# another lottery coincidence

August 29, 2011
By

(This article was first published on Xi'an's Og » R, and kindly contributed to R-bloggers)

Once again, meaningless figures are published about a man who won the French lottery (Le Loto) for the second time. The reported probability of the event is indeed one chance out of 363 (US) trillions (i.e., billions in the metric system. or 1012)… This number is simply the square of

${49 \choose 5}\times{10 \choose 1} = 19,068,840$

which is the number of possible loto grids. Thus, the probability applies to the event “Mr so-&-so plays a winning grid of Le Loto on May 6, 1995 and a winning grid of Le Loto on July 27, 2011“. But this is not the event that occured: one of the bi-weekly winners of Le Loto won a second time and this was spotted by Le Loto spokepersons. If we take the specific winner for today’s draw, Mrs such-&-such, who played bi-weekly one single grid since the creation of Le Loto in 1976, i.e. about 3640 times, the probability that she won earlier is of the order of

$1-\left(1-\frac{1}{{49\choose 5}\times{10\choose 1}}\right)^{3640}=2\cdot 10^{-4}$.

There is thus a chance in 20 thousands to win again for a given (unigrid) winner, not much indeed, but no billion involved either. Now, this is also the probability that, for a given draw (like today’s draw), one of the 3640 previous winners wins again (assuming they all play only one grid,  play independently from each other, &tc.). Over a given year, i.e. over 104 draws, the probability that there is no second-time winner is thus approximately

$\left(1-\frac{1}{2\cdot10^4}\right)^{104} = 0.98,$

showing that within a year there is a 2% chance to find an earlier winner. Not so extreme, isn’t it?! Therefore, less bound to make the headlines…

Now, the above are rough and conservative calculations. The newspaper articles about the double winner report that the man is playing about 1000 euros a month (this is roughly the minimum wage!), representing the equivalent of 62 grids per draw (again I am simplifying to get the correct order of magnitude). If we repeat the above computations, assuming this man has played 62 grids per draw from the beginning of the game in 1976 till now, the probability that he wins again conditional on the fact that he won once is

$1-\left(1-\frac{62}{{49 \choose 5}\times{10 \choose 1}}\right)^{3640} = 0.012$,

a small but not impossible event. (And again, we consider the probability only for Mr so-&-so, while the event of interest does not.) (I wrote this post before Alex pointed out the four-time lottery winner in Texas, whose “luck” seems more related with the imperfections of the lottery process…)

I also stumbled on this bogus site providing the “probabilities” (based on the binomial distribution, nothing less!) for each digit in Le Loto, no need for further comments. (Even the society that runs Le Loto hints at such practices, by providing the number of consecutive draws a given number has not appeared, with the sole warning “N’oubliez jamais que le hasard ne se contrôle pas“, i.e. “Always keep in mind that chance cannot be controlled“…!)

Filed under: R, Statistics Tagged: binomial distribution, coincidence, Le Loto, lottery

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