# Bridge problem

• Dec 1st 2012, 08:01 PM
howlop
Bridge problem
I am sitting at the bridge table and am willing to bid a slam if my partner (holding 13) cards has one or both of two specific cards. What is the probability of success and importantly what is the formula for calculating these odds.
• Dec 2nd 2012, 03:41 AM
BobP
Re: Bridge problem
The basic formula that you need is

$^{n}C_{r}=\frac{n!}{r!(n-r)!}$

which gives you the number of ways in which $r$ items can be taken from a group of $n$ items.

Since you hold 13 cards, the total number of possible hands that partner can hold is $^{39}C_{13}=\frac{39!}{13!(39-13)!}.$

Of these the total number of hands not containing either of your two specified cards will be $^{37}C_{13}=\frac{37!}{13!(37-13)!}$.

The second expression divided by the first gets you the probability that partner holds neither of the two specified cards and that works out to be approximately 0.4386.

Therefore the odds will be roughly 56-44 (slightly better than even money), in your favour.

This assumes of course that you have no information gleaned from any bidding. It's as if you have picked up your hand and are making the opening bid with no other information other that what you see in your own hand.
• Dec 2nd 2012, 04:54 AM
howlop
Re: Bridge problem
We have approached this problem somewhat differently and wonder if is valid.

We considered on any deal only a maximum of 6 cards is pertinent-the 2 significant cards will be distributed out in no more than 2 passes and either paired with each other or an irrelevent card.

Therefore the 6 cards can produce 15 combinations of which 9 contain a "key" card and thus a probability of .60

Is our approsch faulty?

By the way, I bid and made the slam.
• Dec 2nd 2012, 07:26 AM
BobP
Re: Bridge problem
Your method will only get you an approximation to the correct figure, and the problem is that there doesn't seem to be any indication (other than knowing the exact value) just how good an approximation you are going to get. (It could be a very poor approximation).
Why choose six cards ? Is there any particular reason for this choice, am I missing something ?
Had you chosen 5 cards you would have arived at an approximate probability of 0.7
Had you chosen 7 cards you would have arrived at an approximate probability of 0.524.
Choosing 8 cards produces a value of 0.464, etc..
The seven card result is fairly close, but how would you know ?
• Dec 4th 2012, 08:23 PM
babyboo
Re: Bridge problem
Bob, your solution is clearly dependant on the number of cards distributed e.g., a deck of 39 cards with 2 specific gives a different answer than does a deck of 6 cards with 2 specific cards. The question is why should the answer be dependant upon how many cards are in the deck if they are all distributed into 3 piles anyway? It appears that the chance of getting one specific card is 1/3 and the chance of getting the other specific card is also 1/3; since these odds do not account for the possibility that the person may be dealt BOTH cards (1/9) we would have a probability of 1/3 + 1/3 - 1/9 = 5/9 IRRESPECTIVE of how many cards are in the deck provided they are all distributed into 3 piles. Please explain.
• Dec 6th 2012, 01:54 AM
BobP
Re: Bridge problem
As I said in an earlier post, the assumption is that all that you know about the hand is contained in the 13 cards in front of you, (there has been no earlier bidding, there are no exposed cards etc.). There are 39 other cards and partner can hold any 13 of them.

The total number of possible hands for partner will be the number of ways in which 13 cards can be taken from the remaining 39 cards. That will be equal to $^{39}C_{13}$ and is equal to 8122425444. (The formula is stated earlier, if you wish confirmation find an elementary stats book and look in the combinations and permutations section).

Assume now, for sake of argument, that the two relevant cards are the Ace of Clubs and the Ace of Hearts. The pack remaining, excepting the 13 cards that we can see, consists of these two cards plus 37 others, (which are of no interest).
We can separate partner's hand into four possibles.
(1) Partner holds the Ace of Clubs but not the Ace of Hearts.
(2) Partner holds the Ace of Hearts but not the Ace of Clubs.
(3) Partner holds both aces.
(4) Partner holds neither ace.

The number of hands falling into each category will be,

(1) (Any 12 cards from 37) $^{37}C_{12}=1852482996,$
(2) (Any 12 cards from 37) $^{37}C_{12}=1852482996,$
(3) (Any 11 cards from 37) $^{37}C_{11}=854992152,$
(4) (Any 13 cards from 37) $^{37}C_{13}=3562467300.$

As a check,the sum of those four numbers should equal the overall total given in the second paragraph,and dividing each one by that number and multiplying each by 100 gives us the percentage of hands falling into each category.
That works out to be,

(1) 22.807 %
(2) 22.807 %
(3) 10.526 %
(4) 43.860 %

So, you could expect to make your slam 56.14 % of the time.

Working with 6 cards will only ever give you an approximation and you will not know just how good that approximation will be, 7 or 8 cards would get you better approximations.