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Uncertainty Over Time (Lords) November 25, 2013

Posted by tomflesher in Micro.
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The previous post introduced a problem that arose in the Doctor Who 50th Anniversary special in which uncertainty was a core element. To explore that problem in depth, we’ll need some tools to work with uncertainty. Also, that problem was a little grim, so let’s take the edge off.

Let’s say that a friend of mine, Matt, is visiting and I need to order food. Matt’s favorite food is fish custard, so it makes sense that I should order him a dish of fish custard. Simple enough. Of course, if my other friend Peter were coming over, and I knew Peter didn’t like fish custard and much preferred haggis, it wouldn’t make sense for me to order him fish custard. Obviously I’d order haggis for him. To make this work mathematically, let’s say that each friend of mine likes his preferred dish about the same, so we could say that Matt gets utility of uM(fish custard) = 1 from eating fish custard and uM(haggis) = 0 from eating haggis, and Peter gets the reverse – uP(fish custard) = 0 and uP(haggis) = 1. If I want to maximize my friend’s utility, then I should do so by buying each friend his favorite dish. But what if I don’t know who’s coming over?

Intuitively, it makes sense – if Matt’s more likely to come over, I should order fish custard. If Peter’s probably on the way, haggis should go on the menu. If they’re equally likely, flip a coin. More mathematically, if the probability that Matt is coming is πM and the probability that Peter is coming is πP (and πM + πP = 1), the expected utility of my guest when I order fish custard is

E[u(fish custard)] = πMuM(fish custard) + πPuP(fish custard),

which reduces to

πMuM(fish custard) + πP(0).

Since uM(fish custard) = 1, then the expected utility of my guest from fish custard is just πM. For haggis, of course, the same logic applies – Matt gets 0 utility, so we can ignore him, and the expected utility ends up being πP. So, whichever friend of mine is more likely to show up should get his favorite dish ordered, and if I don’t know who’s coming, I might as well just draw names from a hat.

This gets a little bit more complicated if both of my friends like each dish. So, let’s consider what happens if uM(fish custard) = 1 and uM(haggis) = 0, but if uP(fish custard) = 0.5 and uP(haggis) = 1. Then, our calculation for the expected u(haggis) stays the same (it’s still πp) but now our E[u(fish custard) ends up being

πMuM(fish custard) + πPuP(fish custard) = πM*(1) + πP*(.5)

So now, in order to figure out what dish to order, we need to know what the probabilities are! If it’s fifty-fifty, then E[u(haggis)] is .5, but E[u(fish custard)] is .5 + .5*.5 = .75! In that case, we should order fish custard, even though the guys are equally likely to show up, since Peter likes fish custard a little bit, too. We don’t get to flip a coin (mathematically, we don’t reach our indifference point) until

πMuM(fish custard) + πPuP(fish custard) = πMuM(haggis) + πPuP(haggis)

or

πM(1) + πP(.5) = πM(0) + πP(1) => πM + πP(.5) = πP => πM = .5*πP

That means we don’t get to the indifference point until Matt is twice as likely to arrive as Peter! Funny how these probabilities influence the choices we’ll make. Sometimes things get a bit more complicated, but that’s a post for another time.

… lord.

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