# Thirteen cards and 635,013,559,600 possibilitie (geek out, fun with numbers)

Is the name Patrice has given to his bridge section on his web page.

Somehow, this got my imagination running and I wondered… is this true?

Lets see…

A deck has 52 cards, which is split evenly between 4 people (13 cards each).
Then people bid on how many tricks they’ll win.
Then, everone has a turn playing cards, trying to win tricks and make contract.

I decided to not take into account the bidding part. Just the playing of the cards.

My first idea was to try and determine how many hand combinations were possible, then determine how many ways a combination could be played.

That was the idea that is actualyl fun to calculate.

If I remember my stats mathds well, drafting 1 card out of a deck has 52 possibilities. Then, I have 51 cards left, so there’s 51 possibilities for drafting my second card, and so on. So, number of possibilities = 52x51x50x…

1 If I draft the whole 52 cards , it means I have 52! (factorial notation, if my mind serves me well) possibilities, which gives 8.06e67 possibilities to split the 52 cards.

But I don’t want to take into account the order of the cards. A hand with all hearts is the same, no mather in what order you have them!

2 I figure, for a given hand (13 cards), there is 13! ways (orders) for it to be found.

3 So, 52! deck distributions possiblities / 13! hand distribution possibilities = 1.2952932e58 ways to split the deck.

Good. Now, I want to know how many possibilities to play these hands.

1st trick, there’s 13 options for each player. 13x13x13x13 possibilities (13^4).
2st trick, 12 options left. 12x12x12x12 (12^4).
And so on.

4 So I figure there’s 13!^4 ways to play these hands. Wich means roughly 1.5035617e39 possibilities.

So if there’s (3) 1.2952932e58 ways to split the deck in 13 card hands and there’s (4) 1.5035617e39 ways to play the hands,

5 There should be about 1.9475533e97 ways a game could be played. (Or 25% of that, if you don’t want to take into account the possible rotation of the hands around the table).

Then I had another way to calculate it.

1a 52 cards, played one after the other. 52!. 8,0658175e67 possibilities.

Could be as simple as that.

Either way, I’m not even close to what I wanted to reach (and I tried a few other ways which all end up with humongous numbers that vary wildly).

How would you go about calculating this?

## 6 thoughts on “Thirteen cards and 635,013,559,600 possibilitie (geek out, fun with numbers)”

1. I’m lazy.. I googled the number, and this page makes a lot of sense to me:

so the number is the number of possible opening hands.. interesting 🙂 so you can stick that number to any game that distributes the whole pack to 4 players, but only bridge players are nerdy enough to actually do it *grin*

2. Also your #1 ain’t right.. your starting hand isn’t (52!). Your first card comes from a pool of 52 cards, the next from a pool of 51, next from a 50 card pool… for 13 cards. So it’s more

a = 52x51x50x49x48x47x46x45x44x43x42x41x40

Then for the 13 cards you get, you don’t get about the order.. There is where you factor something.. 13 cards combinations = (13!)

b = 13x12x11x10x9x8x7x6x5x4x3x2x1

Calculate your chances of getting a ‘b’ from the ‘a’ pool:

a / b = magic number 🙂

This is how I understand it from the google results 😉

3. Hmm… I’ll have to study the site you pointed to… I’m not catching all of it at this moment…

You’re right – my starting hand ain’t 52!. But I’ve taken that into account. And there are 4 hands total. To be more verbose about it:

52! / 39! * 39! / 26! * 26! / 13! * 13!

The rationale is that 52x51x50x […] x41x40 is equal to 52! divided by 39!
If you simply all of that, you end up with 52! / 13!… (which I mixed up with #3 when I typed this)

But in these number of possible 4 hands, there are multiple equivalent possibilities, which I equated with a 13! factor. So, by my rationale, #3 should be (52! / 13!) / 13! which would end up giving 2.08012e48… which still leaves me in the left field.

Oh, I get it! The 635,013,559,600 is for the probabilities of a single hand while I’m trying to evaluate the number of possible different games. That’s why!!

Total different ball game, then. It should simply end up being 52!/39!/13! (by the rationale explained up there). Which gives out 6.350114e11 – which is the big-ass number I was looking for! AAAAaaaah!

4. Yeah, the label on the ptaff.ca/ page could express the total number of possible games. Taking into account the bidding stage, it would bring a number needing a minimal 1600×1200 resolution at 8pt to see I guess 🙂

5. Powerlord dit :

"You’re right – my starting hand ain’t 52!. But I’ve taken that into account. And there are 4 hands total. To be more verbose about it:

52! / 39! * 39! / 26! * 26! / 13! * 13!"

There’s a slight problem in your logic. Card hands are a combination rather than a permutation. Thus each hand should have an additional 13! in the divisor, like so:
52! / (13! * 39!) * 39! / (13! * 26!) * 26! / (13! * 13!) * 1

The 1 is actually 13! / 13!, because once the other three hands are dealt, there’s only one remaining combination of cards.

6. It’s been awhile since this post, so I don’t remember too well. But the expression you are writing seem to be the equivalent of the one I gave in the previous answer. Can’t you simplify out the two first 13! and the 1 from there?

From:

52! / (13! * 39!) * 39! / (13! * 26!) * 26! / (13! * 13!) * 1

To:

52! / (39!) * 39! / (26!) * 26! / (13! * 13!)

Which would end up exactly like the one you quoted…

If you were looking for some fun wit the actual 635013559600 number, you might want to check out my other post about 635013559600