I'm a little stuck on the size of my sets. Here is what I have so far.
Proof.
Let $A$ denote the set of possible hands in Texas Hold'em. Since order doesn't matter and repeats are not allowed, $|A| =$ $52 \choose 2$ $= 1,326$.Let B denote the set of hands dealt when playing every minute for one full day. This means that you will play 60 hands per hour for 24 hours, so $|B| = 60*24 = 1,440$.
This is where I am stuck. To utilize the pigeonhole principle I will need that $|A| > |B|$, which is not the case. I need to define a function on these sets, so I can't simply switch them.
I am wondering if my issue lies within $A$. Do I want to use 1,326 possibilities here, or 2,652 (the possible pairs for your hand). For example,
Proof.
Let $A$ denote the set of possible card pairs in a hand of Texas Hold'em. By the Multiplication Principle, there are $52 * 51 = 2,652$ possible card pairs for your hand. So $|A| = 2,652.$Let B denote the set of hands dealt when playing every minute for one full day. This means that you will play 60 hands per hour for 24 hours, so $|B| = 60*24 = 1,440$.
In that case, I meet the criteria, but I'm not sure if my set definitions are okay. I already know how to conclude the proof once my set cardinalities are correct.
Proof.
Let $A$ be the set of hands dealt when playing every minute for one day. This means you will play 60 hands per hour for 24 hours, so $|A| = 1,440.$
Let B denote the set of possible hands in Texas Hold 'Em. Since order doesn't matter and repeats are not allowed, $|B| =$ $52 \choose 2$ $= 1,326.$
Define $f:A \to B$ so that $f(x)$ yields a possible Texas Hold'Em hand for a hand dealt within the playing time, $x$.
Since $|A| > |B|$, $f$ is not one-to-one, by the Pigeonhole Principle.
Now let $x$ and $y$ be distinct elements of $A$ for which $f(x) = f(y)$. This means that $x$ and $y$ are two different hands dealt within the playing time, but are the same possible Texas Hold 'Em hand.