using pigeonhole principle for a hand of thirteen cards Say I shuffle and deal a hand of thirteen cards. How can I apply the pigeonhole principle in these cases:


*

*The hand has at least four cards in the same suit

*The hand has at exactly four cards in some suit

*The hand has at least five cards in some suit

*The hand has exactly one suit containing four or more cards


I have problems in which to assign holes and pigeons to, can anyone help?
 A: You can apply the pigeonhole principle to case 1: can you fit thirteen pigeons into four holes each with space for three pigeons? 
A: For the first question, let the four suits be the holes, and the thirteen cards be the pigeons. This is because each time you deal a card of a certain suit, it is like putting a pigeon in a certain hole. The hole fills up with each pigeon, just as the number of cards fill up with each dealt card of a certain suit. By the pigeonhole principle, you are assigning $13$ discrete objects into $4$ containers, so one container must hold at least $\lceil 13/4\rceil=4$ objects. 
This makes intuitive sense, for suppose you consciously tried to assign your cards so that no suit had at least four cards in the hand. You could have at most $3$ cards of the same suit for each suit, but in doing so, you would only be able to assign $3\cdot 4=12$ cards. Then the $13^\text{th}$ card must push one of the suits over, so that you have at least $4$ four of the same suit.
For the other cases, the principle doesn't apply. Take the second option for example. You could be dealt $5$ hearts, $3$ diamonds, $3$ clubs, and $2$ spades. You can show the pigeonhole principle does apply in the other cases be finding counterexamples.
