## Saturday, 4 August 2012

### Road cycling and an explanation for cooperation.

I'm a huge Tour de France fan. I would probably follow more road cycling if I could but I just concentrate on the Tour de France which turns up once a year. First of all it is a ridiculous feat of endurance. Secondly (and this holds for road cycling in general) there's so much game theory involved.

The biggest element of this is the effect wind resistance has on cyclists. I heard Chris Boardman commenting on some cycling recently that 80% of a cyclists effort is expended on moving air. There's obviously a lot of research that goes in to this and I found an interesting paper on the subject when planning on writing this blog:

In general road races have many more than just two cyclists but I'm only going to talk about two cyclists for now. This is for two reasons:
1. Firstly this occurs fairly often. At some points in road races groups of cyclists "breakaway" from the main field and have to work together to stay away.
2. Secondly: the math is nicer with 2 "players".
Let's assume that two cyclists have broken away. Available to each cyclist is two strategies: {Lead, Follow}.  We make the following assumption:
• If both cyclists Lead then it is assumed that they basically work alone (they both suffer the effects of air drag but make the speed of the slowest cyclist).
• If both cyclist Follow then it is assumed that they don't work at all and are caught by the peleton (their escape is basically over).
• If one cyclist Leads and the other Follows then they go at the speed of the Leader and the Leader fatigues.
We'll now write out some utilities that correspond to the above assumptions.
• Let  $v_1,\; v_2$ be the speed for each cyclist.
• Let $f_1,\;f_2$ be the fatigue felt by each cyclist when suffering the effects of air drag.

We assume the following utilities:
• If both cyclists Lead then cyclist $i$ receives utility:
$$\min(v_1,v_2)-f_i$$
$$0$$
• If one cyclist Leads and the other Follows then the Leader (w.l.o.g. let it be cyclist $1$) receives utility:
$$v_1-f_i$$

$$v_1$$

This can all be represented in the following bi-matrix:

$$\begin{pmatrix}(\min(v_1,v_2)-f_1,\min(v_1,v_2)-f_2)&(v_1-f_1,v_1)\\(v_2,v_2-f_2)&(0,0)\end{pmatrix}$$

If we assume for now that $v_1=v_2=2$ and $f_1=f_2=1$ (we'll not get our hands messy with units) than our bi-matrix becomes:

$$\begin{pmatrix}(1,1)&(1,2)\\(2,1)&(0,0)\end{pmatrix}$$

Let $p$ be the probability with which cyclist 1 Leads and $q$ be the probability with which cyclist 2 Leads. It can be shown (using basic game theory and/or using this Sage interact (http://interact.sagemath.org/node/49)) that there are 3 equilibria for this game:
• $(p,q)=(1,0)$ (Cyclist 1 always Leads)
• $(p,q)=(0,1)$ (Cyclist 2 always Leads)
• $(p,q))=(.5,.5)$ (Both cyclists Lead)
Using 1 of the pure strategies ($(1,0)$ or $(0,1)$) the utility to the Leader is 1 and the utility to the Follower is 2. Using the mixed strategy the expected utility to both cyclists is 1. Let us assume for the rest of this that cyclists use mixed strategies (i.e. neither takes advantage of the other).

In reality cyclists that break away cooperate, they talk and share the load as if they were a team. The mixed strategy above does not correspond to cooperation. It corresponds to a random strategy during which the cyclists will be randomly Leading and Following. Cooperation would be that both cyclists equally share the load (in a coordinated manner), the expected utility to both would then be 1.5 (both cyclists are better off if they cooperate - this is a notion quite close to my research, some screencasts of my talks are available here).

We'll continue to consider the situation in terms of a non-cooperative game as it gives an insight as to why cyclists cooperate.

We now assume that the cyclists are different. Perhaps they have escaped on a mountain stage and one of them is a better mountain rider. Let's take a look at the probabilities of Leading in a mixed strategy equilibria for varying $v_1$ (i.e. how fast the 1st cyclist is):

Basic game theoretical arguments give:

$$p=\begin{cases}1/2,&\text{ if }v_1\leq 2\\1/v_1,&\text{otherwise}\end{cases}$$

and

$$q=\begin{cases}1/2,&\text{ if }v_1\leq 2\\1-1/v_1,&\text{otherwise}\end{cases}$$

We see that as soon as $v_1$ is larger than $v_2=2$ the first cyclist has less of an incentive to lead. He/she basically "brings more" to the 2 cyclists and so the second cyclist must make more of an effort.

This further emphasises the need for cyclists to work together when they breakaway.

In reality cyclists would cooperate and the cyclist capable of helping the group stay away would perhaps work more.

A good example of a breakaway was the 2012 Olympic men's road race. The British team has one of the best sprinters of all time: Mark Cavendish. In road cycling, a sprinter is someone who is very good at going very fast over a shirt distance. Mark Cavendish is so good (he has recently been voted the best ever Tour de France sprinter) that no other team wants to be in a bunch finish with him: they'd lose. During this years race, the other teams all tried to put riders in breakaways. Normally other teams left in the peleton would at some point cooperate to accelerate the pace and bring the breakaway back in. In this particular occurrence no other team was good enough to beat Cavendish in a sprint and so no other team wanted to help (i.e. share time riding at the front) team GB. This is a slight over simplification as the Germans have a good sprinter who could have perhaps had a chance but it's not far off the truth. From a game theoretical point of view this make a lot of sense and in essence is exactly what happened. 3 riders formed a final breakaway and they contested the finish (Alexander Vinokourov won gold).

There's so much game theory in road cycling it really is a great sport to watch.

I've written a similar post applying game theory to rugby. For other recent blog posts with an Olympic flavour take a look at Paul Rubin's blog and Laura McLay's blog.