# How to resolve a hostage crisis: Part II

Last week, I showed that if the villains are perfectly rational, and this is “common knowledge,” then the villains will be best off by just surrendering immediately without killing any hostages. But the rationality assumption is a big one – let’s see what happens if we drop it.

Once there’s a good chance that the villains are “crazy,” even criminals who are perfectly sane will pretend to be crazy so they can extract money as ransom. We can then model this as a Bayesian game, with N periods corresponding to the $N$ hostages that are holed up in the bank. Each period will consist of two stages: in the first, the SWAT team decides whether to pay up or not; in the second, the villains choosing whether or not to kill a hostage; if they don’t, it’s tantamount to surrender. If they run out of hostages to kill, they are forced to surrender, and the surrender outcome gets progressively worse as they kill more hostages.

The payoffs are described as such:

Each dead hostage: gives $-H$ to the SWAT team, and $-C$ to the villains if they end up surrendering.

Surrender: gives $-S$ to the villains.

Pay up: gives $-P$ to the SWAT team, P to the villains, where $H > P$ (1)

Finally, we assume that there is a probability $k$ (initially) that the villains are nuts.

We’re going to construct a mixed strategy solution to this problem, which will yield a perfect Bayesian equilibrium. That is, in a given period $i$, the SWAT team pays up with probability $p_{i}$, and the villains execute a hostage with probability $k_{i}$. For those who are unfamiliar with mixed strategies, the basic idea is that both sides are indifferent between (at least) two options, so they may as well flip a coin as to which one to do. The trick is that the coin is weighted so as to make the other side indifferent as well. Thus they will flip their coins so as to make the first side indifferent. Hence this forms a Nash equilibrium – neither side can benefit by unilaterally changing their strategy.

The derivation is a little drawn-out, so I’m going to break it into steps:

(i) If the SWAT team always pays up in some period $i$, then the villains may as well execute their hostages until they reach that stage; after all, this guarantees them an automatic victory. But if they will do that, then the SWAT team should pay up immediately (i.e. in period 1): since they’ll lose anyway, they may as well save the lives of the hostages.

(ii) Conversely, if the villains will always execute a hostage at a certain stage, then the SWAT team should always pay up then. But then we run into the same issue as in (i), so the SWAT team will end up paying at the beginning. Combined, (i) and (ii) limit the types of equilibria we have to analyze.

(iii) If there is a period i at which it is known that the SWAT team will not pay up, no matter what, then the normal (not crazy) types, knowing this, will surrender in period $i-1$. But this possibility leads to an inconsistency. Knowing that the normal types will do this, the SWAT team will believe that anyone who hasn’t surrendered must be crazy. If so, it is a best response to pay up in order to avoid more casualties.

(iv) Thus, from (iii), there cannot be a perfect Bayesian equilibrium in which the SWAT team will maintain the siege to the bitter end – they must always cave in to the villains’ demands with some positive probability. Indeed, we see from (i) and (iii) that either they pay up immediately, or they pay up in period $N$ with some nonzero likelihood.

(v) There is no benefit to the villains in executing the $N^{th}$ hostage. Thus, if they are normal, they will not execute, and so this hostage will be killed if and only if the villains are crazy, which occurs with probability $k_{N}$.

(vi) In period $N$, by (iv), the SWAT team must still be indifferent between paying up and maintaining the siege. This implies that

$-P-(N-1)H=-(N-1)H(1-k_{N})-NHk_{N}$

$k_{N}=\frac{P}{H}$

(vii) In all periods $i$, the normal-type villains must be indifferent between surrendering and killing a hostage. Thus, if their expected payoff, if the game reaches the second stage of period $i+1$, is $\Pi_{villains}^{i+1}$, then we get

$-(i-1)C-S=p_{i+1}P+(1-p_{i+1})\Pi_{villains}^{i+1}$

But we know that since the villains will be indifferent as well in the next period (unless $i=N-1$), $\Pi_{villains}^{i+1}=-iC-S$. In any case, this formula still holds for $i=N-1$, since the normal types surrender in period $N$ anyway. Thus we get, after substitution,

$p_{i+1}=\frac{C}{P+iC+S}$

As expected, this probability goes down as more hostages are executed – after all, there is less of a risk of more future casualties, since there are fewer hostages left – a dark thought, indeed.

(viii) We now derive the probability that the villains execute a hostage in the periods other than the last. The SWAT team is indifferent between maintaining the siege and paying up. Hence if their expected payoff at the beginning of period $i+1$ is $\Pi_{SWAT}^{i+1}$, then

$-(i-1)H-P=-(1-k_{i})(i-1)H+k_{i}\Pi_{SWAT}^{i+1}$

But because the SWAT team will also be indifferent between paying up and maintaining the siege in the next period, $\Pi_{SWAT}^{i+1}=-iH-P$. Hence

$-P=-k_{i}H-k_{i}P$

$k_{i}=\frac{P}{P+H}$

This value is constant! Note that it is decreasing in $H$ – if the hostage is more valuable to the SWAT team, then the likelihood that they will kill another doesn’t have to be as high in order to deter the SWAT team from maintaining the siege. Conversely, if the hostage is less valuable, the villains will have to be more likely to kill the hostage to show they mean business.

(ix) Combining (vi) and (viii) gives us the the proportion of crazy types, at the beginning of any period $i$ (after the first) in the perfect Bayesian equilibrium will be $(\frac{P}{H+P})^{N-i}\frac{P}{H}$. Thus, if at period 1, the initial probability that the villains are crazy, $k$, is greater than this amount, then the risk that they are nuts (or wannabe-nuts) is too high, and the SWAT team should fold immediately. Otherwise, they should maintain the siege in the first period, while the villains execute the hostages with the exact probability so that the likelihood that they are nuts, as of the start of period 2, is exactly $(\frac{P}{H+P})^{N-2}\frac{P}{H}$. Afterwards, the villains follow the strategy given by (vi) and (viii), while the SWAT team follows the strategy given by (vii).

(x) As a final note, we see from (ix) that as $N\rightarrow\infty$, the threshhold for the initial likelihood of craziness necessary to enforce a SWAT team payout goes to 0. This makes sense – there’s a larger potential for more casualties as the number of hostages goes up.

(1) (footnote: though dropping this assumption only slightly changes the outcome. Also, we exclude the possibility of storming the bank, since it’s similar in concept to the other options).

### 3 Comments on “How to resolve a hostage crisis: Part II”

1. macebron says:

It’s pretty amazing that you have explain on the language of mathematics, the idea of resolving a hostage crisis…good job!