# 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.

• My first reaction was: the trivial strategy of always say "spades" would give exactly 13 correct guesses but I presume that you are requiring that your specified strategy be used. – badjohn Sep 27 '17 at 11:42
• Am I right in that the total number of correct guesses will be $13$ + (number of times I make a change of suit)? – AnotherJohnDoe Sep 27 '17 at 12:18
• @AnotherJohnDoe. I think you know the answer now. – Dan Sep 27 '17 at 14:16

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?

• I get the idea but it does not seem to be true with specifically hearts. Suppose that the 13 hearts all come first. You will never be in a situation with all 13 hearts but fewer of the other suits. Also "other four suits"? How many suits do you have in your pack? – badjohn Sep 27 '17 at 12:09
• Why switch suits? If you stick with $Heart$ all the way down it will surely give you 13 right guesses. – Derk Jan Hulsinga Sep 27 '17 at 12:18
• You $have$ to change suits – AnotherJohnDoe Sep 27 '17 at 12:20
• @DerkJanHulsinga The question assumes you always guess the most probable suit. – 5xum Sep 27 '17 at 12:22
• @5xum.The number of correct guesses is the number of times $13\to12\to11\to\ldots\to0$ (13 times) plus the number of times that there are more than one suits with the most cards, am I correct ? – Dan Sep 27 '17 at 13:12

Hint: The last suit to get its first card picked will necessarily be guessed correctly.

• Do I understand your answer correctly, repeatedly guess the suit that has 13 cards until the guess is correct ? Then, repeatedly guess the suit that has 12 cards until the guess is correct and so on until the last card is correct. There will be at least 13 correct guesses. The extra correct guesses occur when there are more than one suits with the maximum number of cards. – Dan Sep 28 '17 at 1:03
• The extra correct guesses can occur when there are multiple suits that are "tied" for longest suit. It is not guaranteed that you will get extra correct guesses even then, however. Consider the first card drawn: you have four suits all tied at 13 cards each, but there is no guarantee that you will guess the first card correctly. – David K Sep 28 '17 at 2:16
• @Dan You say "until the guess is correct", which goes against the rules of the game. Instead, you guess a suit that has 13 cards left until no sit has 13 cards, the you guess a suit with 12 cards left until the are no suits with 12 cards left, and so on. Every time, you have to be correct about the last one. – Arthur Sep 28 '17 at 4:49
• @David K. I see my error now. Thank you. – Dan Sep 28 '17 at 5:38
• @Dan That is correct. For each number between 13 and 1, the last suit with that many cards left must be guessed correctly, while in the case of ties, you may get extra correct guesses. – Arthur Sep 28 '17 at 5:52

Show that before and after each card is drawn, the sum of number of correct guesses already made and the length of the longest suit remaining in the deck is always at least $13.$

• Sorry, I don't get it, will you please explain it more clearly ? – Dan Sep 28 '17 at 0:57
• For example, if the longest suit had 13 cards in it before you drew the last card, and now the deck has no suit longer than 12 cards, is it possible that your last guess was incorrect? (Other answers have already pointed out the answer is "no.") – David K Sep 28 '17 at 2:11
• I understand your explanation. Thank you. – Dan Sep 28 '17 at 5:21

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.