Card guessing game 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.

This is based on 5xum's answer but I think a better explanation...
Initially there are four suits that have equal number of cards. Let us assume we actually are very unlucky and guess incorrectly until there is only one suit left with 13 cards in. We will then keep guessing that suit until the first card in that suit is drawn and we have one correct guess.
We then have at most 12 cards in any one suit. We then repeat the logic... If there are multiple suits with 12 cards assume we guess incorrectly until only one suit has 12 cards left in it. We will then be guessing that suit until a card is drawn from it. We then have 2 correct guesses and at least one suit still has 11 cards in it.
We then keep repeating that logic until we get down to 12 guesses and at most one card in any suit. We then guess wrong until the last card which we guess correctly which gives us our 13 correct guesses.
This is obviously the worst case scenario since we assumed incorrect guesses unless we were guaranteed a correct guess. Obviously we could have done a lot better if some of those guesses were correct.
