F1

Reverse-engineering the team principals' top ten drivers of 2014

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  • #289216

    Autosport published its regular team principals’ top ten drivers last week. Who voted for who is kept anonymous but I thought it might be fun and instructive to try to work out what scores each driver got.

    Each team principal (apart from strife-hit Caterham’s) was invited to pick their top ten drivers who were then scored using the current points system (25, 18, 15, 12, 10, 8, 6, 4, 2, 1).

    For example, Hamilton received a score of 194. One way of making this would be five team principals rating him as the best, three putting him second, one putting him third and one leaving him out of their top ten.

    How else could Hamilton’s score have been made up? And what about the other drivers? And which drivers outside the top ten took the remaining 72 points? Over to the maths geeks…

    #289222
    andae23
    Participant

    Very long post warning!!!

    Wrote a small Matlab code that calculates all possible combinations. The format is as follows: ‘1 0 0 0 2 0 0 0 0 3’ means that one 1st place, two 5th places and three 10th places would get the score.

    For Hamilton (194 points), there are 29 possibilities:
    2 8 0 0 0 0 0 0 0 0
    4 3 2 0 1 0 0 0 0 0
    4 4 0 1 1 0 0 0 0 0
    4 5 0 0 0 0 0 1 0 0
    5 0 3 2 0 0 0 0 0 0
    5 1 1 3 0 0 0 0 0 0
    5 1 3 0 0 0 1 0 0 0
    5 2 1 0 1 1 0 0 0 0
    5 2 1 1 0 0 1 0 0 0
    5 3 1 0 0 0 0 0 0 0 (here’s the one Keith pointed out)
    6 0 0 2 2 0 0 0 0 0
    6 0 0 3 0 1 0 0 0 0
    6 0 2 0 0 1 1 0 0 0
    6 0 2 0 1 0 0 1 0 0
    6 0 2 1 0 0 0 0 1 0
    6 1 0 0 1 2 0 0 0 0
    6 1 0 0 2 0 1 0 0 0
    6 1 0 1 0 1 1 0 0 0
    6 1 0 1 1 0 0 1 0 0
    6 1 0 2 0 0 0 0 1 0
    6 1 1 0 1 0 0 0 0 1
    6 2 0 0 0 0 0 2 0 0
    6 2 0 0 0 0 1 0 1 0
    6 2 0 0 0 1 0 0 0 0
    7 0 0 0 1 1 0 0 0 1
    7 0 0 1 0 0 1 0 0 1
    7 0 1 0 0 0 0 0 2 0
    7 0 1 0 0 0 0 1 0 0
    7 1 0 0 0 0 0 0 0 1

    For Alonso (155 points), there are 334 possibilities.. I’m not going to list them all for obvious reasons. However, if you assume every team principal had Alonso in his top 5, there are sixteen possibilities:
    0 5 3 0 2 0 0 0 0 0
    0 6 1 1 2 0 0 0 0 0
    1 0 8 0 1 0 0 0 0 0
    1 1 6 1 1 0 0 0 0 0
    1 2 4 2 1 0 0 0 0 0
    1 3 2 3 1 0 0 0 0 0
    1 4 0 4 1 0 0 0 0 0
    1 5 0 0 4 0 0 0 0 0
    2 0 3 5 0 0 0 0 0 0
    2 0 5 0 3 0 0 0 0 0
    2 1 1 6 0 0 0 0 0 0
    2 1 3 1 3 0 0 0 0 0
    2 2 1 2 3 0 0 0 0 0
    3 0 0 5 2 0 0 0 0 0
    3 0 2 0 5 0 0 0 0 0
    3 1 0 1 5 0 0 0 0 0

    For Ricciardo (135 points), there are 825 possibilities. I guess from now on I will only mention the number of possibilities.

    For Bottas (119 points), there are 1424 possibilities.

    For Rosberg (114 points), there are 1593 possibilities.

    For Massa (55 points), there are 1272 possibilities.

    For Vettel (53 points), there are 1165 possibilities.

    For Bianchi (44 points), there are 840 possibilities. If you assume no one put Bianchi in his or her top 6, there are 13 possibilities:
    0 0 0 0 0 0 2 8 0 0
    0 0 0 0 0 0 3 6 1 0
    0 0 0 0 0 0 4 4 2 0
    0 0 0 0 0 0 4 5 0 0
    0 0 0 0 0 0 5 2 3 0
    0 0 0 0 0 0 5 3 0 2
    0 0 0 0 0 0 5 3 1 0
    0 0 0 0 0 0 6 0 4 0
    0 0 0 0 0 0 6 1 1 2
    0 0 0 0 0 0 6 1 2 0
    0 0 0 0 0 0 6 2 0 0
    0 0 0 0 0 0 7 0 0 2
    0 0 0 0 0 0 7 0 1 0

    For Button (40 points), there are 654 possibilities. If you assume no one put Button in his or her top 6, there are 20 possibilities:
    0 0 0 0 0 0 0 10 0 0
    0 0 0 0 0 0 1 8 1 0
    0 0 0 0 0 0 2 6 2 0
    0 0 0 0 0 0 2 7 0 0
    0 0 0 0 0 0 3 4 3 0
    0 0 0 0 0 0 3 5 0 2
    0 0 0 0 0 0 3 5 1 0
    0 0 0 0 0 0 4 2 4 0
    0 0 0 0 0 0 4 3 1 2
    0 0 0 0 0 0 4 3 2 0
    0 0 0 0 0 0 4 4 0 0
    0 0 0 0 0 0 5 0 5 0
    0 0 0 0 0 0 5 1 2 2
    0 0 0 0 0 0 5 1 3 0
    0 0 0 0 0 0 5 2 0 2
    0 0 0 0 0 0 5 2 1 0
    0 0 0 0 0 0 6 0 0 4
    0 0 0 0 0 0 6 0 1 2
    0 0 0 0 0 0 6 0 2 0
    0 0 0 0 0 0 6 1 0 0

    For Kvyat (28 points), there are 244 possibilities. If you assume no one put Kvyat in his or her top 6, there are 38 possibilities:
    0 0 0 0 0 0 0 4 6 0
    0 0 0 0 0 0 0 5 3 2
    0 0 0 0 0 0 0 5 4 0
    0 0 0 0 0 0 0 6 0 4
    0 0 0 0 0 0 0 6 1 2
    0 0 0 0 0 0 0 6 2 0
    0 0 0 0 0 0 0 7 0 0
    0 0 0 0 0 0 1 2 7 0
    0 0 0 0 0 0 1 3 4 2
    0 0 0 0 0 0 1 3 5 0
    0 0 0 0 0 0 1 4 1 4
    0 0 0 0 0 0 1 4 2 2
    0 0 0 0 0 0 1 4 3 0
    0 0 0 0 0 0 1 5 0 2
    0 0 0 0 0 0 1 5 1 0
    0 0 0 0 0 0 2 0 8 0
    0 0 0 0 0 0 2 1 5 2
    0 0 0 0 0 0 2 1 6 0
    0 0 0 0 0 0 2 2 2 4
    0 0 0 0 0 0 2 2 3 2
    0 0 0 0 0 0 2 2 4 0
    0 0 0 0 0 0 2 3 0 4
    0 0 0 0 0 0 2 3 1 2
    0 0 0 0 0 0 2 3 2 0
    0 0 0 0 0 0 2 4 0 0
    0 0 0 0 0 0 3 0 3 4
    0 0 0 0 0 0 3 0 4 2
    0 0 0 0 0 0 3 0 5 0
    0 0 0 0 0 0 3 1 0 6
    0 0 0 0 0 0 3 1 1 4
    0 0 0 0 0 0 3 1 2 2
    0 0 0 0 0 0 3 1 3 0
    0 0 0 0 0 0 3 2 0 2
    0 0 0 0 0 0 3 2 1 0
    0 0 0 0 0 0 4 0 0 4
    0 0 0 0 0 0 4 0 1 2
    0 0 0 0 0 0 4 0 2 0
    0 0 0 0 0 0 4 1 0 0
    ___

    Conclusion: it’s pretty much impossible to apply reverse engineering here, there are just too many combinations. Maybe with a better code we can see some possibilities for what the team principals voted for, but I guess there will be far too many to make some sense out of it.

    While we’re on it, here’s the (very clunky) Matlab code if you’re interested:

    clear all
    clc
    p = [25 18 15 12 10 8 6 4 2 1]; % points
    s = 194; % score, for instance 194 for Hamilton
    r = []; % results, the output matrix
    t = 1; % counter, just ignore this one
    for p1 = 0:10
    for p2 = 0:(10-p1)
    for p3 = 0:(10-p1-p2)
    for p4 = 0:(10-p1-p2-p3)
    for p5 = 0:(10-p1-p2-p3-p4)
    for p6 = 0:(10-p1-p2-p3-p4-p5)
    for p7 = 0:(10-p1-p2-p3-p4-p5-p6)
    for p8 = 0:(10-p1-p2-p3-p4-p5-p6-p7)
    for p9 = 0:(10-p1-p2-p3-p4-p5-p6-p7-p8)
    for p10 = 0:(10-p1-p2-p3-p4-p5-p6-p7-p8-p9)
    if p1*p(1)+p2*p(2)+p3*p(3)+p4*p(4)+p5*p(5)+p6*p(6)+p7*p(7)+p8*p(8)+p9*p(9)+p10*p(10) == s
    r(t,:) = [p1 p2 p3 p4 p5 p6 p7 p8 p9 p10];
    t = t+1;
    end
    end
    end
    end
    end
    end
    end
    end
    end
    end
    end

    I might make a possible distribution tonight or tomorrow, if I don’t forget :)

    #289223
    PhilEReid
    Participant

    Is there anything you can’t do @andae23 ? :P

    #289224

    @philereid Making it rain in the Middle Eastern Grands Prix, staying in the top 3 for 2 years in a row in @robocat’s F1FVGPC, and (as he said himself) calculate beyond reasonable doubt the principals’ top 10, perhaps among others :P

    Anyway @andae23 I suggest a (said-to-be) free-software alternative to MATLAB in GNU Octave (if not a total equivalent – I don’t know as I can use neither of them).

    #289228

    So when I wrote ‘over to the maths geeks’ I might as well have just written ‘over to @andae23…’. Terrific work!

    I guess one starting point might be to ask which of those 29 possible combinations for Hamilton do we think is the most realistic, and then we have ten variables we can remove from the other drivers.

    Alternatively we could make certain fixed assumptions about which drivers got which scores from different team principals – e.g. Toto Wolff definitely put Hamilton number one, Christian Horner definitely put Daniel Ricciardo number one etc… But I appreciate this may be rather tricky to programme!

    What I do find interesting, though, is that at least one team principal must have put Hamilton outside the top three, unless he got two firsts and eight seconds, which seems a bit unlikely.

    #289229
    Stuart
    Participant

    @andae23 Would it not be possible to filter the list down to only combinations which would for *ALL* drivers at the same time. e.g. one of HAMS options is ‘2 8 0 0 0 0 0 0 0 0’ however that would not be possible with ALO’s ‘0 5 3 0 2 0 0 0 0 0’. I still doubt it would give us the absolute answer but might create a more reasonable set of options.

    #289238
    andae23
    Participant

    @s5l5 Yes, but I couldn’t do that during my lunch break :P

    If we look at the F1F drivers’ poll, about 40 % voted for Hamilton, 40% for Ricciardo and 20% for other drivers, mostly for Alonso. So if we translate that to our problem, you can say that 4 principals voted for Hamilton, 4 for Ricciardo and, say, 2 for Alonso. If we assume that, then the resulting combinations are very weird, especially for Ricciardo: if he got four 1st places, then at least 3 team principals didn’t include him in their top 5. So that’s maybe not the best starting point.

    Looking at the combinations for Hamilton, I guess ‘4 3 2 0 1 0 0 0 0 0’, ‘4 4 0 1 1 0 0 0 0 0’, ‘5 0 3 2 0 0 0 0 0 0’ and ‘5 1 1 3 0 0 0 0 0 0’ are the most likely candidates. I did a bit of work and it seems like ALO and RIC can’t have more than 2 1st places to have ‘realistic’ combinations (i.e. at least 9/10 team principals had them high up their list), so I’m guessing that ‘5 1 1 3 0 0 0 0 0 0’ is the most likely one. Thoughts?

    #289247

    @andae23 I think that’s an interesting way of looking at it.

    Is it possible that all ten team principals put Ricciardo, Alonso and Hamilton in their top five?

    #289292
    Mal Cockburn
    Participant

    Since there is only one unique solution to the equations, then it should be possible (or even trivial) to calculate the votes. By considering only the top few drivers and their scores, there should emerge the set of votes that the team principles made. Too much for me to do, but I think the heavy lifting has already been done by andae23.

    #289275
    Keith Campbell
    Participant

    I actually already had a go at this – more of a trial and error than mathematical solution but this is one possible solution. I had one significant difference in my understanding of the votes though – i was under the impression team principles couldn’t vote for their own drivers… is this wrong? I can’t remember if i read that from a previous year or made it up – either way my solution assumes each driver receives a maximum of 9 votes only, leading to this:

    +----------+--------+-----------+--------+---------+
    |    1     |   2    |     3     |   4    |    5    |
    +----------+--------+-----------+--------+---------+
    | Hamilton | Alonso | Ricciardo | Bottas | Rosberg |
    | 25       | 25     | 25        | 18     | 18      |
    | 25       | 25     | 25        | 18     | 18      |
    | 25       | 18     | 18        | 15     | 18      |
    | 25       | 18     | 15        | 12     | 15      |
    | 25       | 15     | 12        | 12     | 10      |
    | 18       | 15     | 10        | 12     | 10      |
    | 18       | 15     | 10        | 12     | 10      |
    | 18       | 12     | 10        | 10     | 8       |
    | 15       | 12     | 10        | 10     | 8       |
    |          |        |           |        |         |
    | 194      | 155    | 135       | 119    | 115     |
    +----------+--------+-----------+--------+---------+
    +-------+--------+---------+--------+-------+
    |   6   |   7    |    8    |   9    |  10   |
    +-------+--------+---------+--------+-------+
    | Massa | Vettel | Bianchi | Button | Kvyat |
    | 15    | 15     | 8       | 15     | 8     |
    | 8     | 12     | 8       | 12     | 6     |
    | 8     | 6      | 6       | 8      | 6     |
    | 8     | 6      | 4       | 2      | 2     |
    | 6     | 4      | 4       | 2      | 2     |
    | 6     | 4      | 4       | 1      | 2     |
    | 4     | 4      | 4       |        | 1     |
    |       | 2      | 4       |        | 1     |
    |       |        | 2       |        |       |
    |       |        |         |        |       |
    | 55    | 53     | 44      | 40     | 28    |
    +-------+--------+---------+--------+-------+

    Obviously i used points rather than placings but you can see what it means fairly clearly. This is just one possible solution that fitted neatly, going for what i took to be likely results for the first 3 drivers and then just piecing together the rest (using the ‘easiest’ to find solution rather than the most likely one).

    Many thanks to @BenH for the sensefulsolutions link for the tables.

    #289300

    @keithedin I’ve edited your comment to put your tables within the the code tags to make them easier to read.

    #289391

    @keithedin

    I was under the impression team principles couldn’t vote for their own drivers

    I’ve had a look back over the previous five years and can’t see anything which indicates that is the case.

    In 2012 Alonso scored 269 and eight of the twelve team principals put him first. The only way this can have happened as far as I can tell is he must have been given eight firsts, three seconds and one third. Therefore all 12 team principals must have been able to vote on him.

    So I believe team principals can vote for their drivers.

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