I don’t mean breaking and entering. It’s New Years Eve — 2PM on New Years Eve — and after heading out for a quick lunch I return to find The Jacobs Center locked for the weekend. There is a separate electronic key to the building and I have one somewhere but I never need it so I don’t carry it around with me. So I have to stand in the cold and wait for somebody to enter or exit the building and let me in.
There are two entrances so the question is which one to stand by and wait. I wait for a while at the main entrance and then decide to try my luck at the next one on the other side of the building, about a 2 minute walk. Of course on the way I am imagining that someone must be leaving from the first entrance just as it passed out of sight. When I get to the other entrance I find that there’s just as little activity there as at the first one. After a while I give up again and go back to the first.
I have a sinking feeling as I am walking back that I am violating some basic rationality postulate to have dropped the first alternative only to switch back to it again. But it’s not hard to rationalize switching, even indefinite switching with a simple model of uncertain arrival rates.
At each entrance there is a random arrival process, say Poisson, which produces a comer or goer with some given flow rate. It’s random so even if the arrivals are frequent on average its still possible that there is a long wait just because of bad luck. Because it’s an unusual day I don’t know for sure what the arrival rates are at the two entrances so the best I can do is form a subjective distribution.
As time passes I learn only about the door I am watching and what I am learning is that the arrival rate is slower than I thought. Every moment that passes and I am still out in the cold the current door’s expected arrival rate is continuously dropping. There comes a point in time when it drops low enough that I want to switch to the other door. The expected arrival rate at the other door hasn’t changed becuase I haven’t learned anything about it. I give up and walk to the other door once the estimated rate at the current door drops far enough below that it is worth 2 minutes of walking (and no chance of getting in during that time.) In fact, this may happen before the current door’s expected arrival rate drops below that of the other door. (Due to option value. See below.)
Once at the other door I start to learn about it and I stop learning about the first door. Again, as time passes its estimated arrival rate drops while that of the first door remains constant. There is again another threshold after which I return to the first. Etc. Until I finally give up and throw a brick through the Kellogg student lounge window.
Observation: Consider the threshold at which I switch from door 1 to door 2. That is based on a comparison of the value of staying put versus the value of switching. The value of switching has built into it the option value of being able to switch back. You can see the role of this option value by considering a truncated problem where once I switch doors I am unable to switch back. Relative to that problem, the option of switching back makes me switch more frequently. Because without the option to switch back, I want to hold on to the current option until I am certain that it’s a loser before giving it up for good.
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January 3, 2011 at 5:23 am
chiao
I believe this corresponds directly to the multi armed bandit problem: http://en.m.wikipedia.org/wiki/Multi-armed_bandit
January 3, 2011 at 11:45 am
jeff
yep, with a cost of switching arms. In fact the building has four entrances. With a cost of moving from one to the other this adds a real novelty to the multi-armed bandit problem. Suppose I am at door A and currently door C, on the opposite side of the building, is the arm to switch to absent switching costs. I have three options: stop at B on the way to C, stop at D on the way to C, or walk straight to C. Any of these could be optimal depending on distributions and switching costs.
January 3, 2011 at 8:50 am
Fmb
Surely arrival rates have nonzero correlation and you learn something about the other door
January 3, 2011 at 12:52 pm
Alex
or you could post a note at each door saying “please call me when you open the door because I am in dire need of entering this building” (without sounding creepy)
January 3, 2011 at 4:39 pm
dan
so how many times did you go back and forth? 🙂 (sorry if you said and i missed it)
at some point it probably becomes optimal to stick with 1 place (or give up and go home)
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February 26, 2020 at 7:39 pm
Anonymous
you’re an idiot