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March 12th, 2012, 09:57 PM   #1
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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.
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March 12th, 2012, 10:02 PM   #2
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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?
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March 12th, 2012, 10:52 PM   #3
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Re: how to solve this card game probability problem

4 / (52 C 13)
I think!
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March 12th, 2012, 11:06 PM   #4
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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
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March 13th, 2012, 09:03 AM   #5
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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).
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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


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. )
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March 14th, 2012, 03:52 AM   #7
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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
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March 14th, 2012, 08:36 AM   #8
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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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