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The only thing I remember from courses I had in probability a few years ago is that we also have to clearly defined the event we want to calculate the probability. On the Freakonomics blog, last week, the Israeli lottery was mentioned (here, see also there where I mentioned that, and odds facts from the French lottery),Yesterday, Andrew Gelman claimed (here) that there was a probability error... Well, since Andrew is really a statistician (and a good one... while I am barely an economist), I tried to do the maths.... and to understand where the error was coming from...Since 6 numbers are drawn out of a pool of numbers from 1 to 37, the total number of combination at each lottery is&#62; (n=choose(37,6))[1] 2324784Over 8 lotteries (since there are two draws per week, we can assume there 8 draws per month), the probability of no identical draws isHere is the R code for those who want to check, again,&#62; prod(n-0:7)/n^8[1] 0.999988Each month, the probability of "coincidence" (I define "coincidence" the event "over 8 draws, at least two times, we obtained the same 6-uplet" or more precisely (as mentioned here) "over one calendar month, at least two times, we obtained the same 6-uplet") is&#62; (p=1-(prod(n-0:7)/n^8))[1] 1.204407e-05The occurrence of a coincidence each month as a Geometric distribution, with probability p. And it is classical, following Gumbel's definition (here), to consider 1/p, called the "return period", i.e. the number of months we have to wait until we observe a coincidence (i.e. a repetition in the same month), since for a geometric distribution&#62; 1/p/(12)[1] 6919.034Here, the (expected) return period is 6919 years.From my point of view, this is “the incident of six numbers repeating themselves within a calendar month”, and this is an event of once in 6919.034 years. On the other hand the median of a geometric distribution is&#62; -log(2)/log(1-p)/(12)[1] 4795.88which means that we have 50% chance to get such a coincidence over 4796 years.Of course, if instead of looking at a longer period, say 100 draws, i.e. one year (here I define "coincidence" the event "over 100 draws, at least two times, we obtained the same 6-uplet"), we have in red the expected return period, and in blue the median of the geometric distribution,&#62; M=E=rep(NA,100)&#62; for(i in 2:100){+ p=1-exp((sum(log(n-0:(i-1)))-i*log(n)))+ E[i]=1/p/(100/i)+ M[i]=-log(2)/log(1-p)/(100/i)+ }&#62; plot(1:100,E,ylim=c(0,10000),type="l",col="red",lwd=2)&#62; lines(1:100,M,col="blue",lwd=2)&#62; abline(v=8,lty=2)&#62; points(8,E[8],pch=19,col="red")&#62; points(8,M[8],pch=19,col="blue")or below of a log-scaled versionAs Xi'an did (here), assume now that there is a lottery over 100 countries. Here I define "coincidence" the event "over k lottery draws over 100 around the world, at least two times, we obtained the same 6-uplet", and then the previous graph becomes (with on the x axis the level of k)Here I have a 12% chance if we consider probability to have identical numbers over a month...But here, we can have one 6-uplet in Israel, and the other one in Egypt, say... If we want to get the same 6-uplet in the same country, the graph is nowi.e. each month there is a chance over one thousand...&#62; i=8&#62; p=1-exp((sum(log(n-0:(i-1)))-i*log(n)))&#62; 1-(1-p)^100[1] 0.001203689Note: actually, Xi'an mentioned that the probability that this coincidence [of two identical draws over 188 draws] occurred in at least one out of 100 lotteries (there are hundreds of similar lotteries across the World) is 53%! And I got the same,&#62; 1-(1-P[188])^100[1] 0.5305219