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On a random hand, what is the probability that you and your partner have an 8-card or greater fit in some suit?
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Bridge Probability Question
#2
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Posted 2013-July-05, 10:49
I will leave this to others to calculate. But it seems to me that the basic problem is how many random assortments of 26 cards will have at least 8 in one suit? Obviously, the percentage will be quite high, as the only assortments of 26 cards that do not have at least 8 in one suit are 7-7-6-6 and 7-7-7-5 (in any order).
#3
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Posted 2013-July-05, 11:43
I'm lazy, so I wrote a dealer script which estimates that the probability is approximately 15.73%
Quote
seven =
spades(north) + spades(south) <= 7 and
hearts(north) + hearts(south) <= 7 and
diamonds(north) + diamonds(south) <= 7 and
clubs(north) + clubs(south) <= 7
action
average seven
spades(north) + spades(south) <= 7 and
hearts(north) + hearts(south) <= 7 and
diamonds(north) + diamonds(south) <= 7 and
clubs(north) + clubs(south) <= 7
action
average seven
Alderaan delenda est
#4
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Posted 2013-July-05, 11:52
Yes, about 15.736% that you don't.
You may as well shuffle your hand with partner's as it makes no difference who has what.
There are combin(13,7)=1716 different ways to have 7 spades. Likewise there are 1716 different ways to have 6 and 1287 ways to have 5. There are 6 different patterns for 7,7,6,6, and 4 patterns for 7,7,7,5. So the total number of ways is 6*1716^4+4*1287*1716^3. Then divide by combin(52,26) which is the number of possible sets of 26 cards the two of you can have.
The chance that neither pair has an 8-card fit is about 10.49%
You may as well shuffle your hand with partner's as it makes no difference who has what.
There are combin(13,7)=1716 different ways to have 7 spades. Likewise there are 1716 different ways to have 6 and 1287 ways to have 5. There are 6 different patterns for 7,7,6,6, and 4 patterns for 7,7,7,5. So the total number of ways is 6*1716^4+4*1287*1716^3. Then divide by combin(52,26) which is the number of possible sets of 26 cards the two of you can have.
The chance that neither pair has an 8-card fit is about 10.49%
#5
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Posted 2013-July-05, 12:29
# Trying to learn Dealer Script and confirming Hrothgar's result NStrumps = 8 EWtrumps = 8 NSfit = spades(north) + spades(south) >= NStrumps or hearts(north) + hearts(south) >= NStrumps or diamonds(north) + diamonds(south) >= NStrumps or clubs(north) + clubs(south) >= NStrumps EWfit = spades(east) + spades(west) >= EWtrumps or hearts(east) + hearts(west) >= EWtrumps or diamonds(east) + diamonds(west) >= EWtrumps or clubs(east) + clubs(west) >= EWtrumps action average NSfit, average EWfit, average NSfit and EWfitYes
About 84% of deals: you have have an 8+ fit.
About 79% of deals: both sides have an 8+ fit.
Test: changing NStrumps to nine
About 38% of deals: you have have an 9+ fit.
If you have a 9+ fit, then opponents have an 8+ fit.
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