Probability of drawing all hearts before all spades Let's say we have a 26 card deck(13 hearts and 13 spades), and we want to find the probability that all hearts are drawn before all spades. What is this value? What if we have 52 cards?
For the first case, the way I thought about it was the following. If we have 26 cards, we can arrange them n 26! ways, and if we wanted all the hearts to come before all the spades, this can be done in 13!13! ways since there are 13! ways to arrange each suit of 13 cards. But the answer does not match the probability of 1/2 that is given in the book. I get, 13!13!/26!
The book also mentions that the second probability is also the same. This, I don't understand at all. If someone could please explain to me the answer to this question, as well as why my approach doesn't work, I would be grateful! Thank you so much!

They’re not asking for the probability that all of the hearts are drawn before any spades are drawn: they’re asking for the probability that all of the hearts are drawn before the spades have been completely exhausted, i.e., before the last spade is drawn. Do you see now why the probability is $\frac12$ regardless of how many other cards are in the deck?
Added: Suppose that you have deck of $n$ cards that consists of $13$ hearts, $13$ spades, and $n-26$ other cards that are neither hearts nor spades. Let $H$ be the set of permutations of the deck in which the last heart comes before the last spade, and let $S$ be the set of permutations of the deck in which the last spade comes before the last heart. Every permutation of the deck is in exactly one of the sets $H$ and $S$, and we want to know the probability that a randomly chosen permutation is in $H$.
Let $p$ be a permutation in $H$. Go through the deck and interchange the first heart with the first spade, the second heart with the second spade, and so on, until you’ve completely interchanged the hearts and spades. Call the new permutation $p'$; it will be in $S$. And if you perform the same operation on $p'$, you’ll get $p$ back. In other words, we can pair each permutation $p$ in $H$ with a unique permutation $p'$ in $S$: we have a bijection between $H$ and $S$. $H$ and $S$ must therefore be the same size, so exactly half of the $n!$ permutations of the deck are in $H$, and the probability that a randomly chosen one is in $H$ must be $\frac12$.
