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 March 12th, 2012, 09:57 PM #1 Senior Member   Joined: Apr 2008 Posts: 194 Thanks: 3 how to solve this card game probability problem In the game of bridge, four players are dealt 13 cards each from a well-shuffled deck of 52 playing cards. What is the probability that one of the players holds a hand that is made up of only one suit? my attempt let C represent combination There are 52C13 ways to choose any 13 cards. Since there are 4 players, each player has an equally likely chance to get only on suit. So, I multiply 1/52C13 by 4. Thus, my answer is 4/52C13. My friend says that my answer is wrong. So, how do you do it. Please help.
 March 12th, 2012, 10:02 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: how to solve this card game probability problem What about the 4 ways for each player to get only one suit, since there are 4 suits?
 March 12th, 2012, 10:52 PM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: how to solve this card game probability problem 4 / (52 C 13) I think!
 March 12th, 2012, 11:06 PM #4 Senior Member   Joined: Jan 2012 Posts: 131 Thanks: 0 Re: how to solve this card game probability problem 52C13 ways, but 4 ways are similar suit. Each draw has probability of 4/52C13 of similar suit . 4 draws(4 persons)= 16/52C13
 March 13th, 2012, 09:03 AM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: how to solve this card game probability problem Hmmm...isn't it same as player being dealt 13 cards consecutively? S0: 1 * 12/51 * 11/50 * ..... * 2/41 * 1/40 = ~.000000000006299078 Which is same as 4 / (51C13).
March 13th, 2012, 09:17 PM   #6
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Re: how to solve this card game probability problem

Quote:
 Originally Posted by davedave In the game of bridge, four players are dealt 13 cards each from a well-shuffled deck of 52 playing cards. What is the probability that one of the players holds a hand that is made up of only one suit? my attempt let C represent combination There are 52C13 ways to choose any 13 cards. Since there are 4 players, each player has an equally likely chance to get only on suit. So, I multiply 1/52C13 by 4. Thus, my answer is 4/52C13.
In my humble opinion, I think the probability that one of the players holds a hand that is made up of only one suit should be very, very small.

I think we should also make sure that the other three players must not hold cards of same suits as well.

My very rough calculation gives an approximate answer of
$1.1786\times10^{-18}$

I'll explain more if you think my answer looks right to you.

(Edit: I'm sorry, my figure seems wrong (but it must be an extremely small value)! Again, if you think I'm on the right track, I can give you the overall concept and idea because I don't think I have time to do it now. )

 March 14th, 2012, 03:52 AM #7 Senior Member   Joined: Jan 2012 Posts: 131 Thanks: 0 Re: how to solve this card game probability problem This is the same question taken from a book. In the game of bridge, each of 4 players is dealt 13 cards from an ordinary well-shuffled deck of 32 cards. Find the probability that one of the players(say, the eldest) gets a complete suit. ans: 4/52C13. Here the question emphasized on ONE. Thus the probability= 4C1/52C13= 4/52C13. Let the eldest player no2. xHxx|xxxx|.... xSxx |xxxx|x.. xDxx|xxxx|x... xCxx|xxxx|x.... ->52C13 points The probability of all players get a complete suit must be 4C4/52C13=1/52C13
 March 14th, 2012, 08:36 AM #8 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: how to solve this card game probability problem Aziz, can you show that "clearly" by using only 12 cards: 3 of each suit.

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