Looking for theories or formula to help express this game/competition phenomenon. I am Looking for theories or formula to help express this game/competition phenomenon.
Thank you in advance!
Note to mods: Question marks are rhetorical, this is not a homework problem.
My idea is that significant advantage of one class of competitor over another class of non-advantaged competitor will result (in a large enough size competition with a set of rules) in the top performers of that competition being comprised exclusively/almost exclusively of advantaged competitors.
The larger the advantage, the greater the certainty that the top performing group will be comprised almost entirely of the advantaged class of competitor. Assuming all other factors are about equal. There probably is a threshold number for advantage, where the effect of the advantage becomes very pronounced, and is not likely to be significantly impacted by highly improbable extreme variances/anomalies.
For example, if we have a 5K marathon race with 5,000 contestants, and 4,500 contestants are equipped with running shoes, and 500 contestants are equipped with motorcycles.  The running shoe class can travel at most about 6mph (sorry) sustained for 5K, and the motorcycle class can travel at about 90mph sustained over the 5K distance.
What is the probability of a running shoe class contestant making it in to the top 500 fastest finishers or let alone take the first place time? Scenarios such as a running shoe contestant’s commandeering a motorcycle, etc are not permitted.
The only luck variable being that a motorcyclist crashes and is incapacitated.  But what is the probability of all 500 motorcyclists becoming incapacitated before crossing the finish line, and a natural runner takes first place?  I do not think there is a way to luck your way into first place in this competition scenario.
The effect of the significant advantage of the motorcycle class over the running shoe class, will result in the top 500 fastest finishers being comprised almost entirely, if not entirely, of motorcycle class advantaged competitors.
Maybe this effect of advantage is obvious.  If so, than what theories, or formula can you suggest that I can use to express what I have outlined above.  Thank you very much.
 A: If a motorcyclist that does crash is guaranteed to finish before any runner (reasonable assumption), and each motorcyclist has an equal and independent (one crash doesn't affect any other racer) probability of crashing, $p$, then the probability of any runner making it into the top 500 by at least on cyclist crashing is $$1-(1-p)^{500}$$
To see where we get this, consider the probability that none of the cyclist crash. Each has probability $p$ of crashing, so $1-p$ for not crashing. The probability of multiple independent events happening is the product of the individual probabilities of those events. So, the probability of 500 cyclists not crashing is the product of $(1-p)$ 500 times, or $(1-p)^{500}$. And the probability of that not happening, or put differently, at least one crashing, is $1-(1-p)^{500}$
The chance of a runner finishing first is a bit simpler. In order for this to happen, every cyclist must crash. So if each has probability $p$ of crashing, the chance is $p^{500}$.
For something in between, the formula is a little more complicated. For example, if we ask for the probability that a runner gets 500th place, that would require that 1 cyclist crashes. But we don't care which cyclist crashes, only that there is exactly 1 crash. So, there are 500 possible cyclist that could crash, and the probability that one of them crashes and no one else does is $p\times(1-p)^{499}$ for each that could crash. So there are 500 mutually exclusive outcomes with that probability, one for each cyclist that could crash, and the final probability is
$$500\times p\times(1-p)^{499}$$
In general, the formula for the chance that exactly $n$ crash is ${500 \choose n}\times p^{n}\times(1-p)^{500-n}$ where ${x \choose n}$ is the number of combinations of $n$ items from a set of $x$. More generally, this is the  binomial distribution.
