# How many different place outcomes from 4 cars each running in 4 heats and how many result in ties?

Cars A-D run four heats and the 1st place winner of each heat get a point, 2nd place gets 2 pts, etc. I then sum up each car's points across all heats and the fewest points win. However, when using points and not times, ties can occur. (I'm sure ties can happen when using times but going out to the thousandths place, I doubt it.)

What's the easiest way to figure out the number of all possible race outcomes over the four heats, summing each car's points and then determining how many race outcomes results in a tie?

Thanks,

-bill

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I am assuming you mean tie for top spot. The numbers are small, break up into cases. First observe that any "top" car has at least $10$ points, and no more than $14$. – André Nicolas Jan 22 '13 at 19:52
Doesn't have to be a tie for the top spot. Any tie. – Bill Jan 22 '13 at 20:05
Ouch! Then cases gets unpleasant. – André Nicolas Jan 22 '13 at 20:07

What's the easiest way to figure out the number of all possible race outcomes over the four heats,

It depends on how you define a unique outcome.

Taking $n$ cars, for each heat, there are $n!$ possible orderings of those cars ($n!= 24$ for $n=4$).

Taking $n$ heats, defining a unique outcome based on an ordered list of results for each heat (i.e., caring about the ordering of results) where the specific cars matter (i.e., counting outcomes that swap cars as distinct), there are $n!^n$ unique outcomes ($n!^n = 317,776$ for $n = 4$).

If the order of results for a specific heat doesn't matter, but the specific cars do, the number of unique outcomes is given as $\binom{(n!+n-1)!}{n!(n!-n)!}$, combinations of $n$ from $n!$ allowing repetitions (this equals 17,550 for $n = 4$).

If the specific cars don't matter, you can divide the outcomes for either formula above by $n!$, which is the number of ways to relabel the cars.

and then determining how many race outcomes results in a tie?

Assuming the cars are evenly matched, and that each car had an equal chance of any position in every heat, to estimate the race outcomes resulting in a tie, you could compute the probability distribution to figure out the relative frequency of final scores based on the number of ways those scores can be generated as a sum of four random digits from one-to-four. This creates a discrete probability distribution of final scores around the mean ($2.5 \times 4$), which looks like this:

To find the probabilities of each final score, you just need to divide by the total number of outcomes: 256. Thus, for example, the probability of a car getting a score of 10 is p=$\frac{44}{256}$. Taking four cars, we can then estimate there to be a probability of about 0.14 that at least two or more cars will finish with the score of 10. When completed for each possible output score, you'll have an estimate of how often ties occur at individual scores.

Unfortunately, the above method doesn't consider that the final scores of cars are dependant on each other ... for example, two people cannot be tied at 16, or at 1. It should, however, provide a reasonable estimate.

...

Anyways, taking my tattered math hat off and putting on my shiny programmer hat, if we take the model of there being $n!^n$ unique outcomes ($317,776$ for $n=4$), then precisely $169,272$ have ties (51%). I just used some code to enumerate all outcomes and count those with ties. :)