# What is the probability that only one of the cards will have the matching suit?

A group of 4 friends are playing a card game using a standard deck of 52 cards. Each friend receives one card from the deck, and the next card is flipped up. Only one of the friend has a card that matches the suit of the flipped up card. What is the probability that only one person is able to have a suit that matches the suit of the card that's flipped up?

My work (it's most likely wrong):

Let's say that the turned up card is a hearts. There are 13 hearts in a deck of hearts.

13*39*38*37= 712,842

Only one of the cards is a hearts (13). The rest of the cards are not hearts (39, 38, 37).

The probability that one player has the matching suit -> 13/712,842.

The number is too small, and I feel that I did something wrong.

Three things:

1. With the card being flipped one of the $$13$$ hearts, the player whose card is a heart as well must have one of the $$12$$ remaining hearts, not $$13$$. So you must use a factor of $$12$$ not $$13$$

2. There are $$4$$ possible players whose card is a heart, so you need to have a factor of $$4$$ in there.

3. Most importantly, you should not be dividing by the factors of $$12$$, $$39$$, $$38$$, and $$37$$: these are the possible target combinations that go in the numerator. The denominator are all the possible ways to get $$4$$ cards to the $$4$$ players, so there you get something like $$51$$, $$50$$, $$49$$, and $$48$$

OK, using these hints, try again!

• @N.F.Taussig Right, thanks! :P – Bram28 Dec 3 '18 at 12:22
• Since one card is placed on the table, there are $51$ cards left from which the four players can each select a card. – N. F. Taussig Dec 3 '18 at 12:27
• Thanks! Would it be instead 51, 50, 49, and 48 since one of the cards is already used up? – Lauren Pablo Dec 3 '18 at 12:28
• @laurenpablo exactly! – Bram28 Dec 3 '18 at 13:01