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There are a few complications in the allocation of Formula 1 points through the season, such as half points for an incomplete race, and the fact that a team can change drivers mid-way through a season.

However, if we assume the same 20 drivers all start the 19 races; that at least ten drivers finish each race; that all races are completed; and points are awarded thus:

  • 1st : 25 points 2nd : 18 points 3rd : 15 points 4th : 12 points 5th : 10 points 6th : 8 points 7th : 6 points 8th : 4 points 9th : 2 points 10th : 1 point

... what is the minimum theoretical number of races that have to be run before we can say that an individual could be champion by that point. That is, that their points lead is unassailable.

The answer (I think) must be at least ten: until you've won half, there's no way you can guarantee being champion. But (given our constraints) other drivers must have been awarded some points by that point.

It's at this point that I'm at a loss to work out what permutation of results minimises the points for second and below, and hence whether winning the championship in ten (or eleven, or twelve) races is a mathematical possibility.

Party Ark
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  • Do we assume that the other competitors competed in the most convenient way possible, or that our solution must work if there is a very competitive runner up? – djechlin Nov 24 '15 at 00:44
  • The most convenient way. If there's a very competitive runner-up, then you're not going to know the champ till the very last race (which is what fans would like to see). But I'm asking the opposite - what's the most un-fan-friendly possibility, so that the season is wrapped up at the very earliest opportunity. – Party Ark Nov 24 '15 at 00:47
  • If you win the first $15$, you will be first, and winning the first $14$ is in principle not enough. – André Nicolas Nov 24 '15 at 00:49
  • I don't think 10 is enough. 10 races could put the leader at 250 pts, leaving 760 divied up among the other 19 racers. The lower limit for 2nd place would be $760/19=40$pts. 9 more races offers 225 pts for a total of 265 for the chaser. The original leader could be out of the top 10 each time and still be stuck on 250. I do think 11 wins could lead to a clinch. That's 275 pts. Perfect distribution of the other $76*11=836$ pts puts 2nd place at 44pts. Eight more races would not be enough to catch 275. – turkeyhundt Nov 24 '15 at 00:49
  • @AndréNicolas yes, that's true, but the questions is, could it be fewer? How would you organise the other cars finish places for the fastest possible way to guarantee you're champion? – Party Ark Nov 24 '15 at 00:56
  • @AndréNicolas The question can be interpreted two ways, I think your answer covered perfectly one interpretation, while my answer turkey's cover the other interpretation. I would have not deleted your answer, as it would be good to cover both meanings :) – N. S. Nov 24 '15 at 01:04

7 Answers7

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Andre covered the guaranteed case no matter what. if you are asking for the best case scenario (i.e. if the results in the first races are ideal for you) the answer is 11.

After 10 there cannot be any winner no matter what The most a driver can have after 10 is 250.

The remaining drivers win on average 4 points per race (76 points for places 2-10, divided by 19 racers). This means that one average a driver will have 40 points after 10 races. As a racer can get another 225 points in the remaining 9 races, 250 are not enough.

After 11 there can be an winner under ideal circumstances If a driver wins the first 11 they have 275 points.

If every other driver has under 74 points, they can get at most 274 points.

As the average of the points obtained by the 19 remaining drivers for the first 11 races is 44, is very easy to distribute the places 2-10 in the first 11 races in such a way that no one has over 74 points. If you want an actual example, I can create such a list.

The list

  • Driver 1: 11 first place finishes, 275 points
  • Drivers 2-3: 4 second place finishes, no other point, 72 points.
  • Driver 4: 3 second place finishes, no other point, 54 points.
  • Drivers 5-6: 4 third place finishes, no other point, 60 points.
  • Driver 7: 3 third place finishes, no other point, 45 points.
  • Driver 8: 6 fourth place finishes, no other point, 72 points.
  • Driver 9: 5 fourth place finishes, no other point, 60 points.
  • Driver 10: 6 fifth place finishes, no other point, 60 points.
  • Driver 11: 5 fifth place finishes, no other point, 50 points.
  • Driver 12: 6 sixth place finishes, no other point, 48 points.
  • Driver 13: 5 sixth place finishes, no other point, 40 points.
  • Driver 14: 11 seventh place finishes, no other point, 66 points.
  • Driver 15: 11 eight place finishes, no other point, 44 points.
  • Driver 16: 11 ninth place finishes, no other point, 22 points.
  • Driver 17: 11 tenth place finishes, no other point, 11 points.

The list if far from ideal, if you want one ideal, you just move points from the last drivers to the ones at the top to make as many drivers as possible with 74 points.

N. S.
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  • No list needed, I think you've explained it very well :) – Party Ark Nov 24 '15 at 01:07
  • Following on, another way of saying what you've said: the least possible points a driver could have and be in second place after 11 races is less than or equal to 74; hence first place at that point could be guaranteed champ. Intuitively that seems right, but is there a way to demonstrate this without listing each race? – Party Ark Nov 24 '15 at 01:31
  • @PartyArk Yes. You can have 3 drivers getting 4,4 respectively 3 second spots, and no points in the remaining races. You can have another 3 drivers getting 4,4 respectively 3 third spots, and no points in the remaining races. 2 more drivers, each getting 6 fourth spots, and nothing else. 2 more drivers, each getting 6 fifth spots, and nothing else. 2 more drivers, each getting 6 sixth spots, and nothing else. Four more drivers always finishing 7th, 8th, 9th respectively 10th. That's 16 additional drivers. – N. S. Nov 24 '15 at 01:41
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I believe the earliest a season could ever be clinched is after 11 races.

11 wins gives the leader 275 pts. There are 836 remaining pts from those first 11 races to be distributed among the other 19 drivers, so there is a lower limit of 44 pts for the next best cumulative score. 8 more races do not provide enough pts for 44 to catch 275.

Similarly, 10 wins gives the leader 250 pts and the lower limit for next in line would be 40. 9 more races offers a maximum of 225 pts for the chaser, for a total of 265. Enough to pass the original leader if he or she was stuck on 250. So 10 races isn't enough to create an insurmountable lead.

turkeyhundt
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  • This sounds wrong or at least incomplete... the contender gets 25 points a game thereafter, so is short 50 points from the winner. you need to prove that there is no contender in the first 11 games who has amassed 50 points. – djechlin Nov 24 '15 at 01:00
  • @djechlin i am probably interpreting it wrong then. I was thinking OP meant, if we could make the perfect case, what is the earliest the season could be clinched. – turkeyhundt Nov 24 '15 at 01:03
  • That is my intention, yes: the perfect case. The very fastest way you could win, and the very fastest way to ensure everyone lost interest in the championship. – Party Ark Nov 24 '15 at 01:05
  • Then I believe 11 is the answer. @N.S.'s answer is a little easier to follow than mine. – turkeyhundt Nov 24 '15 at 01:05
  • I think that what djechlin is saying is that you should emphasize why in the first case you can get a scenario where nobody has 50 points (the actual number should 75) . This is actually very easy to do...+1 – N. S. Nov 24 '15 at 01:06
  • @djechlin There are 19 races in total, so the drivers only can get 200 points in the remaining races. This means that you only need to make sure noone gets 75 points in the first 11 races, this is easy. Even with 50 you can find a scenario... – N. S. Nov 24 '15 at 01:08
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If you win the first $15$, then you have $375$ points. The person in second place has at most $270$, and can get at most $100$ points in the last four races, so cannot catch up.

A similar calculation shows that winning the first $14$ is not enough to guarantee first place. For that produces $350$ points. The person in second place might have $252$, so could end with $377$.

André Nicolas
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    Interesting, this answer is also answering the following different question: Which is the largest number of races you can win without winning the championship :) – N. S. Nov 24 '15 at 01:12
  • Another comment on this different question, sorry for the offtopic, but this looks like an interesting question: A driver can win 14 races, and still lose to a driver which only wins 1 race (and of course 18 second places)! – N. S. Nov 24 '15 at 01:19
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In the first 11 races there are $101\times11=1111$ total points. If one driver takes maximum points ($25\times11=275$) in each of those races then there are $1111-275=836$ remaining points and if that is split equally between each driver then there are $44$ points each.

One optimal arrangement of drivers after 11 races (listing position in each race or 0 for a non-scoring race) is:

Driver 1:  <1,1,1,1,1,1,1,1,1,1,1>    = 275 Points
Driver 2:  <2,0,0,10,0,7,0,0,10,0,2>  =  44 Points
Driver 3:  <4,0,0,0,0,0,0,3,9,0,3>    =  44 Points
Driver 4:  <3,0,0,9,0,2,0,0,0,10,6>   =  44 Points
Driver 5:  <6,10,0,0,0,5,0,0,0,3,5>   =  44 Points
Driver 6:  <5,0,0,0,0,4,0,0,5,0,4>    =  44 Points
Driver 7:  <7,9,0,3,0,3,0,0,0,0,7>    =  44 Points
Driver 8:  <8,0,0,4,10,6,3,0,0,0,8>   =  44 Points
Driver 9:  <0,0,8,2,9,9,9,4,0,9,9>    =  44 Points
Driver 10: <9,0,0,5,0,0,4,9,0,2,0>    =  44 Points
Driver 11: <0,2,0,8,8,0,5,0,6,0,0>    =  44 Points
Driver 12: <0,3,10,0,6,0,0,6,4,0,0>   =  44 Points
Driver 13: <0,4,9,6,0,0,8,0,2,0,0>    =  44 Points
Driver 14: <0,5,0,0,0,8,6,5,0,4,0>    =  44 Points
Driver 15: <0,6,7,0,7,0,7,2,0,0,0>    =  44 Points
Driver 16: <0,7,6,0,5,0,0,7,8,5,0>    =  44 Points
Driver 17: <0,0,5,7,4,0,0,10,3,0,0>   =  44 Points
Driver 18: <0,8,4,0,3,10,0,0,7,7,0>   =  44 Points
Driver 19: <10,0,3,0,2,0,10,8,0,8,10> =  44 Points
Driver 20: <0,0,2,0,0,0,2,0,0,6,0>    =  44 Points

It is not possible for any other driver to get more than $200$ points in the remaining 8 races and all the other drivers are $231$ points adrift so Driver 1 cannot be caught.

A similar arrangement can be made for 10 races with a split of points $250:40$ between the 1st driver and each of the rest. However, with the extra race remaining then there are $225$ points still available and any of the other drivers could win the championship if they win all of those races. So it is not possible to determine the chamionship after 10 races.

The answer is a minimum of 11 races are required to determine the championship.

MT0
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I have an alternate solution to this, which I feel has a strong mathematical backing to it. I shall explain this below. If anything is wrong with my explanation, please let me know.

I am assuming that there are 20 races in the season, 20 drivers and no double points scheme. The standard points scheme 25-18-15-....-2-1 is followed. Let's number the drivers from 1 to 20. Driver 1 wins all races, and is the title contender. He gets 25 points per race. There are other 19 drivers, and the points 18-15-12-...-4-2-1 must be equally distributed among them, so that no one driver is closer to driver 1.

So there is a set of 9 points per race, that must be equally distributed among 19 drivers. If we try to do some kind of summation followed by division, we end up with drivers having fractional points, which is not allowed. Hence, we should follow this technique. Drivers numbered 2-10 get points 18-1 in race 1. In race 2, 3-11 get points 18-1 and so on. When we hit the bottom, the points cyclicly wrap around. This image describes what I mean. It has drivers on rows and races as columns. https://i.stack.imgur.com/gQDA3.jpg

So I wrote a program that will do a summation and predict at which point the driver 1's lead is unattainable, and it showed 13. The program is here http://ideone.com/aKpEWX

My predicted answer is 13. Please let me know is this is correct or not.

Anirudh Srinivasan
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The minimum number of races for a guaranteed win is 16/20 when a 1st place is 25 points and 2nd is 18 points.

For a guarantee we must assume that Driver 1 comes 1st until the guaranteed point and then gets zero points thereafter. And that driver 2 comes in 2nd place up until that point and then comes in 1st thereafter.

If Driver 1 wins 15 races and thereafter gets zero points their total at season end is 375. With Driver 2 coming second up until that point and winning every race thereafter their total would be 395.

If Driver 1 were to win 16 races and thereafter get zero points their total at season end would be 400. With Driver 2 coming second up until that point and winning every race thereafter their total would be 388. Therefore the minimum number of races a driver must win for a guaranteed championship win is 16.

DashOfAsh
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Take only two drivers. Driver 1 wins minimum number of races for championship and no points for the others. Driver 2 comes second each time driver one wins and wins the times driver one gets no points.

       Wins     Points(1st)     Points (2nd)    Total

Driver 1 14 350 350 Driver 2 5 125 252 377

        Wins    Points(1st)     Points (2nd)    Total

Driver 1 15 375 375 Driver 2 4 100 270 370

Minimum number of wins is 15 (375 points) without any possibility of being caught.

Namunswa
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