What are the odds of being dealt a monochromatic hand?
#1
Posted 2013-May-15, 19:13
South's hand:
♠ KJT853
♥
♦
♣ AKT8764
North had the 3 missing Aces and a singleton Club Q
#2
Posted 2013-May-15, 20:39
FM75, on 2013-May-15, 19:13, said:
South's hand:
♠ KJT853
♥
♦
♣ AKT8764
North had the 3 missing Aces and a singleton Club Q
One night at the club one hand (spots approximate) was
♠AQJT953
♥
♦
♣AK8764
One friend of mine opened 1♠, heard a 4♣ splinter, and shot straight to the ice-cold 7♠.
Another time I had a 3=0=0=10 hand. That's the oddest one I've held.
#3
Posted 2013-May-15, 21:27
#4
Posted 2013-May-15, 21:49
#5
Posted 2013-May-16, 00:16
#6
Posted 2013-May-16, 01:04
If you played 24 hands a day, every day, you would expect to be dealt a monochromatic hand every 3 1/2 years.
#7
Posted 2013-May-16, 05:45
I held ♠AQJxxxxxxx,♣xxx
Partner showed me an 0454 18 ish and I bid 6♠ as did most of the room. At only 2 tables was this doubled, once against me, and the other time by somebody I partner occasionally. We emerged without a matchpoint between us after his partner led his singleton trump picking up his Kx, and against me they didn't lead a club, but -,AQxx,AQxxx,KQJx didn't help with both red suit finesses wrong.
#8
Posted 2013-May-16, 06:46
♠♥♦♣
Since you now need thirteen of a suit.
#9
Posted 2013-May-16, 08:11
EricK, on 2013-May-16, 01:04, said:
Agreed. Alternatively, there are 62,403,588 totally 2-suited hands (35,335,872 7-6-0-0, 19,876,428 8-5-0-0, 6,134,700 9-4-0-0, 981,552 10-3-0-0, 73,008 11-2-0-0 & 2,028 12-1-0-0); 2/6ths of these are monochromatic (in the sense of the OP), ie 20,801,196; add in the 4 1-suited hands for a total of 20,801,200, which is EricK's proportion of the total number of hands (635,013,559,600).
#11
Posted 2013-May-16, 09:18
Fluffy, on 2013-May-16, 00:16, said:
The chance of exactly two players having monochromatic hands is 12*combin(26,13)^2*(combin(26,13)-2)/(combin(52,13)*combin(39,13)*combin(26,13)) which is about 1 in 4000000. The chance of all four players having monochromatic hands is 6*combin(26,13)^2/(combin(52,13)*combin(39,13)*combin(26,13)), which is about 1 in 80000000000000.
Taking those into account, the probability that someone will have a monochromatic hand is 0.000130777... or about 1 in 7647.
#12
Posted 2013-May-16, 09:25
EricK, on 2013-May-16, 01:04, said:
If you played 24 hands a day, every day, you would expect to be dealt a monochromatic hand every 3 1/2 years.
This is the clearest method for determing the correct probability (which, if you reduce it, is about 1 in 30,528). Congrats.
Reminds me of a story from Jerry Machlin's book, Tournament Bridge: An Uncensored Memoir (a must for any tournament bridge player). He recounted a conversation between his uncle, Al Sobel, a legendary tournament director, and Ozzie Jacoby, who, among other things, was the youngest actuary at the time he became one. Al walks up to Ozzie and asks him how many matches have to be played in a single-elimination KO event with an original field of 64 teams to determine a winner. Ozzie rattles off "32, 16, 8, 4, 2 and 1 - 63 matches." Al says, "Correct, but it took you too long. Since all the teams but one have to lose, it stands to reason that there have to be 63 matches played." Ozzie countered with "That is a nice solution, but if you were directing the event, the number of matches played could be any number between 55 and 120!"
#14
Posted 2013-May-16, 14:22
yes my opponents let him play 4♠X
#15
Posted 2013-May-16, 15:45
FrancesHinden, on 2013-May-16, 13:56, said:
9-3-1-0 or 5-3-3-2 are both odder
Even so ...
#17
Posted 2013-May-17, 00:40
PhilKing, on 2013-May-16, 06:46, said:
♠♥♦♣
Since you now need thirteen of a suit.
I was like WTF? Then I remembered I was colorblind. Are the BBF symbols 4 different colors?
#18
Posted 2013-May-17, 01:00