There is a pile of $52$ cards with $13$ cards in each suit (diamonds, clubs, hearts, spades). The cards are turned over one at a time. At any time, the player must try to guess its suit before it is revealed. If the player guesses the suit that has the most cards and if there is more than one suit with the most cards, he guesses one of these, show that he will make at least thirteen correct guesses.
Attempt: At first, the probability for each suit is $\frac{1}{4}$. If the first card is, say, a diamond then, for the second card, the probability of diamonds, clubs, hearts, spades are $\frac{12}{51}, \frac{13}{51}, \frac{13}{51}, \frac{13}{51}$. So the player should guess the suit that has most cards, but I don't know how to show that he will make at least thirteen correct guesses.
Think of it this way:
Let's say there's a situation when there are still $13$ hearts in the deck, but there are fewer than $13$ cards of the other three suits (such a situation will surely arrive - you can prove it). But that means the player will guess "Heart" every time until the heart is actually drawn, so he will certainly guess at least one correct card, and there will, after that, still be $12$ hearts in the deck.
Now, you are left with $12$ hearts in the deck, and an unknown number of other suits. If there are fewer than $12$ spades, clubs and diamonds, the same argument from above applies. If not, keep drawing cards until only one suit has $12$ representatives in the deck.
Can you see the pattern?