## Mr Meyrowitz’s glasses

Posted in Statistics, University life with tags , , , , , , , on October 23, 2011 by xi'an

Today, I found a site entitled Mr Meyrowitz’s Class that links to my first post on coincidences in lotteries as an example of “fatal error”. This seems to be part of a student’s assignment, apparently for the CollegeBoard programme, with 10 minutes allocated to students to find my “fatal error with decimals and probabilities”… As there is no hint, I wonder where my fatal error stands: I could not find it after those 10 minutes of intense searching and recomputing. Maybe Mr Meyrowitz actually needs new glasses to spot the difference between a 1‰ chance and a 1% chance… (Which actually misled a few other readers of the post.)

Question 6) in this assignment also sounds very much inspired from another of my posts on coincidences in lotteries [although not acknowledged in the assignment] since the question refers to the same original France Soir article in French. The question is however rather vague: “do you suspect him of cheating?” and it shows a lack of knowledge about French loto where cheating is [close to] impossible. It is certainly not recommended as an exercise for beginning students in probability or statistics. [Actually, in my opinion, the whole assignment is poor, being either imprecise, e.g question 7), useless, as for question 4) "Pick one topic that you understand very well and one that you do not understand well" (!), or plain wrong, as for question 2)...]

## Another coincidence…

Posted in Mountains, pictures, Travel with tags , , , , , , , on September 9, 2011 by xi'an

After the coincidence of bumping into Marc Suchard in an Edinburgian Indian restaurant on Tuesday night, I faced another if much less pleasant coincidental event: for the third time in a row, my bag went missing on a Air France flight to Scotland… This happened first for the mixture meeting in 2009, costing me an attempt at Tower Ridge on Ben Nevis, then again when I took part in the colloquium celebrating Mike Titterington last May. Since having three independently lost luggages on three (or six) trips is very unlikely, there must be a reason to this pattern! Besides a conspiracy theory about the airline pushing me towards other companies because of my grumpy patronage or my unsuccessful requests for Irn Bru, possible reasons are late checking-ins (even though this does not apply in the last two cases) and use of a [small] backpack that is always turned into an “oversized” piece of luggage by the airline ground personal (does not apply to the first occurrence), but I do not carry particularly suspicious items, not even haggis on the way back… There must be a better reason than that!

## another lottery coincidence

Posted in R, Statistics with tags , , , on August 30, 2011 by xi'an

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 are thus two chances in 10 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“…!)

## Common ancestors

Posted in Statistics with tags , , , , on June 8, 2011 by xi'an

In conjunction with President Obama’s visit to Ireland two weeks ago and in particular to his ancestral Irish town, I happened to glance at his family tree and saw that he shared a common ancestor with George W. Bush. (They are 11th cousins, meaning that a 12th-order ancestor is common to both their  family trees.) This sounds at first amazing, but it is another occurrence of the (von Mises) birthday problem. (The fact that it is not that amazing is demonstrated by the simultaneous presence of [French!] ancestors of Dick Cheney in the same tree.) If we consider President Obama’s mother side, the probability that all of her 11th-order ancestors differ from all of George W. Bush’s 12th-order ancestors is

$p=\dfrac{(M-2^{12})(M-2^{12}-1)\cdots(M-2^{12}-2^{11}+1)}{M(M-1)\cdots(M-2^{11}+1)}$

where M denotes the whole population of potential ancestors at this period. If we consider all those ancestors as coming from the British Isles, then about 1650, the population was about 8 million. This would lead to a probability of p=0.35, i.e. there is a 65% chance that they share a 12th-order ancestor. If instead we consider the whole European population at that time (if only to include German and French ancestors to President Obama), M is about 100 million and the probability increases to p=0.92, so there is then an 8% probability for them to share an ancestor. Obviously, this rough calculation relies on simplifying assumptions, avoiding the issue of inbreeding which means that the potential 2¹¹ ancestors are in fact much less than 2¹¹, and the fact that their ancestors are necessarily emigrants, which reduces the value of M(This post appeared yesterday on the Statistics Forum.)

## Another coincidence in lotteries

Posted in Statistics with tags , , , on January 10, 2011 by xi'an

Here is a lottery story that appeared in the LA Times (and in numerous other journals and news reports):

On the television show Lost, the character Hugo “Hurley” Reyes played the numbers 4, 8, 15, 16, 23 and 42 and ended up winning the $114-million jackpot. In real life, the lotto’s selections for its$355-million prize — 4, 8, 15, 25, 47 and the crucial “Mega ball,” 42 — included four of Hurley’s numbers. According to the Mega Millions website, which reported receiving “unprecedented traffic” after the drawing, 41,763 people matched those four numbers, earning \$150 apiece.

Matching at least four numbers out of six is not that unlikely. Assuming there are 48 possible numbers in the lottery, and not bothering about specifics for this lottery like “Mega ball” and specific order, a basic probability computation for this match is

${6\choose 4}{44\choose 2}/{48\choose 6}=0.001156,$

that is a 1‰ chance when taking this lottery and this sequence of numbers in isolation. This probability does not account for the parallel and subsequent runs of many other lotteries where many players will have undoubtedly used those “lucky” numbers, since they seem to have been around for several years, nor for the fact that other series have also proposed lottery numbers (just a few weeks ago my daughter was watching an episode of the series Cold Case which involved a lottery draw).