## alone in Napoli

A combinatorics puzzle from The Riddler about a Napoli solitaire where 4 x 10 cards numbered from 1 to 10 are shuffled and the game is lost when a number (1,2, or 3) is equal to its position modulo 3 (1,2 or 3). A simple R code shows that the probability of winning is around 0.00831:

N=40
for(t in 1:1e6)F=F+!sum(!(sample((1:N)%%10)-(1:N)%%3))


ChatGPT bends over backward to achieve this figure! Now, the exact probability can be found by combinatorics. While there are 40! ways of permuting the 40 cards, those missing the coincidences are

$\sum\limits_{k=0}^4\sum\limits_{j=0}^4\sum\limits_{i=0}^4{13\choose i}{13\choose 4-i}{14\choose j}{13-i\choose 4-j}{14-j\choose k}{9+i\choose 4-k}$

multiplied by 4!4!4!28! (which I initially forgot), resulting in 0.00831:

for(i in 0:4)for(j in 0:4)for(k in 0:4)
F=F+exp(lchoose(13,i)+lchoose(13,4-i)+3*lfactorial(4)+
lchoose(14,j)+lchoose(13-i,4-j)+lfactorial(28)+
lchoose(14-j,k)+lchoose(9+i,4-k)-lfactorial(40))


### One Response to “alone in Napoli”

1. […] article was first published on R – Xi’an’s Og, and kindly contributed to R-bloggers]. (You can report issue about the content on this page […]

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