## Problem of the Day #16: ‘Sid Nils’
*April 4, 2011*

*Posted by Aziz in : potd , trackback*

The legendary Indian national hero Sid is a master of Spades. His patented strategy of “N-I-L” is world renowned. However, due to a recent streak of consistently failed nils, he has hired you to help him figure out why he is losing.

In the game of Spades, there are four players, and each player is randomly dealt $13$ cards from a standard deck of $52$ cards. The game consists of 13 rounds, where each player has to play one of their cards. A person wins a “trick” by playing the highest card of the leading suit or the highest spade in a round. Before play begins, each person has to bid a certain amount, depending on how many tricks they think they can win. Nil is British for zero, so a player who bids “nil” expects to win $0$ tricks.

Since Spades is the trump suit, having a certain set of Spades will guarantee a failed nil. For example, if you have the Ace of Spades, you are guaranteed to win a trick (there is no card higher than the Ace, and Spades trumps). The same principle applies to you if have the King and Queen of Spades: If another person plays the Ace of Spades, you can play either the King or Queen that round, but not both. Hands with these sets of Spades are called “impossible nils”.

Sid wants you to figure out how likely it is that he will be dealt an “impossible nil” hand.

You do not need to consider the distribution of the spades not in your hand among the remaining players. Simply determine the probability that the set of spades in your hand cannot be “defeated” by distinct cards from the set of spades not in your hand.

## Comments»

This problem is pro.

Arjun, it’s also very very hard to do by hand

Hadn’t seen this before, gj guys. Also, I think there’s an ambiguity in the problem statement – is the state of the hand (i.e. impossible nil or possible nil) a function of just the hand, or also the hands of others? That is, suppose I get a Q,J of spades. This is not necessarily an impossible nil, but if somebody has a singleton K or something, it might be, right?

We’re interested in the probability that the hand you’re dealt contains a set of spades that cannot be “defeated” by distinct cards from the set of spades not in your hand.

We’ve added the statement above to the original post to clarify any ambiguity concerning whether to consider the hands of others (you do not have to). However, I don’t think that someone else having a singleton K when you have QJ is necessarily an issue due to the fact that spades can be played on a non-spades trick if you don’t have cards of the initial suit.