Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
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.
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…
R-bloggers.com offers daily e-mail updates about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.