Time to Broaden our Minds with The Dark Knight

Gentlemen, lets broaden our minds So I had an article all planned out that was going to be along similar lines to my last one where we could try to explore some of the mythological roots in modern comic books. Then I went to see the Dark Knight.

I know, I know, I’m months behind everyone else – but thats because I hate going to the cinema and having to sit amongst the great unwashed as they eat nosily and check their cellphones for new texts.  However I decided to make an exception and wow, just….wow. What a great film.  So many cool things it’s hard to know what to mention first.  Heath Ledger’s Joker, Christian Bale’s sore throat, the ludicrous use of cellphones, the mathematics… Mathematics?  Hell yes!  In fact there are two major parts of the movie where the plot has been lifted, almost without any modification, from very well known mathematical problems.  They both come from Game Theory.  Anyone who has seen ‘A Beautiful Mind’ will be familiar with game theory since Russell Crowe played John Nash, the father of the field.  In this article I’ll take a detailed look at one of the scenes and then, depending on the response, I might take on the other one. Prisoner’s Dilemma Towards the end of the movie (spoiler alert) the Joker plant’s explosives on two vessels out in Gotham Harbour, lets call them Ship A and Ship B.  One is filled with convicts and the other just ordinary people.  The occupants of ship A are given the controls that will blow up ship B and vice versa.  They are told that in a certain amount of time (let’s say 100 minutes to make the analysis easier) BOTH ships explode.  They are unable to contact the other vessel and confer.  So the question is, what do you do?

This is almost identical to the classic game theory problem called Prisoner’s Dilemma, which says this: Two suspects are arrested by the police. The police have insufficient evidence for a conviction, and, having separated both prisoners, visit each of them to offer the same deal. If one testifies (“defects”) for the prosecution against the other and the other remains silent, the betrayer goes free and the silent accomplice receives the full 10-year sentence. If both remain silent, both prisoners are sentenced to only six months in jail for a minor charge. If each betrays the other, each receives a five-year sentence. Each prisoner must choose to betray the other or to remain silent. Each one is assured that the other would not know about the betrayal before the end of the investigation. How should the prisoners act? I’ve copied this statement of the problem from Wikipedia, but anywhere you look will give you a similar definition.  Despite being very simple to state the problem is deceptively complex, especially when played multiple times.  In so called iterated prisoners dilemma the game is played over and over and the aim is to minimise your total jail time. Sound familiar?  Well placed in a different context it’s precisely the problem that the ship passengers face at the end. So as crazy as this might sound is there an optimum strategy if you were to find yourself on one of the boats? What cost death? So in order to study this further we need to put some sort of a cost on pressing the button and the cost of dying.  Let’s first say that the cost of dying is ‘100’.  A ‘100’ what doesn’t really matter we just want a figure that expresses the relative cost of dying again the relative cost of pressing the button. Well on first reflection it seems that the cost of pressing the button and blowing up the other ship is zero, so you might as well press it immediately.  Well if you think that then, frankly, you’re a damn dirty psychopath.  Obviously there’s a cost to pressing the button, you have to live with the guilt and shame of killing several hundred people so that you can live, there’s the social consequences of your friends and family knowing what you did, and there’s also possibly legal and criminal implications. It’s reasonable to assume though that the cost of pressing the button diminishes over time.  At the start of the trial there’s still plenty of time until the end, you might get rescued, you might escape – any number of things can happen – so the cost of pressing the button should be fairly high (though still less than the cost of getting blown up).  Towards the end the cost should be much lower.  The chances of being rescued have started to look very slim, if nobody pressed the button then you’re about to get blown up anyway, so it seems sensible that the cost associated with pressing the button at the end should be correspondingly lower. Let’s assume the cost diminishes over time in just a straight line.  So if we say the cost of pressing the button immediately is 60 then halfway through the trial the cost is 30 three-quarters of the way through the cost is 15 etc. So what we want to do is minimise the cost to ourselves, by picking an optimum time to press the button.  Do it too soon and the cost is very high.  The longer we wait the lower the cost of pressing the button but the higher the chance we ourselves will get blown up (and thus pay the ultimate price). So the question is now what is the optimum strategy?  We can expect it to be dependent on how much value we put on our lives compared to the lives of others.  So lets look at some results.  Every graph that follows is based upon ten million simulations of the dilemma and the results presented are the average costs of pressing the button at each respective minute. Initial cost of pressing the button is 50 So this is for you if you think the initial cost to you of pressing the button immediately is half as much as if you were to be killed.  The average cost of pressing the button at each minute is as follows figure1 This shows that the cost always goes up the longer you wait.  What this means is that if you think the consequences of pressing the button are half as bad (or less) than the consequence of dying then you’re better off not waiting (and hence risking that the other boat might press first).  So the moment you get your hands on that button you should press it without hesitation (you heartless bastard). Initial cost of pressing the button is 75 Now things start to get a little more interesting.  If you think that the cost to yourself of pressing the button straight away is 3/4 as bad as actually dying then there’s an optimum time to press the button that isn’t right at the very start or the very end of the trial.  Take a look at this figure: figure2 So now the optimum strategy for you to pick is pressing the button about 30 minutes after getting hold of that big red button.  Over a sufficient number of attempts this is long enough for you to convince yourself you held out long enough so you can assuage your guilt, but it’s still fast enough to make it relatively unlikely that you yourself will get blown up. Finally, what if we’re so charitable that we think the cost of pressing the button and killing everyone straight away is as bad for us as actually dying. Initial cost of pressing the button is 100 So say you’re very level-headed about the whole thing and you’re used to putting everyone else first so pressing the button straight away is just as bad as being killed.  Well this is what the cost graph looks like: figure3 There’s still an optimum point to press the button, and this time it turns out that it’s exactly halfway through the Joker’s dastardly scheme.  What’s interesting about this is that even if you value the lives of those in the other vessel as much as yours then there’s no incentive for you to wait more than halfway into the trial before pressing the button and blowing up the other ship. Pop quiz, hotshot So what would you do?  Press the button straight away?  Not press the button at all?  However much you value others compared to yourself at least having read this you’re not going to have to sit down with a pen and paper and work it all out! Yarrrrrr

So let me know what you think of the article, and particularly what you yourself would do.  That way I know who to trust when we find ourselves in this exact situation at Bristol Comicon.  It will also give me an indication of whether to discuss the other mathematical problem in The Dark Knight (it’s all about pirates!) -Scott (who would press the button straight away…just out of boredom). – Scott Grandison E-mail: scott (AT) comicbookoutsiders.com Scott, along with co-host Stephen Aryan presents the Comic Book Outsiders podcast, part of the UK Comic Book Podcast Group.  Check out their podcast and blog at: http://www.comicbookoutsiders.com Also available through iTunes, Podcast Alley and other podcast indexing sites.

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One comment

  1. comicbookoutsiders /

    There’s some severe formatting problems in this post. I’ll work with Barry to get it sorted as soon as he comes back from holiday. Sorry about that!

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