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The $1989$ Tour was won by Greg Lemond (USA, $1961$ - ), who beat Laurent Fignon (France, $1960$ - $2010$) by $8''$. Yes, eight seconds! The closest tour in history.

Let me recall a few rules concerning the time measurement. At each stage, a measurement is made for every group of riders (for safety reasons, riders arriving together should not fight for winning seconds). Each time is a natural number of seconds. Therefore, each measurement involves a round-off and can be viewed as the integral part (closest integer) of a real (random) variable. For instance, if a rider arrives alone and $X$ is his real time, then his official time is the integer $n$ such that $X\in(n-1/2,n+1/2]$. At the end of the race, each rider has a total official time $T$, and his real time (which nobody knows) belongs to the interval $(T-m/2,T+m/2]$ with $m$ the number of stages ($21$ in $1989$).

When comparing two riders (say Greg and Laurent), we should only consider the stages where they arrived in different groups (they are not suppose to fight at the end of the irrelevant stages otherwise). This happened $11$ times to Greg and Laurent. Disregarding the irrelevant stages, we can say that the real time of Greg belongs to $(T_g-11/2,T_g+11/2]$ and that of Laurent belongs to $(T_\ell-11/2,T_\ell+11/2]$. With $T_\ell=T_g+8$, we see that these intervals overlap! There is thus a possibility that Laurent rode faster than Greg but lost because of the round-offs. Of course, this has a very tiny probability, because the event needs that the round-offs be close to maximal at each stage and always in favor of Greg. Unlikely, but

What is the probability that Greg won just because of round-offs ?

Let us assume that the round-offs are independent random variables, uniformly distributed over $(-1/2,1/2]$.

By the way, I am not chauvinist, and I admire Greg.

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  • $\begingroup$ Just to clarify, are you sure that they don't take the floor instead of the roundoff? $\endgroup$
    – Tony Huynh
    Commented Jan 6, 2011 at 12:34
  • $\begingroup$ @Tony. No, I'm not. But whatever they took (floor, roundoff, top), the mathematical problem is the same. $\endgroup$
    – Denis Serre
    Commented Jan 6, 2011 at 12:36
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    $\begingroup$ There may be some calls for closure, since this is a standard homework-type problem in elementary statistics courses. @Tony: so that they are uniformly distributed in [0,1) ... It doesn't matter for the question. $\endgroup$
    – Gerald Edgar
    Commented Jan 6, 2011 at 12:57
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    $\begingroup$ 2 out of 100000000 simulations of the difference between the sum of 11 samples from two U(0,1) distributions were greater than 8. $\endgroup$
    – B R
    Commented Jan 6, 2011 at 16:40
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    $\begingroup$ If a sports journalist find this post, this last comment will be rendered as "Several simulations, run by the eminent mathematician Prof. BR, prove that L. Fignon should have won the Tour de France in 1989 over G. LeMond." $\endgroup$
    – Maxime Bourrigan
    Commented Jan 6, 2011 at 17:09

2 Answers 2

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The sum of independent identically distributed random variables follows the Irwin-Hall distribution. For $n$ variables that are uniformly distributed on (0,1), the probability density function is $$f_X(x)=\frac{1}{2\left(n-1\right)!}\sum_{k=0}^{n}\left(-1\right)^k{n \choose k}\left(x-k\right)^{n-1}\text{sgn}(x-k).$$

So for $n=11$ you will have to integrate an explicit piecewise polynomial function.

Edit: Tony is right, we need not quite the integral of the Irwin-Hall distribution. Rather, we need the difference distribution of two random variables that are distributed according to the above distribution (each of the RVs representing the total round-off error for each cyclist). So we need $$\int_0^3\left(f_X(x)\int_{x+8}^{11}f_X(y)dy\right)dx.$$ This is the probability that the difference between the round off errors is between 8 and 11. It is obtained by integrating over all permissible combinations (one error, other error).

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  • $\begingroup$ Mathematica says (with $n=11$) that $\int_8^{11} f_X(x)\,dx$ is $\frac{8593}{2217600} \approx 0.004$. $\endgroup$
    – Kevin O'Bryant
    Commented Jan 6, 2011 at 13:52
  • $\begingroup$ @Kevin: I'm not sure that this is the integral that we want though. The respective roundoffs for Greg and Laurent are assumed to be uniform, but the difference between the roundoffs is certainly not uniform. See my answer below. $\endgroup$
    – Tony Huynh
    Commented Jan 6, 2011 at 13:58
  • $\begingroup$ Isn't $f_X$ the self-convolution of $n$ copies of the uniform probability over $(0,1)$ ? $\endgroup$
    – Denis Serre
    Commented Jan 6, 2011 at 14:08
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    $\begingroup$ Actually, that $0.004$ was the probability Greg's real time was at least $8/2=4$ seconds more than the sum of his rounded times. The actual value is $P(X_{22}\gt 19).$ If I calculate correctly, that is $122221817/4390627842881280000$ $=2.7837 ~10^{−11},$ although I am concerned that this does not agree with BR's simulation. The sum of the errors would have to be about $5.9$ standard deviations away from the mean, and the normal approximation would overestimate the size of the tail in this region. $\endgroup$
    – Douglas Zare
    Commented Jan 6, 2011 at 16:51
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    $\begingroup$ I wouldn't be too concerned: naive simulations of rare events should not be taken too literally. $\endgroup$
    – B R
    Commented Jan 6, 2011 at 18:09
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Caveat: this is not a complete answer, just a call for Matlab gurus to jump in here.

Let $Y=Y_1+\dots + Y_{11}$, where each $Y_i$ is the uniform difference distribution. Note that the uniform difference distribution is a nice piecewise linear function with domain $[-1,1]$. We simply want $P(Y \geq 8)$.

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  • $\begingroup$ Instead of taking the convolution of 11 distributions which are each convolutions of 2 identical distributions, you can just take a convolution of 22. $\endgroup$
    – Douglas Zare
    Commented Jan 6, 2011 at 16:10
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    $\begingroup$ Which means that you take a unit cube in 22 dimensions and want the volume of the part given by $x_1+\dots+x_{22}\le 3$. I'm not sure what MATLAB has to do with it, but the answer is $\frac 1{22!}(1^{22}+(2^{22}-22\cdot 1^{22})+(3^{22}-22\cdot 2^{22}+231\cdot 1^{22}))$. Since $(3/2)^22>2^{11}>2000$, we can reduce it to $3^{22}/{22!}\approx \frac 1{\sqrt 44\pi}\left(\frac{3e}{22}\right)^{-22}$. Now that is almost $0.1\cdot 8^{22}/22^{22}\approx \frac 8{27}10^{-10}$ give or take a factor of $1.2$. Not a fat chance, if you ask me :). $\endgroup$
    – fedja
    Commented Jan 6, 2011 at 17:13
  • $\begingroup$ Impossibility to edit or preview comments is really irritating. Of course, it should be $(3/2)^{22}$ and $\left(\frac{3e}{22}\right)^{22}$. Sorry for stupid typos :). $\endgroup$
    – fedja
    Commented Jan 6, 2011 at 17:16
  • $\begingroup$ @Douglas: I think the issue is the domain of the uniform distribution. Indeed, it is more convenient to work with $U(-1/2,1/2)$ instead of $U(0,1)$, because as you say we can just take the convolution of $U(-1/2,1/2)$ 22 times (because it is symmetric about zero). I only used $U(0,1)$ to be consistent with Alex's answer. @fedja: If you don't mind to post your comment as an answer, I would vote it up. $\endgroup$
    – Tony Huynh
    Commented Jan 7, 2011 at 13:28

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