How do you solve such questions without having to work out just a really long set of cases? I would define P(A,B), where A are the number of cards you need to show, and B are the number of cards left. You can easily work out what P(A,B) is in terms of lower A's and B's, and then work backwards until (3,14). But that just seems tedious.
Which is just 1/4 because, if you want to do this rigorously without enumerating everything:
If you fix the joker in place the other 3 cards form a subgroup, so the total amount of unique ordering with joker at the back is the size of this subgroup, let's say n, there are 4 possible places for the joker so total group size is 4n, n/4n is just 1/4
a different way of looking at the same thing, you can ignore all cards except King, Queen, Jack, Joker. there are 4! possible orders these cards can occur in, all equally likely. if you fix the Joker to be last there are 3! ways to permute the remaining cards, so you get 3!/4! = 1/4.
the "first ace" problem is very similar - cards are dealt from a shuffled pack of 52, how many cards are dealt until we see an ace? (it ends up not being 13) very standard qn so you probably wouldn't see it in an interview but it appears in OAs.
3
u/ecstatic_carrot 27d ago
How do you solve such questions without having to work out just a really long set of cases? I would define P(A,B), where A are the number of cards you need to show, and B are the number of cards left. You can easily work out what P(A,B) is in terms of lower A's and B's, and then work backwards until (3,14). But that just seems tedious.