Sunday 20 October 2013

How not to play Poker

Demonstrating that behavioural economists are no more rational than anyone else I committed the same mistake playing poker that I did about a year ago...


I acted irrationally by seeing different colour chips as qualitatively different as well as quantitatively different. Despite the fact that one black chip was worth five white chips, I preferred betting five white chips than one black chip because I saw the higher value black chips as in some way better. This led me to be more risk averse when I had lost my white chips and had only black chips to bet.

Because I saw equivalent amounts of money as different I acted in a non-fungible manner. I'm not blaming this violation of fungibility for my poor performance (I lost, badly) but I am surprised to see myself commit the exact same mistake that I even wrote a blog about before. Some people never learn...

Thursday 10 October 2013

Little Jimmy

The 4G auction in the UK used a highly complex mechanism called the combinatorial clock auction. Here is  a simple example of how an combinatorial auction works (using the the Vickrey-Clarke-Groves rule).

Little Jimmy has two chickens to sell. His friends are little Dave, little Fred and big Phil. Little Dave wants one chicken and is prepared to pay £5 for it. Little Fred wants one chicken and is prepared to pay £3 for it. Big Phil wants both chickens for a total of £7, but it not prepared to buy just one on its own.

How do we decide who gets the chickens? How do we create an auction where Dave, Fred and Phil have an incentive to bid their maximum valuation?


Answer: Maximise the value of the bids and make the winners pay the opportunity cost of winning. (Price = total bids of all winning bids if winner didn't bid - total bids of all other winning bids)...

We maximise the sum of the bids by having little Dave and little Fred winning one chicken each. But how much do they pay?
 
Dave: Fred is the other winner with utility £3. If Dave was removed Phil would win both with £7. (7-3=4)

Fred: Dave is the other winner with utility £5. If Fred was removed Phil would win both with £7. (7-5=2)

Therefore...

Little Dave wins one chicken and pays £4

Little Fred wins one chicken and pays £2

Big Phil wins nothing. Big Phil is sad. Big Phil is slightly less big now.

Friday 4 October 2013

The Stable Marriage Problem

How does one go about matching husbands with wives?

Is there a way to do it that will result in stable marriages? 



If we define stable marriages as the situation where no couple would prefer each other to their current partner, then the answer is yes: a stable matching does exist. We can create a situation where no new couple would be created at the cost of a current couple.

Let's say there is a finite number of singles, who all go to a dance together.

We now use something called the deferred acceptance algorithm (Gale & Shapely, 1962):

Step 1. Each man proposes to his preferred woman.

Step 2. Each woman records all the men who propose to her.

Step 3. Each woman then rejects all the men except her most preferred suitor.

Step 4. Each rejected man then proposes to his next most preferred woman.

Step 5. Repeat Step 2 onwards until all men and/or all women are matched.


Ta da! We have a stable matching. There is no new couple that could be created. If a man preferred a different woman to his partner, well, the woman would already have rejected him.

Romance versus efficiency: which will win??