**Freakonometrics - Tag - R-english**, and kindly contributed to R-bloggers)

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

> (n=choose(37,6))

[1] 2324784

Over 8 lotteries (since there are two draws per week, we can assume there 8 draws

per month), the probability of no identical

draws is

Here is the R code for those who want to check, again,

> prod(n-0:7)/n^8

[1] 0.999988

Each 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

least two times, we obtained the same 6-uplet

p=1.204407e-05.

> (p=1-(prod(n-0:7)/n^8))

[1] 1.204407e-05

The 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

period

we observe a coincidence (i.e. a repetition in the same month), since

for a geometric distribution

>

1/p/(12)

[1] 6919.034

Here, 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

> -log(2)/log(1-p)/(12)

[1] 4795.88

which 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,

> M=E=rep(NA,100)

> 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)

+ }

> plot(1:100,E,ylim=c(0,10000),type=”l”,col=”red”,lwd=2)

> lines(1:100,M,col=”blue”,lwd=2)

> abline(v=8,lty=2)

> points(8,E[8],pch=19,col=”red”)

> points(8,M[8],pch=19,col=”blue”)

or below of a log-scaled version

As 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 now

i.e. each month there is a chance over one thousand…

> i=8

> p=1-exp((sum(log(n-0:(i-1)))-i*log(n)))

> 1-(1-p)^100

[1] 0.001203689**Note**:

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,

> 1-(1-P[188])^100

[1] 0.5305219

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