Ratio of people you overtake on a bike vs. times you are overtaken and relationship to average speed This question came to me while commuting to work on my bike. 
The ratio of people I overtake vs. people who overtake me is (let's say) $1:2$ - so if on my $5$ km ride I overtake $5$ people, $10$ would overtake me.
What does that say for my average speed, assuming the general average speed is $20$ km/h?
How would that change if the ratio is $3:1$, $1:1$ or $1:10$?
I assume if the ride is longer, say $30$ km, then the error margins would decrease?
For simplicity's sake we can assume that there are no traffic lights (thereby discounting different acceleration rates) and no wind, or uphill/downhill stretches.
If you tell me the relevant statistics branch, I would happily update the tags.
 A: If everybody rode the same very long route you would catch all the slower people who started before you and be caught by all the faster people who started after you.  If we assume the distribution in speeds is independent of starting time, you would know you are faster than $1/3$ of the people and slower than $2/3$.  
Because the distance is short there are slower people in front of you that you won't catch because they get done before you catch them.  Similarly, there are faster people behind who don't catch you because you arrive before they do.  
There is a bias toward catching slower riders because they spend more time on the course.  If you ride $20$ km/h it takes $15$ minutes to finish.  A rider who goes $15$ km/h takes $20$ minutes, so you can spot them $5$ minutes and still catch them.  For a faster rider to spot you $5$ minutes they have do finish in $10$ minutes, so do $30$ km/h.  
If a slower rider starts before you at speed $v$ km/h, you will catch them if you spot them less than $\frac {300}v-15$ minutes.  If we define $\rho(v)$ as the density of riders per minute at speed $v$, the number you catch is $\int_0^{20} \rho(v)(\frac {300}v-15)\; dv$.  Similarly you are caught by $\int_{20}^\infty  \rho(v)(15-\frac {300}v)\; dv$
