What is the mathematical probability of dealing a Spades hand in which each player has only one suit?

Question

Meaning, one hand is all Spades, one hand is all Hearts, one hand is all Clubs, and one hand is all Diamonds.

Answer 1

I believe Spades is played by dealing out all 52 cards (13 each) just like bridge hands.

First let's figure the chance of getting one hand all the same suit. The total number of possible spade hands is 52C13 = 52·51·50·49·48·47·46·45·44·43·42·41·40 / 13! = 635,013,559,600, which is the number of ways of being dealt 13 cards from a deck of 52. The number of ways of being dealt 13 cards from the 13 spades in the deck is 13C13 = 13·12·11·10·9·8·7·6·5·4·3·2·1 / 13! = 1. That is, there is only 1 hand consisting entirely of spades. Thus, the probability of being dealt a hand of 13 cards consisting entirely of spades is 1 / 635,013,559,600.

Now let's figure the chance of dealing 13 hearts to the second player. This would be 39C13 combinations (with only one being all hearts). This is a probability of 1/8,122,425,444.

Next, the odds of dealing all clubs to the third player. Again 26C12 combinations results in a probability of 1/10,400,600.

Finally the last player can't help but get the remaining diamonds.

Putting this all together you have 1 / 635,013,559,600 * 1/8,122,425,444 * 1/10,400,600. But this only accounts for one arrangement of the suits. There are actually 4! ways to distribute the suits, so this would all be multiplied by 24. It's not a lot to offset this really tiny number.

The final odds would be about:
4.4738 * 10^-28 or approximately 1 in 45 octillion... seriously a small probability!

Answer 2

Obviously the odds are pretty small. You can figure this by dealing the hands one after the other instead of in rounds. The probability is the same but the visualization is much easier.

First deal any card - the probability for this is 1.
The next card must be of the same suit. There are 12 of them left in the remaining deck of 51, so the probability is 12/51. Each additional card reduces both the number of cards in that suit and the number in the deck by now so the next card has an 11/50 probability. After this it is 10/49, 9/49, 8/47, 7/46, etc. down to 1/40 for the last card of that suit. Since each event must happen, all the probabilities are mutlipied together. In factorial terms, the probability of dealing out this hand is (39!/51!)*12!

On to hand number two. Again, any card will suffice for the first one since any suit is OK. Of the remaining 38 cards, as before 12 will have the correct suit. Using the logic above, the odds of getting subsequent cards are: 12/38, 11/37, 10/36..... down to 1/27. Mulitplying the odds together and converting to factorial this is: (26!/38!)*12!

By now there is a pattern in the probability of subsequent hands. The next hand probability is: (13!/25!)*12!

And the last hand is: (0!/12!)*12!

The proability of the last hand is 1 as it should be. If the other three hands are each one suit, the last one must be. For the final answer, we mulitply the three probabilities together:

P = (39!/51!)*12! * (26!/38!)*12! * (13!/25!)*12!

You can just multiply but I will simplify first:

P = (12!)^3 * (39!/38!) * (26!/25!) * (13!/51!)
P = (12!)^3 * 39 * 26 * 13! / 51!
P = 4.474 x 10^-28

To put it in perspective, this means, on average, you would get 1 such hand during the age of the universe (about 12 billion years) if everyone on Earth (6 billion of us) dealt 1 hand per second.

Answer 3

The total ways 52 cards can be ordered is 52!, or 52 x 51 x 50 x ... x 2 x 1.

The total ways that the first 13 cards can be ordered is 4 x 13 (the first card can be anything!) x 12 x 11 x 10 x ... x 2 x 1. or 4 x 13!. The second set 13 cards can be of any remaining suit, so 3 x 13!. The third 13 cards can be any of two suits, so 2 x 13! The last set can only be 13!.

Multiplying this all out, the probability is...

4 x 3 x 2 x 1 x 13! x 13! x 13! x 13! divided by 52!

That comes out to about 1 chance in

2 235 197 406 895 370 000 000 000 000.

Answer 4

The answer would be (4! * 39! * 26! * 13!^4) / (52! * 39! * 26!), which simplifies to (4! * 13!^4) / 52!. Crunch those numbers and I think you'll find it ain't likely.

* * * *

OK, I got curious and did a rough calculation on the PC calculator. 52! is over 10^67, while the numerator is over 10^40. So, I think you are talking about 1 in 10^27, give or take an order of magnitude.

By the way, if you had this happen in an actual game, then you are playing with a practical joker. You can rig this yourself by sneaking a pre-set deck into the game, then doing a false shuffle, i.e. pretending to shuffle without actually altering the cards. Then let someone cut the cards, which doesn't change the outcome. Naturally, the other players will be aghast when they see their hands (heh heh).

Answer 5

Well let's see...

I'm assuming poker hands which are 5 cards each.

1*39/51*26/50*13/49
*12/48*12/47*12/46*12/45
*11/44*11/43*11/42*11/41
*10/40*10/39*10/38*10/37
*9/36*9/35*9/34*9/33

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Oh are all 52 cards are dealt?
1*39/51*26/50*13/49* (12!)^4/48!