## Thursday, November 01, 2007

### Expected Loss = E[(a)(Tricks)**2 + (1-a)(Leftover Candy)**2]

The asymmetric risk inherent in Halloween is unhealthy. Which is more costly in expected value terms, running out of candy and having to turn out the lights and pretend you aren't home to try to avoid getting tricked, or having too much candy on hand and having to eat the leftovers? (This is a priori, before you are sick from eating the candy, and it also assumes you already have an ideal amount of candy on hand, i.e. the term leftovers refers to deviations from this optimum. It also assumes that bringing the leftovers to work and dumping them on others won't be appreciated, even if the candy does disappear rather quickly.) For me, the answer is clear, the loss from tricks is weighed more heavily than the loss from leftover treats. I think I even subconsciously overestimate the size of the safety cushion, i.e. the value of the parameter a is near one.

As the night went on and it was clear I had way too much candy, I tried to solve the problem by increasing the number of items each trick or treater got. But not enough so there was any chance of stocking out. If I could have determined who the last trick or treater was, they would have done very, very well. But by increasing the payoff as the night wears on, I have to worry that each year thereafter the last trick or treater will arrive later and later as they game each other to increase the payoff per house visited, so maybe that isn't such a good idea after all.

I do know this. Right now the incremental value of one more 'Fun Size' Snickers bar is less than zero, though of course the value will reset itself in due time. I might be able to stand one more bag of peanut M&Ms though.

Posted by on Thursday, November 1, 2007 at 12:33 AM in Economics | Permalink  TrackBack (1)  Comments (12)