6

A Brutal Logic

An equilibrium is not always an optimum; it might not even be good. This may be the most important discovery of game theory.

—IVAR EKELAND, The Best of All Possible Worlds: Mathematics and Destiny

If interrogational torture is to work the way its proponents claim, then it must follow a brutal cost–benefit, pain–information logic. Now that we have a more realistic model, but one still faithful to the pragmatic model in Chapter 3, it is time to trace out this brutal logic, solving for the RIT model’s equilibria. In following this logic, systematically and relentlessly, we will use bloodless and sterile language to reconstruct a bloody and dirty work. This, however, is the logic of the pragmatic model and it must be faced squarely by those advancing it. The equilibria we find tell us what can happen, and under what conditions, when torture along the lines of the pragmatic model is admitted into an interrogation room.

In one sense, this chapter and the next constitute the core of the argument. The deductive argument about what happens when interrogators torture is actually located in the nitty-gritty mathematical derivation of the equilibria and the investigation of their properties. As you’ve already seen, the math is really just an accounting device to keep track of the logic, but we will all but eliminate it here. For those interested, the derivation of all the equilibria can be found in Appendix A.

We will solve RIT the same way we solved the BIT model, with one additional step. The Interrogator will use the move made by the Detainee (“no information,” for example) to see if that gives her a better idea of what type of Detainee she is facing. Updating your beliefs in this way requires a new tool, Bayesian belief updating. Once we have this tool in hand, we’ll check for two equilibria of the RIT game. Having walked through an example of how to solve for equilibria, we leave the derivation of the other equilibria for Appendix A and turn to their properties in Chapter 7.

UPDATING YOUR BELIEFS

Suppose you’ve returned to my casino and I offer you the opportunity to play a new game. I’ll give you $10 if you can guess the suit (club, diamond, heart, or spade) of a card I’ll draw randomly from a regular 52-card deck. If you guess wrong, you give me $2 (in other words you lose $2). Let’s say that I charge you $3 to play this game.

Would you pay the $3 to play this game?

In order to figure this out, you will want to consider the likelihoods of winning the $10 and losing the $2. At this point, before I’ve drawn the card, what are the chances you’ll pick the right suit and win the $10? One in four, or 25%, because the total of 52 cards are divided equally among the four suits (13 cards each). Of course this also means that your chances of guessing wrong and picking one of the other three suits are three in four, or 75%. Not great odds, admittedly.

Still, you can use this information to calculate the expected value of this bet in exactly the same way we did in the last chapter. Just as before, the expected value is just the weighted average of the two outcomes, where the weights come from the probabilities or chances that each outcome occurs. For this game it is . If you were to play this game over and over again, your average per game winnings would be $1.

Assuming you’re still risk neutral, as in the last chapter, this isn’t a profitable game for you (though, like a real casino, it would be for me). You would pay $3 and expect, on average, to win $1.

Now suppose that, as in real life, I’m not very handy with cards, and as I draw the card in front of you, you notice that it is red. It flashed too fast for you to tell the particular card or even the particular suit, but you did notice that it was definitely red and not black.¹

So there we are: I’ve drawn the card, you haven’t paid yet, and I ask you whether you want to put down the $3 and play. What would you do? Would you reason as before and decline my offer? Or would you change your mind? (We’ll keep assuming you’re risk neutral through the rest of this example.)

If you stick to your guns, that means that you would ignore the new information, the fact that you know the card is red and not black. Does that sound reasonable? Is it rational to continue to insist that there is, say, only a 25% chance the card I’m holding is a diamond? Or that there is still a 25% chance that the card is a spade?

No. It’s not rational or reasonable. Given that the card is red, you know it cannot be either a club or a spade (which are both black) and that it must be either a diamond or a heart. You don’t know which of the two, but you do know it has to be one of them. There is no good reason for you to stick to your old beliefs; you should revise or update them given this new information.

So what are your updated beliefs? Well, you know now you’re going to guess either “diamond” or “heart” because it has to be one of these two and there is an equal chance of either one. With no other information, you have a 50/50 chance of picking the right one. Still not great, but your chances have improved significantly from the .25/.75 chances before I drew the card.

How does this affect your expected value calculation? Substitute in the .5 and .5 for the .25 and .75 from before, so that you have . This changes things pretty dramatically. You can now expect to win $4 on average, not $1. With the $3 price tag on this game, you can expect to come out ahead by $1 and it now makes sense to take the bet and put down your $3. If I keep drawing cards like this, it won’t be long before I’m out of the casino business.

The point of all this is that it is irrational to ignore relevant new information when deciding on a course of action. Of course, the key here is relevant; had I drawn the card more expertly so that you only saw the back of it, you would have no new information, and your estimates should stay as they were before. But if there is truly new relevant information, as in my card draw, then you would be irrational to ignore it.

We will assume that the Interrogator is as reasonable as you are and apply the same principle to the RIT model. The Detainee moves first and that move may tell the Interrogator something about his type. Under objective questioning, for example, the move “information” tells the Interrogator that she is facing a Cooperative Detainee because Innocent Detainees cannot provide the information and Resistant Detainees will not provide any information. Knowing the Detainee’s type is important to the Interrogator because the action she wants to take in response depends upon which type the Detainee is as well as upon the action the Detainee has taken (and her other beliefs about whether all the information has been divulged, etc.).

Bayes’ Rule is the mathematical formula for updating beliefs in this way, and we will use it to calculate expected payoffs and solve for perfect Bayesian equilibria (PBE), an extension of subgame perfect equilibria (SPE) from the previous chapter.² The only difference is that players update their beliefs using Bayes’ Rule whenever possible and use those beliefs to calculate their expected utilities.

We will give the pragmatic interrogation model the full benefit of the doubt and assume the Interrogator updates her beliefs rationally (that is, according to Bayes’ Rule) whenever possible, uses those beliefs to estimate which type of Detainee she is facing, and then calculates her expected payoffs in our game.³

FINDING EQUILIBRIA

The three general steps for solving for PBE in RIT are as follows:

1. Posit a set of possible strategies (moves) by each of the three Detainee types.

2. Figure out each Interrogator type’s optimal response to both possible moves of the Detainee, given the assumed Detainee strategies and updating beliefs according to Bayes’ Rule where possible.

3. Go back and check to see if any of the Detainee types would want to change his strategy given the Interrogator’s anticipated responses.

If the answer to 3. is yes and one or more of the Detainee types would want to switch his strategy, then the strategy profile (the Detainee candidate strategies and the Interrogator responses) is not an equilibrium. If the answer is no, and all the Detainee types would stick to the candidate strategies, then we have found an equilibrium. Let’s see how each step works in practice.

Step #1: Candidate Strategies

The first step is to trot out a candidate equilibrium by simply imagining a move by each player and/or player type. For example, in the RIT game, we might imagine the set of strategies (no information, no information, information). Another would be (no information, no information, no information). In the first example, the Cooperative and Resistant types of Detainee both provide no information, while the Innocent Detainee falsely confesses information (under leading questioning, obviously). In the second example, all three Detainee types fail to provide useful information. We will always assume this order in listing the strategies for the Detainee types: (Cooperative, Resistant, Innocent).

Why begin by assuming a set of strategies and why do we have to have one for each of the Detainee types? After all, there is only one actual Detainee sitting across from an Interrogator.

Remember that acting rationally is acting systematically, considering all the possibilities, and game theory reflects that idea by having the players consider all the possible combinations of their moves. Just as a football coach tries to have a defensive alignment for any offensive setup, the Interrogator wants to have a plan of what to do for any possible move by the Detainee. And because there are different types of Detainee, this means the possible moves of each Detainee type. Each possible combination could be an equilibrium and we need to check them all.

The Interrogator must think something like the following. “I don’t know which type of Detainee I’m facing or what he will do, but one possibility is (no information, no information, information). Another possibility I must consider is (no information, no information, no information).”

Still, the situation is not that bad; not every combination is possible and some can be ruled out. The first candidate, for example, (no information, no information, information), can only happen under leading questioning, since the Innocent Detainee cannot even make this move under objective questioning. The Interrogator can also rule out any set of strategies that has the Resistant Detainee playing “information” since the Resistant Detainee’s dominant strategy is “no information.” As it turns out, this eliminates four sets of strategies the Interrogator must consider. Abbreviating the move “information” with , here is one eliminated combination, for example: (info, info, info).⁴

A rational Interrogator will systematically examine what she wants to do for any possible strategy combinations that remain and, since we are modeling her behavior, we will do the same. Considering both objective and leading questioning and using the same notation as above, this gives us the following list of pure strategies, keeping in mind that “no info” can mean false and misleading information or nonvaluable information in addition to no information at all⁵:

1. Objective Questioning

(a) (info, no info, no info)

(b) (no info, no info, no info)

2. Leading Questioning

(a) (info, no info, info)

(b) (info, no info, no info)

(c) (no info, no info, info)

(d) (no info, no info, no info)

Step #2: The Interrogator’s Response

Once the Pragmatic Interrogator has the list of strategy combinations she must consider, she then figures out what her best response is to each, taken one at a time.⁶ She does this in three steps:

1. First, she updates her beliefs about the likelihood of each Detainee type using the Detainee’s candidate moves as her data for updating her beliefs according to Bayes’ Rule.

2. Second, she uses these updated beliefs to calculate the expected utility of each of her moves.

3. Third, she chooses the move with the higher expected utility after “information” and the move with the higher expected utility after “no information,” giving her a complete plan of action for every Detainee move.

Take, for example, the first set of strategies under objective questioning: . Assuming the Detainee types are playing these strategies, what are her updated beliefs about the likelihood she is facing a Cooperative Detainee after observing the move “information”?

You don’t even need to get out a pencil and paper for this one: In this candidate set of strategies, only the Cooperative Detainee plays “information,” so the probability is 1. In my casino example, this is equivalent to my drawing the card so slowly that you actually could see it was a diamond, not just that it was red. This means that after she sees “information,” she knows she is playing against a Cooperative Detainee, and this simplifies things a bit for her (not completely, because there will still be uncertainty about the clarity and completeness of the information). In calculating her expected utility for which action is better, “torture” or “not torture,” she needs only to consider her payoffs in the upper part of the tree, after the “Detainee is Cooperative” node.⁷

What if she observes “no information”? Here things are not as clear. “No information” is consistent with the candidate strategies of both the Resistant and the Innocent Detainee. She can rule out the Cooperative Detainee type, but she doesn’t know which of the other two sits across from her.⁸ In the casino example, the color red ruled out spades and clubs, but still left the possibility that it could be either a diamond or a heart. This remaining uncertainty is important because while she prefers to torture a Resistant Detainee, she does not want to torture an Innocent Detainee. This time when she calculates her expected utility for which action is better after “no information,” either “torture” or “not torture,” she needs to weight the payoffs by the likelihood that she is playing them against a Resistant Detainee versus an Innocent Detainee.

The weights come from Bayes’ Rule. We’ll calculate them in a moment, but before we do, consider the other possible set of strategies under objective questioning: (no info, no info, no info). In this case, all three Detainee types play “no information.” What does the move “no information” tell us about the Detainee’s type?

Nothing at all. In my casino example, this is equivalent to my drawing the card expertly so that you only see the back of the card and no new information is provided. When this happens, you stick to your original estimate that the card is a diamond (.25). Our Interrogator will do the same. Since there is no new information, there is no reason to change her beliefs, and she sticks with her initial or prior beliefs (p_C, p_R, and p_I).⁹

Return now to the first candidate set of strategies, (info, no info, no info).

The Interrogator needs to figure out her optimal (best) response after both “information” and “no information” so she has a complete plan of action regardless of which action ends up being played. So she needs to calculate and compare her expected utilities of “torture” and “not torture” after “information” and after “no information.”

AFTER “INFORMATION”

Take “information” first. Since only the Cooperative Detainee plays “information” in this candidate set of strategies, by Bayes’ Rule (common sense!), she knows (probability = 1) she is in the topmost branch of the tree after the nodes “Detainee is Cooperative” and “Interrogator is Pragmatic” (she knows her own type). Accordingly, she knows she will receive the payoffs from that node. All she needs to do is simplify the payoffs to “torture” and “not torture” and choose the one giving her a higher payoff.

The calculation proceeds just as in the previous chapter, we just have more variables. Working through that algebra, the expected utility of “torture” is . The expected utility of “no torture” is . The Interrogator will prefer to torture when the expected value of “torture” is greater than the expected value of “no torture”—that is, when the first quantity is greater than the second quantity: .

You’re to be forgiven if at this point you’re wondering how far all this has gotten you if you’re left with this soup of letters, an uncomfortable flashback to high school, glazed eyes, and increasingly clammy hands. You’d also be right if you’re thinking that which action provides the higher payoff depends on the values of the variables.

To see the next step, consider your quandary about whether or not to take an umbrella with you in the morning. When you stand in the morning darkness of your hallway deciding whether or not to take your umbrella, you don’t know what state of the world you’ll be in at 5:00 pm: whether it will be raining or not. There is an imaginary dotted line connecting your morning choice nodes for these two states of the world: it ends up raining and it doesn’t end up raining. Your morning choice is the same in both cases: take, don’t take your umbrella. You just don’t know at which node you stand when you make that choice.

Your decision will depend on two factors: how likely it is to rain and how you feel about some of the possible consequences of your choices. You’ll want to think about how bad it would be if it rained and you didn’t have your umbrella or how annoying it would be to tote around an umbrella in the sunshine. Your decision is easy if the chances of rain are really high or really low or you’re wearing some new and fragile item of clothing or just had your hair done. Otherwise, for the intermediate cases, you’ll have some threshold, maybe only unconscious, for taking the umbrella versus not taking it.

The situation is actually very similar for the Interrogator. Think about what she wants to know. She is trying to decide whether or not to torture because although she’s received information, she doesn’t know for sure whether she’s received all the Detainee knows. So let’s solve the above inequality for f, the probability that the Detainee has revealed all of his information. This will tell us at what value of f she switches from “no torture” to “torture,” just like you might switch from “no umbrella” to “take an umbrella” when the forecast says 65% chance of rain. A little bit of algebra gives us , where is just a label for this particular value of f, read “hat f,” and just means “is defined as.”

This says that the Interrogator will torture when her confidence (belief f) that the Detainee has revealed all he knows drops below the threshold . To help fix this idea, take a look at Figure 6.1. The horizontal axis is the probability f the detainee has revealed all of his information. Since it’s a probability, it ranges from 0 on the left (no chance he provided all the information) to 1 on the right (he’s provided everything he knows for sure). If f is equal to or above , then she does not torture; she is confident enough that the Detainee has revealed all the information he has. If, however, moved to , then f is below the threshold and she tortures.

Figure 6.1 Threshold Example

Note that this is a real constraint; given that , there are values of between 0 and 1, which is required for a belief. It also makes intuitive sense; at some point the Interrogator, whether she is a CIA operative at a black site or a medieval European magistrate, believes that the Detainee has revealed everything he knows and “switches” from “torture” to “not torture.” This “switching” point is the threshold . So even though our game is “one-shot,” with each player moving one time, the setup captures this intuitive ideal central to the (brutal) logic of torture anywhere and everywhere it has been practiced.

This information hiding threshold means that the Interrogator’s response actually depends on her belief f. We cannot simply write down “play ‘torture’ after ‘information’” as a response any more than we could write “play ‘no torture’ after ‘information’.” The strategies depend on her beliefs and we need to put those down too. So her response after “information” is actually “torture” for and “not torture” for .

AFTER “NO INFORMATION”

Now that she has her optimal response after “information,” the Interrogator must figure out what to do after “no information.” (Remember, she needs a complete plan of action.) What will she do? Remember that in the candidate set of strategies for this possible equilibrium , both the Resistant and Innocent Detainees play “no information” and the Cooperative Detainee never does. Because the Interrogator wants to take different actions depending on whether she believes the Detainee is Resistant or Innocent, she wants to know the chances of each possibility. Bayes’ Rule will tell her those chances.

Rather than trot out an unfamiliar and strange-looking formula, return to the card-drawing example to see the intuition. Once you observed the card was red, you knew it was either a diamond or a heart and couldn’t be a club or a spade. In the same way, for this candidate set of strategies, if the Interrogator observes the move “no information,” she knows it was made by either the Resistant or Innocent Detainee and couldn’t have been made by the Cooperative Detainee, because he plays “information.”

Since the diamond and heart were the only possibilities and were equally likely, the updated probability the card was a diamond given it was red was just the ratio of the probability of a diamond to the sum of the probability of a diamond and the probability of a heart, or . Our case is similar, so that the updated probability the Detainee is Resistant (p_R) given no information is the ratio of the probability the Detainee is Resistant (p_R) to the sum of the probability the Detainee is Resistant (p_R) and the probability the Detainee is Innocent (p_I). We don’t know, however, how likely each is, so we have to stick with the variables to keep it general, or . The exact same reasoning leads to for the updated probability the Detainee is Innocent given no information.

With those updated beliefs in hand, the Interrogator can use them to calculate the expected utilities of her two possible responses to “no information:” “torture” and “no torture.” In the (fair) coin toss example in Chapter 5, the outcomes were weighted .5 and .5. In our case here, we don’t know the numbers; we instead have the probabilities defined by the variables p_R and p_I, but we’ll use them and multiply through in the usual way.

Here’s an overview of what we need to do. The Interrogator must calculate the expected utility of “torture” and compare that to the calculated expected utility of “no torture” to see which she prefers. The expected utility of “torture” after “no information” is the payoffs to torturing a Resistant Detainee, weighted by the (updated) probability the Detainee is Resistant, plus the payoffs to torturing an Innocent Detainee, weighted by the (updated) probability the Detainee is Innocent. Ditto for the expected utility of “no torture.” When we do all that, we get .

As was the case after observing “information,” we are left with a soup of letters (parameters actually) that may appear at first not to tell us very much. Once again though, if we go back to what it is the Interrogator wants to know, the way forward is more clear. The Interrogator is faced with “no information” and she knows the Detainee failing to provide that information is doing so either out of intransigence (he is Resistant) or ignorance (he is Innocent). She wants to torture a Resistant Detainee but not an Innocent Detainee. Given a high enough probability the Detainee is Resistant, she is willing to torture. We can find that threshold where she switches from “torture” to “no torture” by solving for p_I, the probability the Detainee is Innocent, given “no information.”

More algebra, then, results in . This is the Innocent Detainee recognition threshold. For values of below this threshold, the Interrogator tortures; values equal to or above the threshold mean that there is enough of a chance the Detainee is Innocent to keep the Interrogator from torturing. Just as with “information,” the Interrogator’s optimal response depends on her beliefs; she does not have a simple “torture” or “not torture” strategy. Instead, her response after “no information” is “torture” for and “not torture” for .

The multiple uncertainties (about Detainee type, amount of information divulged) have made things more complicated compared to our earlier games. Instead of simply (no torture, torture), representing “play ‘torture’ after ‘information’ and ‘no torture’ after ‘no information’,” we have two responses after “information” and two responses after “no information,” each dependent upon a threshold.

In keeping with the systematic nature of game theory, this means there are four possibilities to consider in terms of whether the beliefs are over or under the two thresholds and : over, over; over, under; under, over; and under, under.

You may not be happy to hear this, but it’s actually a little worse. Notice the u in the information hiding threshold : . Remember that the u stands for the probability the Interrogator understands that the information is valuable. If u were zero, it would mean the Interrogator thinks the information is worthless (equivalent to no information); if , it means the Interrogator fully understands the value of the information.¹⁰

Remember, too, that when we set up the game in Chapter 5, we said that while Interrogators sometimes do not fully understand the value of information (), the Detainee always assumes that the Interrogator does fully understand the value of the information (i.e., ). Torture for information makes no sense if the Interrogator cannot understand the value of the information provided. As a result, the Detainee always assumes for the Interrogator.

But what if ? We need to take into account these misperceptions in the next step, when we check to see if the three Detainee types are willing to stick to the candidate strategies set out for them now that we know the Interrogator’s responses.

Step #3: Detainee Deviation Check

Take another look at the information hiding threshold and u’s position in the denominator. If the Detainee assumes that , then that means he thinks the Interrogator’s information hiding threshold is . If for the Interrogator, then the two thresholds are identical (). If, however, for the Interrogator, then because once u starts dropping below 1, it reduces the value of the denominator, making the value of larger. In other words, the Interrogator’s threshold is always greater than or equal to the Detainee’s version of the threshold, never less.¹¹

This difference complicates the four cases we identified above. We had identified two cases for the threshold : f could be over or under the threshold. Now, however, we must take into account the Detainee’s version of the threshold. Since the latter can be no larger than the former, this does limit the possibilities, however. In fact, there are three: f is less than both thresholds, between and , or greater than both.

Since each case is possible for each of the two possibilities for the innocent detainee recognition threshold ( and ), we have six cases to consider. Wait! Don’t run! We won’t do all of these. Let’s just work through two of them to complete what we’ve started and illustrate how to find equilibria in the game.

CASE 1: AND

In this case, the Interrogator’s belief f that the Detainee has provided full information falls below both her own threshold as well as the Detainee’s version of her threshold . In other words, the Interrogator believes the Detainee is hiding sufficient information to warrant “torture,” and the Detainee thinks the Interrogator believes this as well. There is no misunderstanding or misperception in this case. The means that the Interrogator also believes that, after observing “no information,” the Detainee is sufficiently likely to be Resistant that she plays “torture.” In sum, for this combination of beliefs, the Interrogator plays (torture, torture)—that is, “torture” after “information” and “torture” after “no information.”

You might think we are done. We posited a set of strategies by the Detainee (each Detainee type) and found the Interrogator’s optimal response. The key here is “posit.” We just identified a candidate set of strategies; we don’t yet know if the Detainee types would actually stick to them now that we (they) know the Interrogator’s response. In order to be a (perfect Bayesian) Nash equilibrium, the strategies have to be best responses to each other, and we don’t know that yet. We need to check whether (information, no information, no information) is the best response to (torture, torture); we need to see whether any of the Detainee types has an incentive to deviate from the strategy proposed for him. So let’s check.

We know that the Resistant Detainee never reveals information (“no information” is his dominant strategy), so he will not deviate. We also know that, since this is the objective questioning variant of the model, “no information” is the only move the Innocent Detainee even has, so he will deviate either. The only real question is whether the Cooperative Detainee has an incentive to deviate. Does he?

Since , the Cooperative Detainee knows that the Interrogator “tortures” after “information.” He also knows that the Interrogator “tortures” after “no information.” If he is going to get tortured anyway (and pay ), he doesn’t want to give up information too (and pay ). Put another way, he doesn’t want to give up information and then be tortured anyway; the whole reason to give up information is to avoid torture. In short, he has an incentive to switch his strategy to “no information.” He would, in other words, deviate, and so this set of strategies and beliefs does not constitute a PBE.

Congratulations, you have failed to find your first perfect Bayesian equilibrium. Or, rather, you successfully determined that a set of strategies and beliefs fails to generate a stable outcome. That’s useful, actually.

One down, many to go for me, but we’ll do just one more together.

CASE 2: AND

This case is identical to the first case in terms of the Interrogator’s beliefs (and so actions) after observing “no information”: The Interrogator believes that the Detainee is unlikely to be Innocent, likely instead to be Resistant, so the Interrogator plays “torture.” The difference is f; it has “moved” so that it is “between” the Detainee’s version of the Interrogator’s threshold and the Interrogator’s actual threshold . This means that the Interrogator again believes the Detainee has provided insufficient information to warrant not torturing him. Just as in the first case, she is sufficiently confident he is holding information back () to warrant torture. So, given these two beliefs, the Interrogator plays (torture, torture), that is, “torture” after “information” and “torture” after “no information,” just as in the first case.

This time, however, the Detainee believes that the Interrogator believes the Detainee has revealed enough information not to be tortured (). This time there is misunderstanding, misperception. Consequently, the Detainee believes the Interrogator plays “no torture” after “information” and “torture” after “information.”

Once again we need to check whether (information, no information, no information) is the best response to (torture, torture), keeping in mind that the Cooperative Detainee anticipates (no torture, torture). For the exact same reasons as in the first case, neither the Resistant nor the Innocent Detainees will (can) deviate from their strategies. The question is again whether the Cooperative type has an incentive to switch his strategy, as we found he did in the first case.

Since he believes that , and thus expects “no torture” after “information,” he (we) must check to see whether “information” is an optimal response. Here is where the likelihood of a Sadistic Interrogator comes into play. As we saw in the first case, a Cooperative Detainee is willing to give up information to avoid torture, but does not want to give up information and be tortured anyway. That is exactly what will happen if the Interrogator is Sadistic. The Detainee has to try and figure out, to estimate the likelihood, that the Interrogator is Pragmatic (and so does not torture if sufficient information is provided) or Sadistic (and so tortures regardless of what the Detainee does).

Thus, the Cooperative Detainee has his own expected utility calculation to do, weighting the differing payoffs to each action (“information” and “no information”) by the two possible states of the world (Pragmatic vs. Sadistic Interrogator). Since the Interrogator is Pragmatic with probability q and Sadistic with probability , the expected utility calculation of “information” comes out to . The expected utility of “no information” is just . Thus, the Cooperative Detainee prefers “information” (the strategy called for in our candidate equilibrium) to “no information” for .

Since the Cooperative Detainee’s (and our) task is to find the threshold at which he believes with sufficient confidence the Interrogator is Pragmatic so as to provide “information,” we solve for exactly that, q, which gives us . This is the Cooperative Detainee’s information revelation threshold. If the Cooperative Detainee believes that the probability the Interrogator is Pragmatic is above this threshold , then he has no incentive to deviate from “information” and will play the candidate strategy. We have our first perfect Bayesian equilibrium.

In this valuable information, surprise torture equilibrium, the Cooperative Detainee gives up information thinking the Interrogator will believe it’s all he has and so he won’t be tortured. But the Interrogator thinks he has more and so tortures him. In this equilibrium an Innocent Detainee is also tortured because the Interrogator believes he is simply Resistant.

When all the thresholds are calculated, there are 15 more cases to work through (17 in all) under both objective and leading questioning, but you now know the principles and steps behind solving for equilibria.¹² The rest is just tedious (but important!) slogging through the algebra, which you can find in Appendix A.

The argument thus far has shown that:

1. EITs are torture and the effectiveness of interrogational torture is an open question. (Chapter 2)

2. The Bush program approximates closely the ideal model of interrogational torture and includes limits on torture; the Bush and ideal models provide benchmarks for comparison with the game theory models to come. (Chapter 3)

3. The Bush model generates strange, quixotic outcomes. (Chapter 4)

4. The Bush interrogational torture program is more realistically modeled as objective and leading question variants of an incomplete information game, with three types of detainees, two types of interrogators, and uncertainty about the amount and value of information provided. (Chapter 5)

5. By positing a set of Detainee strategies, calculating the Interrogator’s expected utility using Bayes’ Rule to identify her best response, and checking for incentives to deviate by any of the Detainee types, it is possible to derive a perfect Bayesian equilibrium in which a Detainee is tortured after providing information. (Chapter 6)

The next step in the argument is to describe and examine the properties of this and the other equilibria from the RIT model.