# Cursed numbers ?

**Freakonometrics - Tag - R-english**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

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In *Lost*, Hugo “*Hurley*”
Reyes played the numbers 4, 8, 15, 16, 23 and 42 at the lottery, and
ended up winning the $114-million jackpot. And over the ensuing weeks,
everyone around him seems to suffer increasingly bad luck: Hurley’s
grandfather dies of a heart attack, his brother’s wife walks out, his
mother breaks her ankle while the house Hurley bought her goes up in
flames, and Hurley himself is falsely arrested.

Anyway, last week (here) 4 numbers (out of 6) appeared at the lottery in LA. As pointed out by Xian (here), the odds were not that small, i.e. it is a 1‰ chance,

> loto=read.table("loto.csv",dec=",",header=TRUE,sep=";") > ntirage=nrow(loto) > loto=loto[51:ntirage,] > ntirage=nrow(loto) > N=as.matrix(loto[,c("boule_1","boule_2","boule_3", "boule_4","boule_5","boule_6")]) > P=rep(NA,nrow(N)) > for(s in 1:nrow(N)){ + P[s]=sum(N[s,1]%in%c(4, 8, 15, 16, 23, 42)+ + N[s,2]%in%c(4, 8, 15, 16, 23, 42)+ + N[s,3]%in%c(4, 8, 15, 16, 23, 42)+ + N[s,4]%in%c(4, 8, 15, 16, 23, 42)+ + N[s,5]%in%c(4, 8, 15, 16, 23, 42)+ + N[s,6]%in%c(4, 8, 15, 16, 23, 42)) + } > table(P)/nrow(N)*1000 P 0 1 2 3 4 435.732113 405.366057 137.271215 19.966722 1.663894

But what about the full sequence…? Imagine that in France, at the official lottery, the exact sequence played by Hugo appears. What a coincidence. The probability that the sequence appears, assuming that there are 48 possible numbers in the lottery, is

Now if we look at all official lotteries around the world, say 100 per week, what is the probability to see Hurley’s sequence shows up – at least once – in 25 years (assuming that after 25 years, no one will remember

*Lost*and those cursed numbers) ? It looks like it is a 1% chance…

So let us wait and see…

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**Freakonometrics - Tag - R-english**.

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