 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  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.) solution | Comment 1 of 2

10   dim Valu(40)
20   for A=0 to 4
30   for K=0 to 4
40   for Q=0 to 4
50   for J=0 to 4
60     V=4*A+3*K+2*Q+J
70     P=1:Arem=4:Krem=4:Qrem=4:Jrem=4:Crem=52
80     for I=1 to A
90       P=P*Arem//Crem:Arem=Arem-1:Crem=Crem-1
100     next
110     for I=1 to K
120       P=P*Krem//Crem:Krem=Krem-1:Crem=Crem-1
130     next
140     for I=1 to Q
150       P=P*Qrem//Crem:Qrem=Qrem-1:Crem=Crem-1
160     next
170     for I=1 to J
180       P=P*Jrem//Crem:Jrem=Jrem-1:Crem=Crem-1
190     next
191     Zrem=52-16:Hrem=26-A-K-Q-J
192     for I=1 to Hrem
193       P=P*Zrem//Crem:Zrem=Zrem-1:Crem=Crem-1
194     next
195     P=P*combi(26,A)*combi(26-A,K)*combi(26-A-K,Q)*combi(26-A-K-Q,J)
200     Valu(V)=Valu(V)+P
210   next
220   next
230   next
240   next
250   T=0
260   for I=26 to 40
270     T=T+Valu(I)
280   next
285   print T,T/1,1/T
290   T=0
300   for I=33 to 40
310     T=T+Valu(I)
320   next
325   print T,T/1,1/T
330   T=0
340   for I=37 to 40
350     T=T+Valu(I)
360   next
365   print T,T/1,1/T
370   T=0
380   for I=0 to 40
390     T=T+Valu(I)
400   next
405   print T,T/1,1/T

Division by 1 converts the rational number to a decimal.  Summation over zero through 40 is a check to make sure the probabilities add to 1.

The results:

`   at least:                                prob                             rational                  decimal                reciprocala) 26 points in 26 cards   5813137849/45969459858 0.1264565184571849575   7.907856488541357b) 33 points in 26 cards   1568212/450680979      0.0034796498478361563   287.3852380928088c) 37 points in 26 cards   5744/64382997          0.0000892161015741469   11208.73903203342`

so for example the odds of getting 37 or more points over 26 cards is 1 in about 11,209.

No card higher than 10 has probability

(36*35*34*...*13*12*11) / (52*51*...*28*27) = 11/21460999 ~= 0.00000051255768662 ~= 1/1950999.90909

e) No card higher than 9 has probability

(32*31*...*9*8*7) / (52*51*...*28*27) = 2/1094510949 ~= 0.0000000018273001305 ~= 1/547255474.5

d) So the probability of having no card higher than a 10, but having at least one 10 is the difference between these, or 13/25453743 ~= 0.0000005107303864897 ~= 1/1957980.23

And isn't f) the same as a) ?

Edited on December 27, 2007, 4:07 pm
 Posted by Charlie on 2007-12-27 16:04:48 Please log in:

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