Re: Bridge hands and probabilitiy


Taria wrote:
Hello all,

I am studying for my probability final coming up and aside from my head
getting jumbled up from all these theories, I'm stuck on this bridge
probability problem and I'm not sure how to go about solving it.

A bridge consists of 13 cards in a hand. Determine the probability
that a bridge hand has an 8-4-1 distribution 8 cards of one suit, 4 of
another and 1 of another.

My answer: C(4,1) * C(13,8) * C(4,1) * C(13,4) * C(4,2) * C(13,1)

Or what I think my answer would be.

What begins to confuse me as I try to analyze my answer and think it
through is wouldn't the 2nd "C(4,1)" start to mesh with the first
C(4,1)?
Yes.
In other words, how would you know if the first suit you chose
is not the second one?

Because you go through the problem in a sequential manner. You ask "How
many suits can I have for the 8 cards that are all of the same suit
(ans:4 of course)". Then you ask, "Given that I used up one suit on the
8 cards, how many suits can I choose for the 4 of the same suit (i.e.
3)", etc

Wouldn't it be C(3,1)? (I apologize for the
non mathematical technical language that I use but I'm simply not one a
math professional.)

Am I even right in the way I'm thinking?

Close.
Consider the suit you have 8 of.
You have 4 choices for the suit = C(4,1)
and there are C(13,8) possible ways you could draw the 8 from the 13.

Consider the choices for the suit of 4.
You've "used up" one suit earlier, so now you have a choice of C(3,1)=3
suits left
of which you can draw those 4 C(13,4) ways.

The last suit can be drawn from the remaining 2, so you have C(2,1)=2
choices for the last suit and there are C(13,1)=13 ways that can be
drawn.

So final answer is 4*3*2*C(13,8)*C(13,4)*C(13.1)

Once you see this easily, you'll notice that 4*3*2 is just C(4,3), i.e.
how many ways can we choose the 3 suits from the 4? The
C(13,8)*C(13,4)*C(13.1), then gives the number of ways that scenario
can be drawn for every choice of 3 suits.

HTH,

Matt

.

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