1♥ 2♥ pa 2♠
pa pa ??
We have 3163.
If we fix:
Partner: 5+♥
RHO: 5+♠5+m
LHO: 2-3♠
What is the percentage that RHO holds clubs, as his minor suit?
By intuition, I would say ~85%, but what says the math?
Page 1 of 1
A little bit of math
#2
Posted 2010-May-11, 01:37
I don't know about the actual numbers, but we can do a quick calculation as follows. Assume that partner has 5 hearts, LHO has 2 spades, and RHO has 5 spades.
This means that there are 3 spades, 7 hearts, 7 diamonds, and 10 clubs we don't know about (27 cards). This also means LHO has 11 cards, partner has 8 cards, and RHO has 8 cards we don't know about.
We deal RHO 8 cards from the 27 card "deck". We can count the number of ways to deal him a certain number of a minor. An example calculation for "probability he has 5 clubs" is
binomial(10,5) * binomial(17, 3) / binomial(27,8)
This is the number of ways to choose 5 clubs, multiplied by the number of ways to choose his remaining 3 cards, divided by the total number of possible 8-card hands.
CLUBS
5: .077
6: .0129
7: .0009
8: 0
TOTAL: .0908
DIAMONDS
5: .0108
6: .000599
7: 0
TOTAL: .01679
Given that one of these events occurred then, it is 84.4% that he had clubs rather than diamonds.
The same calculation by assuming RHO has 6 spades (LHO has 2 spades, RHO has 5 hearts still) yields a 90.1% chance that he has clubs.
What is the actual number? In general having more known cards (like knowing LHO has 3 spades or partner has 6 hearts), as well as increasing RHO's number of spades, will increase the probability that he has clubs. The real number is probably hovering around 87-88.
This means that there are 3 spades, 7 hearts, 7 diamonds, and 10 clubs we don't know about (27 cards). This also means LHO has 11 cards, partner has 8 cards, and RHO has 8 cards we don't know about.
We deal RHO 8 cards from the 27 card "deck". We can count the number of ways to deal him a certain number of a minor. An example calculation for "probability he has 5 clubs" is
binomial(10,5) * binomial(17, 3) / binomial(27,8)
This is the number of ways to choose 5 clubs, multiplied by the number of ways to choose his remaining 3 cards, divided by the total number of possible 8-card hands.
CLUBS
5: .077
6: .0129
7: .0009
8: 0
TOTAL: .0908
DIAMONDS
5: .0108
6: .000599
7: 0
TOTAL: .01679
Given that one of these events occurred then, it is 84.4% that he had clubs rather than diamonds.
The same calculation by assuming RHO has 6 spades (LHO has 2 spades, RHO has 5 hearts still) yields a 90.1% chance that he has clubs.
What is the actual number? In general having more known cards (like knowing LHO has 3 spades or partner has 6 hearts), as well as increasing RHO's number of spades, will increase the probability that he has clubs. The real number is probably hovering around 87-88.
#3
Posted 2010-May-11, 05:01
Very nice, Roger.
The world would be such a happy place, if only everyone played Acol :) --- TramTicket
Page 1 of 1