Math Is Fun Forum

EMPhillips1989

Member

Registered: 2008-01-21

Posts: 40

probability

it is assumed that all

bridge hands are equally likely, what is the probability of being dealt 13 cards all the same suit??

does anyone know how to do this question??? please help direct me

Last edited by EMPhillips1989 (2008-01-21 09:34:41)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: probability

There are only four ways of getting a single-suited hand, namely all clubs, all diamonds, all hearts or all spades.

As you said, there are [sup]52[/sup]C[sub]13[/sub] possible hands in total.

So to find the probability, divide the first number by the second one.
Intuitively, I'd guess the answer has to be somewhere around a billionth.

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Why did the vector cross the road?
It wanted to be normal.

pi man

Member

Registered: 2006-07-06

Posts: 251

Re: probability

635,013,559,600 / 4 = 158,753,389,900 .

That's (52 choose 13) divided 4. 

Did that happen to you EMPhillips?   If it ever does, just realize that with you having all of the cards in one suit, the chances of one of the other 3 players  having all 13 cards of one of the other suits drops to a mere 1 in 2,707,475,148!