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)?
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 reciprocal
a) 26 points in 26 cards 5813137849/45969459858 0.1264565184571849575 7.907856488541357
b) 33 points in 26 cards 1568212/450680979 0.0034796498478361563 287.3852380928088
c) 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
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Posted by Charlie
on 2007-12-27 16:04:48 |
solution
