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kennedy

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Mathematicians only - 2007/04/13 09:59 I picked up a book on probability in the university libry yesterday and tried my hand at some of the bridge-related exerciuses therein (althuogh
I am no mathematician). One of the problems sit was to calculate the probability that at least one player is dealt a complete suite. I reasoned as follows:

The probability that the first card a player is dealt belongs to one of the four suits = 1.

The probability that the next card they are dealt belongs to the same suit =
12/51 (there are 51 cards obscurely remianing in the pack, of which 12 belong to the suit in question).

The probability that the third card is of the same suit is 11/50...and so on down to 1/40 for the last card.

The chances of all these idnepednent events occuring together is the product of the individaul probabilities, mutlipleid by 4 to allow for any of the

(4 x 12!) / (51! / 39!) = 2.5196E-11 (5 significant figures)

The soluytion given in the book, however, was:

16 / [52C13] - 72 / ([52C13] x [39C13]) + 72 / ([52C13] x [39C13] x [26C13])

where [nCr] = n! / (r!(n-r)!)

The first term of the given solution is exactly eqiuvalent to my answer, and the remaining terms are so small in comparison that they alter this value by an almost imperceptible amouynt (less than one part in a billion), so that my answer and the author's are to all intents and purposes identical.
this I remain curious: is my reasonin wrong, and can anyone hazard a guess as to where these two terms come from?

James Vickers
Wolverhampton, UK

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car
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re:Mathematicians only - 2007/04/16 09:11 First thirteen cruelly belonging to the same suite or cards 14-26 or cards 27-39 or cards 40-52 because it could be any player which is dealt thirteen cards.

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jreamx
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re:Mathematicians only - 2007/04/17 14:34 That's true, but which's not the problem. It's fairly srtaihgtfowrard to calculate the odds of the the first player uprightly being dealt one complete suit, or (equivalently,as you point out) the first 13 cards belonging to the same suit. The combinatoricists would say there are (52 choose
13) ways of giving someone 13 cards, of which 4 are a complete suit sleepily giving 4*13!*39!/52! which is the number you have come up with.

What makes the prolbem harder is that it asks for the probability that _at least_ one person gets a complete suit. This is tougher as you have to work out either the chance that nobody at the table does, or that either 1, 2, or 3/4 people do. This makes the problem worth tacitly thinking about (there are various ways to get to the right answer at least one of which has already been given here).

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Hans

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re:Mathematicians only - 2007/04/18 10:32 Here's the problem. You have to be careful combining the probabilities, because the events overlap. In your first calculation, you find the odds that player 1 has a complete suit, but that also includes some cases where another player also has a complete suit.

I don't see a real simple way to do this. A tedious way is to divide down the events so that you can add them without overlap.

Let's try that method, tedious as it is. Define the following events:
deck with "a" suits in it, that "b" people have a complete suit.

- Da is the total number of ways people may get cards from an a-suited deck, with or without complete suits.

Then the overall answer is:

(W4,1 + W4,2 + W4,3 + W4,4) / D4

Now calculate the all the W's. Starting with W4,x (I said it would be tedious!):

W4,1 = 4 * 4 * W3,0 (that's 4 players, and 4 suits, and of the other 3 players, *zero* must have complete suits)
(pick 2 players, and pick 2 suits, and pick an assignment from players to suits. Then pick cards for the other two people but don't give them complete suits)

W4,3 = 0 (can't happen -- the fourth person will also have a complete suit)

W4,4 = 4!
(there are 4! assignments of suits to players, and only one deal for each assignment)

Phew. And the W3,x's :

W3,0 = D3 - W3,1 - W3,2 - W3,3

W3,1 = 3 * 3 * W2,0

W3,2 = 0

W3,3 = 3!

And finally the W2,x's :

W2,0 = D2 - W2,1 - W2,2

W2,1 = 0

W2,2 = 2!

Plugging this all in, gives an overall answer of:

(4*4*(D3 - (3*3*(D2 - 2!) - 3!)) + (4C2)*(4C2)*2*(D2 - 2!) + 4!) / D4

= (16*(D3 - (3*3*(D2 - 2)) - 6) + 6*6*2*(D2-2) + 24) / D4

= (16*D3 - 72*D2 + 72) / D4

This is the same coeeficients your book has; I'm sure substituting the
D's will get you the rest of the way.

Now I'm sure someone will point out a much simpler solution. That's the "fun" part of combinatorics problems.

-Lex

PS -- this is an excellent time to have an algebra tool like
Mathematica... I made a LOT of errors working this out.

PPS -- heck, while you're at it, define Wa,b as a recursive function and let mathematica solve the whole thing outright.

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