[shevek]

Can someone work out the chance of picking up an exactly average hand, one of everything?
So 1 ace, 1 king, etc .... 1 two.

Or point me to a resource ...

TAI

[barmar]

The first card dealt can be anything, so it doesn't affect the probabilities. Let's assume it's an Ace.

The second card can be anything other than an Ace. Of the remaining 51 cards in the deck, there are 48 cards, so it's 48/51 that it will be something else.

Next, of the remaining 50 cards in the deck, there are 44 possibilities.

And so on.

So it's 48/51 * 44/50 * 40/49 * 36/48 * ... * 4/40. This simplifies to 4^12*(12!)/(51!/39!).

I think the value of this is .0106%. If you play one session of bridge every day, you could expect to see about one a year.

[nige1]

Same as Barmar but slower

[banrock]

I'm sure your maths is correct but I can't help thinking your logic is flawed. on dealing the 2nd card to oneself there are only 48 left in the pack as, (assuming a normal deal and not a goulash), partner and ops have three between them

[shevek]

 barmar, on 2014-April-12, 19:58, said:

The first card dealt can be anything, so it doesn't affect the probabilities. Let's assume it's an Ace.

The second card can be anything other than an Ace. Of the remaining 51 cards in the deck, there are 48 cards, so it's 48/51 that it will be something else.

Next, of the remaining 50 cards in the deck, there are 44 possibilities.

And so on.

So it's 48/51 * 44/50 * 40/49 * 36/48 * ... * 4/40. This simplifies to 4^12*(12!)/(51!/39!).

I think the value of this is .0106%. If you play one session of bridge every day, you could expect to see about one a year.

Thanks for that

[psyck]

Another was of approaching this would be...

The number of ways of selecting one Ace = 4
The number of ways of selecting one Ace and one King = 4*4
The number of ways of selecting one Ace, one King, and one Queen = 4*4*4
The number of ways of selecting one of each denomination = 4^13

The number of ways of selecting 13 cards = 52C13 = 52!/13!(52-13)!

The possibility of being dealt one of each denomination = 4^13/52C13 = 67,108,864/635,013,559,600

Which is approximately a 1 in 9462 chance, or roughly 0.0106 probability that Barmar derived.

Similarly...
the chance of being dealt a 9 high hand (Yarborough) is 32C13/52C13 or 1 in 1828
the chance of being dealt a 8 high hand is 28C13/52C13 or 1 in 16960
the chance of being dealt a 7 high hand is 24C13/52C13 or 1 in 254397
...
the chance of being dealt a 5432 (any) 432 432 432 hand is 158,753,389,900 to 1

[dave_beer]

 barmar, on 2014-April-12, 19:58, said:

The first card dealt can be anything, so it doesn't affect the probabilities. Let's assume it's an Ace.

The second card can be anything other than an Ace. Of the remaining 51 cards in the deck, there are 48 cards, so it's 48/51 that it will be something else.

Next, of the remaining 50 cards in the deck, there are 44 possibilities.

And so on.

So it's 48/51 * 44/50 * 40/49 * 36/48 * ... * 4/40. This simplifies to 4^12*(12!)/(51!/39!).

I think the value of this is .0106%. If you play one session of bridge every day, you could expect to see about one a year.

It is easier if you include the first term which is 52/52. Then it simplifies to (4^13)*(13!)*(52!/39!) or (4^13)/(52!/13!*39!) which are the two terms in Psych's answer.

Dave

[madhu1955]

Mathematically the answers are right but in practice it happens more often (especially where Bridge has become a pastime of old men and cards are dealt manually). Many times I have seen a new deck being dealt without being shuffled and all 4 players being dealt A 10 6 2, K 9 5,Q 8 4,J 7 3 in different suits. Many times the hand gets passed out and some later table realises that the cards are not shuffled and inform the director

[barmar]

 banrock, on 2014-April-13, 03:24, said:

I'm sure your maths is correct but I can't help thinking your logic is flawed. on dealing the 2nd card to oneself there are only 48 left in the pack as, (assuming a normal deal and not a goulash), partner and ops have three between them

For purposes of determining the probability of cards in one hand, you can ignore the fact that we're simultaneously dealing to 3 other hands. The cards that were dealt to those hands aren't known, so any of the 51 other cards in the deck could have been dealt to this hand. And as far as probilities are concerned, there's no difference between dealing NESW-NESW-NESW... and NNNNNNNNNNNNN-EEEEEEEEEEEEE-... (in pratice it makes a difference because shuffling isn't perfect, but when calculating the probailities we assume the cards are truly random).

[Siegmund]

Nothing to add to the maths... but can tell you that I learned about the game of Chinese Poker, and the "dragon hand," in the summer of 2004. My partner was dealt one, and explained what Chinese Poker was to the rest of us at the table afterward.

I have had my eyes open for them at the bridge table ever since. I had one within a year, and one in 2010.

Given that I was playing two club sessions a week when I was in Alaska, and a bit less at the club but a lot more at sectionals since I moved - that gives me 2500 or 3000 live hands a year, so it looks like I am "right on schedule" for seeing them about once in 9000 trials.

Unlike madhu, I have never seen an unshuffled pack dealt out, though I have seen a lot of new packs shuffled and then put into boards. (Admittedly in recent years I've mostly been either at tournaments with hand records or at a club with a dealing machine.)

I have, however, seen an 8-high hand more recently than I have seen a dragon hand.