Is there an unbeatable Cross-Country score?

Dumb Foreigner Question......

What are the rules, and do all XC races in the USA work the same way?

I'm working on formalising all the common systems.... There are a TON of variants out there

- n to score, add up the raw positions.

- n to score, add up the scoring positions after eliminating guests and non-scorers

- n to score, total the times (e.g. sum of best 3 times is a team at London Marathon, or Euro Track 10,000)

- invert the positions, high points wins (easier on incomplete teams in junior races, 3 good guys can beat a team of 4 worse ones)

- age adjust the times (equivalent to age graded rankings), n to score

- mob match rules, various systems so a bigger team has an advantage

Anyway, in the good old Surrey League with 10 per team to score, I think my old club Boxhill has the record when we closed 10 in 23 (out of a 200-person race with 10 clubs...)

Thanks!

Top 5 score. Add up the finish places. Low score wins.

For championship races, they will throw out the non-scoring runners. So 3rd place could be 1 point if the first two runners were non-scoring.

Lowest possible score is 15: 1,2,3,4,5.

Skeeple wrote:

math nerd wrote:

Now that this has been solved, what is the best score that guarantees last place?

My guess is (Number of teams * 35) - 40

Or what is the worst score that can still win?

My guess is (Number of teams * 14) + 1

This would be a more math intensive question. But lets go of off NCAA DI championships- 31 teams are listed on the results from 2017 so for the TEAM competition with individuals thrown out, thats 217 places, 155 of which are counted (top 5).

What is the worst possible score that could win? Without putting too much thought into it..................I'm drawing a blank, I have no idea how to do this and have to get back to work

Worst possible score that "could" win is hard to calculate because of the #6 and #7 runners factor. Higher than 389 points could win it.

Theoretically, the best potential score that could still get last place is 391 points. I have not fully checked it the math on it however. 31 teams * 5 scoring runners with no teams' #6 & #7 displacing yields 155 places. The summation of 1-155 is 12090. That divided by 31 is 390. Take one place from 390 and make the winner have 389, 2nd thru 30th has 390, and 31st has 391. This is the answer to a different non-asked question

To answer your question specifically, there are 13 possible unbeatable scores; 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, or 27.

Hard to believe some of your responses!

...and the next two on your team can displace another team's scoring runners. So if your team's 6th runner displaces the other team's 5th runner than a point is added to the 5th runner of the other team.

malmo wrote:

What the hell does 1-4 plus 24 have to do with anything? It's not relevant to the question being ssked.

It is the scenario in which:

1. The 2nd place team performs the best it possibly can, given the placing of your team (all 7 runners finishing 5-11)

And

2. You score the highest amount of points possible without losing.

You cannot deny that there is a scenario where your team can score 34 points, and is still unbeatable. Which is my point.

Jsksoekcnjdoamx wrote:

[quote]malmo wrote:

You cannot deny that there is a scenario where your team can score 34 points, and is still unbeatable. Which is my point.

You also cannot deny that there is a scenario where your team can score 100 points and is still unbeatable. That scenario would be if all the other teams scored at least 101 points. FFS, let the 34 points notion rest

I don't know if this helps, but I found a way to prove that a score of 375 can win with a score 405 being last place. For this to happen, all the top 5 runners from the 31 teams would have to take the top 155 spots, with no one's 6th or 7th runner ahead of another teams 5th. The simplest way for the math to work is to have the sum of every team's 1st, 2nd, 4th, and 5th runner equal to 312. Like this:

Team 1: 1, 2, 154, 155

Team 2: 4, 5, 152, 153

...

Team 31: 61, 62, 94, 95

So places 1 through 62 and 94 through 155 are covered by all teams' 1, 2, 4, and 5, runners. This leaves places 63 through 93 for everyone's 3rd runner. You assign each team a number 63-93 for their 3rd runner (no repeats obviously) and you'll get this:

Team 1: 1, 2, 63, 154, 155 = 375

Team 2: 4, 5, 64, 152, 153 = 376

...

Team 31: 61, 62, 93, 94, 95 = 405

Since all teams have the same score between their 1st, 2nd, 4th, and 5th runners, whichever team has the best 3rd runner will win.

So it is possible for a team to score 375 and still win. It may be possible to prove that a score higher than that can still win, but I don't know how to begin solving that.

I messed up my Team 2 example, the top 2 for Team 2 should be 3rd and 4th place, not 4th and 5th place. The sum is still correct though.

You could "reverse" my example and have Team 31 win and Team 1 lose by reversing the distribution of 63-93 for the 3rd runners:

Team 31: 61, 62, 63, 94, 95 = 375

Team 30: 59, 60, 64, 96, 97 = 376

...

Team 1: 1, 2, 93, 154, 155 = 405

So Team 31 could win even though they have worst 1st and 2nd runners. Of course, they also have the best 3rd, 4th, and 5th runners.

Sorry for so many posts, but my "reversed" example made me realize something. If a team has the best 3rd, 4th, and 5th runners, it is impossible for them to lose.

Conversely, if a team has the worst 3rd, 4th, and 5th runners, it is impossible for them to beat any of the other teams.

Just shows how important depth is in cross country.

math nerd wrote:

Citizen Runner wrote:

The worst case would be all members of the two worst scoring teams bringing up the rear. This makes the competition for last essentially a dual meet between the last two places.

The first finisher in this "dual meet" would score after the runner in place (#teams - 2) * (5+2) where so we can just add that value to each runner in the "dual meet". The (5+2) represents the 5 scorers and 2 pushers for all but the last two teams.

Thus a team scoring 5*(#teams -2) *7 + 28 would always be last.

I had the same logic except I added 30 instead of 28.

Then reduced: 5*(#teams -2) *7 +30 = 35*(#teams -2) + 30 = 35*#teams -70 +30 = 35*#teams -40

I think that you've got it right. Considering pushers in the "duel meet" for last place, a score of 29 to 30 seems to be the limit with team A = [1,2,3,11,12] = 29 and team B = [4,5,6,7,8,(9),(10)] or team A = [3,5,6,7,8,(9),(10)] = 29 and team B = [1,2,4,11,12].

This thread is fantastic for multiple reasons.

snow blow wrote:

Sorry for so many posts, but my "reversed" example made me realize something. If a team has the best 3rd, 4th, and 5th runners, it is impossible for them to lose.

Conversely, if a team has the worst 3rd, 4th, and 5th runners, it is impossible for them to beat any of the other teams.

Just shows how important depth is in cross country.

Actually I just realized that's not quite true. To guarantee a win the team would have to have the best 3rd, 4th, 5th runners AND their 5th runner would have to be ahead of all other teams' 4th runner.

ThatAverageRunner wrote:

This thread is fantastic for multiple reasons.

Its fast becoming as stupid as the "airplane on the treadmill" and the "how long is lane(x)" threads. But not as stupid as the "flat earth" and "faked moon landing" threads.

Non-scoring runners defined as runners not on a team or a team with less than 5 runners.

To answer your question specifically, there are 13 possible unbeatable scores; 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, or 27.

Hard to believe some of your responses!

_________________________________________________________________________________________________________________________________

This has been the correct answer already given numerous times. The question was: What is the HIGHEST LOW SCORE that can not be beaten? The answer is 27.

What about T&F scoring? In High School you can finish last in your league meet and still win the State Meet. You have the two fastest 100-200 meter runners in the state, but not another athlete on your team. You'll never win a league meet, but you'll score 36 points in your State Meet and probably win.

old guy 72 wrote:

What about T&F scoring? In High School you can finish last in your league meet and still win the State Meet. You have the two fastest 100-200 meter runners in the state, but not another athlete on your team. You'll never win a league meet, but you'll score 36 points in your State Meet and probably win.

You could win the whole State Meet with one really good sprinter/jumper. Possible combinations:

100m, 200m, 400m, Long Jump

100m, 200m, 400m, 110m Hurdles

100m, 200m, 110m Hurdles, 300m Hurdles

100m, 200m, Long Jump, Triple Jump

There are more combos but you get the idea.

old guy 72 wrote:

What about T&F scoring? In High School you can finish last in your league meet and still win the State Meet. You have the two fastest 100-200 meter runners in the state, but not another athlete on your team. You'll never win a league meet, but you'll score 36 points in your State Meet and probably win.

I'd guess those 4 runners could run a 4x100 fast enough to make the final and score some more points.

I'm 72 too ---yards on cinders for all my (unimpressive) PRs