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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,

Hence, with one lottery per week, the return period is 16 years. Note this percentage is very close to what we did observe on the French lottery (below the statistics in ‰, from here, in a zip file),
```> 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

i.e. the expected number of draws we need before seeing that sequence for the first time is almost a billion.
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…