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 Bridge Probabilities (Posted on 2007-12-27)
In the card game Bridge, 4 players in teams of two play against each other. One pack of cards is dealt out among them, so that each player has 13 cards. Point values can be assigned to the picture cards - Ace is 4, King 3, Queen 2, and Jack 1, so there are 40 points in the deck.

An important part of Bridge is to make contracts (or bids) for a certain number of tricks, and a good indicator of your success is how many points you and your partner have combined. What is the probability that one team will have at least enough for:

a) a "game contract" - 26 points? (bidding to take most of the tricks)
b) a "small slam" - 33 points? (bidding to take all but one of the tricks)
c) a "grand slam" - 37 points? (bidding to take all the tricks)

As bonus questions, what are the odds that just your hand will be:
d) a Yarborough (0 point hand with highest card a 10)
e) a Royal Yarborough (0 point hand with no card higher than a 9)
f) a "game hand" (a hand with 26 or more points)?

 No Solution Yet Submitted by Rob No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: solution -- correction on bonus questions Comment 2 of 2 |
(In reply to solution by Charlie)

I see I misunderstood the bonuses. They involve only one hand of 13 cards.

So the probability of no card higher than 10, and therefore zero points, and therefore the answer to f), becomes

(36!/23!) / (52!/39!) = 48546/13340621

and of no card higher than 9

(32!/19!) / (52!/39!) = 5394/9860459 ~= 0.00054703335818 ~= 1/1828.0420837968

for a Royal Yarborough.

The difference is,

701220/226790557 ~= 0.0030919276766889 ~= 1/323.42283

for a non-Royal Yarborough.

 Posted by Charlie on 2007-12-27 16:21:10

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