always accepts (1,0) and Y always rejects this, so it is accepted (since there are no vetoes). Y always accepts ($1- \epsilon$, $1 - \epsilon$) since it is Y’s best outcome, so it is accepted. X rejects (b,b), but Y accepts (b,b) because b > 1/2 and so it improves Y’s average. (If Y also rejected it, then Y’s payoff would be the average of 0 and $1- \epsilon$.)

Therefore, the expected utility is:

$U(X) = (1- \epsilon)(\epsilon) + (b)(1- 2\epsilon) + (1)(\epsilon)$
$U(Y) = (1- \epsilon)(\epsilon) + (b)(1- 2\epsilon)$

If X has exactly one veto and Y doesn’t, then X would use up the veto when Y accepts b in one round, and we return to the starting point. This makes X slightly better off and Y slightly worse off.

$U^*(X) = (1- \epsilon)(\epsilon) + (U(X))(1- 2\epsilon) + (1)(\epsilon)$
$U^*(Y) = (1- \epsilon)(\epsilon) + (U(Y))(1- 2\epsilon)$

However, if Y has one veto and X doesn’t, then X would need to reject (1,0) (since Y would veto this and X would end up with U close to b). If (b,b) arrives, X and Y both reject, and if $(1- \epsilon)$ arrives, both accept. So, Y’s having the veto is actually better for both candidates than X’s having the veto (and, if X could, he should give the veto to Y). Additionally, since this arrangement guarantees a higher expected payoff for both sides, giving one side a veto can even be Pareto-improving.

Extension: What happens if both sides have a positive, finite number of vetoes?

It’s easy for X to guarantee himself an outcome of $(1- \epsilon)$. He can simply accept (1,0) and $(1- \epsilon,1- \epsilon)$ every time they appear and reject (b,b) up until Y only has one veto left. Then, play as in the previous case. He can’t do any better, since there are no cases where Y has vetoes and (1,0) is accepted, and there are no cases where Y has no vetoes and X’s expectation is greater than $(1- \epsilon)$.

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[1] Mathematicians and other quantitative people talk about $\epsilon$ (epsilon, pronounced either “EP-si-lon” in the US or “ep-SIGH-len” in the UK) a lot. We’ve certainly used it often. You can think of it as an arbitrarily small positive number, or “a number as small as you need it to be, but not 0.”

Example is based on Shmuel Gal, Steve Alpern, and Eilon Solan’s A Sequential Selection Game with Vetoes (2008)

# Hot deficit potato!

I had originally wanted to share some really cool veto math today, but I’m really fascinated by the chicken endgame being played out — it has to be periodic and not simultaneous near the end. Vetoes will have to wait.

Deficit chicken is starting to look a lot like hot potato in these final days before August 2nd. Even though bank analysts believe the Treasury will be able to hold things together for a few days longer, it’s probably in everyone (in the District)’s interest to take the deadline seriously — better the deadline you know than the one you don’t.

Even if default and/or downgrade don’t happen, Congress is quite aware that the situation is FUBAR and that passing the blame is the name of the game. Whichever of the Dems, GOP, or Obama happens to say the last “no” via veto, filibuster, or downvoting by the time the deadline rolls around will be blamed for preventing legislation of any kind from passing. Whereas we have previously modeled the situation as a game of Chicken, by now we can pretty much count the number of days it will take to get anything through the legislature. Now, if you’re unfamiliar with the game, Hot Potato involves passing an object back and forth or in a circle while music plays. The loser is the player holding the potato when the music stops.

In our game, all parties know exactly when the music will stop, and if all threats to defeat, filibuster, and veto are actually carried out, one of the three players is going to have to say the last “no.”

This is why I don’t find Obama’s threat to veto House Speaker Boehner’s bill (if it passes in both houses) credible — if he vetoes, legislature is basically out of time to draft something else that will pass in both houses. Obama would almost certainly get stuck with the hot potato.

No wonder he’s upset — it’s a Catch-22 for him, since he’s already looking weak to the Democrats for offering such a steep compromise. But, “singlehandedly” causing a default would be much much worse for Obama, which is exactly what the Tea Party hobbits seems to want, according to Senator McCain.

So, why is Obama making this threat in the first place, if it’s not credible? If he makes it seem like vetoing the bill is on the equilibrium path for him, it saves the Senate Democrats some face if they kill Boehner’s bill if/when it arrives in Senate. Otherwise, if the bill is stopped in the Senate, the Democrats in Senate will be left holding the hot debt potato at the end of the game.

Meanwhile, Senator Reid is biding his time before putting his version of a compromise to vote in the Senate. If Boehner’s bill fails to pass today, Reid will have the upper hand because he figures there’s just enough time left to pass his bill in both houses. Obama is supportive, so there’s no threat of veto, and even if the Senate Republicans filibuster as they’ve threatened to do, the GOP will certainly be saddled with the blame, since filing for cloture will set Senate back at least a day or two[1]. If it passes in Senate, the House has little choice but to pass the bill, or be blamed. Unless, of course, they really believe there’s more time to continue tossing the potato, or that doing this might actually make Obama look bad.

Quote of the day, from Reid: “Magic things can happen here in Congress in a very short period of time under the right circumstances.” Right circumstances => Not when filibustering is involved.

A very simplified game tree follows:

As we can see, all current threats, if carried out, lead to someone’s becoming a scapegoat. Interestingly, even though it looks like Boehner might get the votes he needs today, the Senate Democrats are using some excellent commitment strategy, by leaking a letter signed by all 53 members (hey, it looks like they didn’t get Joe Lieberman the Independent on the bandwagon) stating that they do not support the Boehner bill.

Next time: how having veto-power can hurt your outcome.

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[1] If the GOP decides to filibuster until the Fed defaults, it’s going to be pretty hard to pin this on Obama. Just saying.
[2] Fun infographic table thingy.

# Preemptive bribes in legislature

Last week, Jeff discussed Deficit Chicken and the consequences of playing out the game. The August 2nd deadline is almost upon us, and up until now pretty much every product of debt-ceiling negotiations in Congress has failed to yield anything that has a shot at passing in both houses. How are the members of these parties “standing strong” together, anyhow?

It’s clear that some incentive prevents the moderates of each party from switching over — that’s why there’s a Gang of Six and not a Gang of Sixteen or Twenty (yet). Whether it’s pride, pork, or pressure, as Ted DiBiase would say, “Everybody has a price.”

This price, for an individual legislator, would be equivalent to the utility of standing his ground. Let’s suppose these utilities look like this for nine representative legislators on a committee. (Notice that I represent utility below the line as a negative utility from the status quo — it is a positive utility for the opposite outcome.)

Suppose the other party has a budget of 20 (say, in billions of dollars) to spend on bribing members of this party to join their coalition. The status quo party (call it X) has the first move and can pre-emptively give the committee members some additional utility for staying on this side, and afterward, the other party (Y) has the opportunity to make counteroffers.

Now, say that Y needs to ultimately win over 5 of these 9 committee members. As it stands, who would Y bribe? Certainly, Mr. I is already on Y’s side, so Y needs to bribe four more people — probably the least expensive of them, E, F, G, and H. Perhaps party X should devote its energy toward bolstering these weak members, possibly even try to win the dissenters back.

But, this isn’t always the case. Tim Groseclose and James Snyder in their celebrated 1996 paper offered a delightful best response strategy for party X in their vote-buying model.

To win over 5 members in the current state, Y just needs to spend 4 + $\epsilon$ on E, 3 + $\epsilon$ on F and G, and 2 + $\epsilon$ on H to win. Party X can make Y’s life harder by preemptively giving E, F, G, and H more money, say 1, 2, 2, and 3 respectively, to make their payoffs 5 each — this exhausts Y’s war chest. But now, there is a less expensive party member (D) for Y to bribe!

So, X must give all of the members (except I) enough to have a payoff of at least 5, so that there are no “soft spots” to target and Y cannot win over any of them. It turns out that the best strategy for X is always this kind of “leveling schedule,” and X often ends up bribing more members than Y needs to win. We can easily test this here: suppose the optimal strategy is not a leveling schedule. If we add x more to anyone’s payoff, it is unnecessary. If we remove x from anyone’s payoff in the leveled coalition (say, we bribe H with 2 instead of 3). Then, Y will certainly target H, and have 20-4 = 16 left to spend on winning 3 more votes. X would need to add 1/3 to each of C, D, E, F, and G, to prevent Y from winning any three of four, which is another leveling schedule (and one that happens to be more expensive than the 1 saved from H.)

It could be, then, that “standing strong” for our status quo party really does mean emulating Aesop’s fable of the bundle of sticks (which, incidentally, is fascio in Latin).

# Creating class divisions as a Marxist strategy

Why are Americans just eating up all of these “tax the top” policy proposals? Easily, because it’s “not us.” It’s a monetary transfer to anyone who isn’t being taxed.

Romer (1975), and later, Meltzer and Richards (1981) came up with an elegant model which demonstrates how societies with majoritarian democracy will continue to vote for more redistribution. Taxation is proportional, which means everyone pays the same percentage of their income, and the tax revenue is redistributed evenly (total tax revenue/population). In this scheme, the average (mean) earner is taxed and receives about the same amount, taking into account the inefficiency of taxation — otherwise he would vote for a tax rate of 100% [1].

As long as the amount an individual is taxed is less than the amount he would get back from redistribution, that voter will vote for more redistribution. Since in all real world distributions, the median income is less than the mean income, the median voter — political economy’s favorite son — will always vote for more redistribution.

Just to do a quick overview: Let voters in a society be uniformly distributed on the interval [0,1], where 0 is the farthest left one could be and 1 is the farthest right. If two policies are competing for votes, and voters vote for the policy closest to their own position, then the Nash equilibrium position for both policies is at 0.5, where the median voter is.

Why? Well, suppose a policy deviates from the median — it will surely lose votes.

The Romer/Meltzer-Richards model predicts that since the median voter prefers to get back more than he pays, he would vote for policies that increase the tax rate. Everyone earning less than the mean gets a positive transfer, and everyone earning more than the mean gets a negative transfer.

In the US, the difference in income between the mean and median household is about \$7,000. For the median voter, voting for redistribution in the RMR model is like voting to increase taxes for your slightly wealthier neighbor, the guy you aspire to be in few years. Moreover, the median voter might not even be the median earner (and almost certainly isn’t — voter turnout drops dramatically as income decreases).

Well, this certainly wouldn’t result in more redistribution. The solution? Devise taxing schemes that only tax the top x% — these are not your neighbors, nor people you think you’ll ever be, and nowhere near the mean earner (at ~65th percentile) or median voter. This will certainly win over the median voter, wherever he or she turns out to be. In fact, Karl Marx had a similar idea, really.

From the foregoing, it is evident that Marx deduces the inevitability of the transformation of capitalist society into socialist society and wholly and exclusively from the economic law of the development of contemporary society. – Vladimir Lenin, 1914

Indeed, RMR corroborates in a very limited fashion Lenin’s sentiment, though Lenin himself seems to have interpreted “economic law of the development of contemporary society” to mean “bloody revolution.”

Marx believed society looked something like this, which was probably not far from the truth in feudal societies but not quite true in capitalist ones:

Since members of the proletariat are pretty much the same and basically have the same preferences, they will act as a bloc and are all equivalent to the median voter. The mean and above mean earners are all among the bourgeoisie, so tax, tax, tax away! (Rabble rabble rabble!) Likewise, when policies are created to tax the top 2%, the proposition creates two somewhat artificial classes: those who are in the top 2% and those who are not.

In reality, radical redistribution doesn’t happen in the US for a slew of mostly sociological and historical reasons[2]. In the status quo of American culture, we don’t like to see these kinds of class divisions.  Americans in general, and especially the more right-wing ones, tend to believe that we have a great deal of social mobility and tomorrow one could end up anywhere on the spectrum (for our Rawlsian friends, this a post for another day.) Along similar lines, Americans tend to see low income as a result of laziness and not bad luck, so relatively limited redistribution schemes are punishment/pity for the lazy (from George H. W. Bush’s inaugural speech, “[we should reach out and help] those who cannot free themselves of enslavement to whatever addiction — drugs, welfare, demoralization that rules the slums.”) For another game theoretic analysis of welfare, be sure to check out Jeff’s “Welfare for lazy bums” post.

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[1] Also, the median voter doesn’t vote for any tax rate above the Laffer curve, since reducing the tax rate would increase revenue.

[2] An excellent overview: Alberto Alesina and Ed Glaeser’s lecture series “Fighting Poverty in the US and Europe”