What are the odds of being dealt a monochromatic hand?

[FM75]

I was kibbing at a bidding practice.

 

South's hand:

 

♠ KJT853

♥

♦

♣ AKT8764

 

North had the 3 missing Aces and a singleton Club Q

[GreenMan]

I was kibbing at a bidding practice.

 

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.

[iamdavej]

Approximately 1 in 61055 (calculated as C(26,13) / C(52,13)) - number of ways of picking 13 cards from a universe of 26 divided by number of ways of picking 13 cards from a universe of 52.

[GreenMan]

Wouldn't you multiply by 2, because there are two colors? What you have there is the number of hands from any specific two suits (or any defined half of the deck) divided by the total number of hands, which accounts for one color, but we want to account for both. Or am I confused?

[Fluffy]

and by 4 players perhaps if you are a kibitzer, but thnen you would have to substract the hands were 2+ players have monochromatic hands because they shouldn't add on wich I would hate to calculate.

[EricK]

To get a monochromatic hand, the first card you pick up can be anything. The second must be the same colour as the first - which is a probability of 25/51. The third must be the same colour again - probability = 24/50. And so on down to the 13th with probabilty 14/40. Multiply all of them together and cancel down common factors, and you get a probability of 19/580,027 for any random bridge hand.

 

If you played 24 hands a day, every day, you would expect to be dealt a monochromatic hand every 3 1/2 years.

[Cyberyeti]

The one I remember comes from one of the early mind sports olympiads.

 

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.

[PhilKing]

The odds are much longer when you use the four-colour scheme:

 

♠♥♦♣

 

Since you now need thirteen of a suit.

[PeterAlan]

To get a monochromatic hand, the first card you pick up can be anything. The second must be the same colour as the first - which is a probability of 25/51. The third must be the same colour again - probability = 24/50. And so on down to the 13th with probabilty 14/40. Multiply all of them together and cancel down common factors, and you get a probability of 19/580,027 for any random bridge hand.

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).

[TylerE]

I was dealt ♠AKQJTxxx ♥ ♦ ♣AKJTx in a hand-shuffled club game once.

[campboy]

and by 4 players perhaps if you are a kibitzer, but thnen you would have to substract the hands were 2+ players have monochromatic hands because they shouldn't add on wich I would hate to calculate.

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.

[ArtK78]

To get a monochromatic hand, the first card you pick up can be anything. The second must be the same colour as the first - which is a probability of 25/51. The third must be the same colour again - probability = 24/50. And so on down to the 13th with probabilty 14/40. Multiply all of them together and cancel down common factors, and you get a probability of 19/580,027 for any random bridge hand.

 

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!"

[Finch]

 

Another time I had a 3=0=0=10 hand. That's the oddest one I've held.

 

That's not very odd at all.

9-3-1-0 or 5-3-3-2 are both odder

[Lord Molyb]

my partner once got ♠KQJxxxxxxx ♥ - ♦ - ♣ xxx and failed in 4♠ by a trick (I opened a weak 2♦)

yes my opponents let him play 4♠X ;)

[GreenMan]

That's not very odd at all.

9-3-1-0 or 5-3-3-2 are both odder

 

Even so ...

[mikl_plkcc]

That's not very odd at all.

9-3-1-0 or 5-3-3-2 are both odder

Are you sure?

[JLOGIC]

The odds are much longer when you use the four-colour scheme:

 

♠♥♦♣

 

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?

[Antrax]

Yeah, Diamonds are orange and clubs are green (and spades are actually blue, rather than black). Just how colorblind are you?

[EricK]

Yeah, Diamonds are orange and clubs are green (and spades are actually blue, rather than black). Just how colorblind are you?

I think he's asking whether you think the chances of a monochromatic hand are 1.