Hello,
First time here and first question. I hope I am proceeding correctly.

My stepdad teaches a bridge class. He would like to know what the odds or
probability are of drawing ANY ( not a particular hand ) 13 cards out of 52.

Would the odds (or is the right word probability) simply be 52 X 13
or is it more likely some more obscure formula ?
Or to put it another way - how many different 13 card hands could be dealt from a deck of 52 ? In practical terms - you deal 13 cards - then you return the cards to the deck and deal another 13 - Is there a way to figure out how many different 13-hand cards are possible without repetition of any of the hands?

Thank you for your help. I would appreciate the formula as to how you
derive your answer.

Blurij my email is [email protected]

How many 13 card hands can be dealt from 52?

It is quite large. C(52,13)=635,013,559,600

This is based on combinations. \frac{n!}{k!(n-k)!}

where n=52 and k=13.

Are you familiar with the mathematics of this? That is, what a factorial is and so forth?

Thank you Galactus OK so it is a bit more than 52 x 13 lol
Actually I had thought of the factorial solutions but had no idea of the formula
I do understand factorial operations Thank you for the formula.
I have to admit I don't understand why the formula works for example
I don't understand why you have to multiply k! Times (n-k) !

But I got a stratospheric answer and I do thank you for that and your time

#1. Choose k elements from some set. In this case, choose 13 cards from 52.

#2. Arrange the chosen k elements in some order.


The number of ways to carry out #1 is, by definition, C(n,k).

The number of ways to carry out #2 is P(n,k)=k!*P(n,k).

By the multiplication principle, we have P(n,k)\cdot k!

Use the formula P(n,k)=\frac{n!}{(n-k)!}

So, we get \frac{n!}{k!(n-k)!}=C(n,k)

P stands for permutation. C stands for combination.

Google and you'll find lots.

Or perhaps you are used to the notation:

^n C _r? That's the same thing.

It's easier to start with why a permutation works. It's the same equation except without the k! In the denominator. So it only has (n-k)!

Ooopps... accidentally hit submit instead of preview... I'll be back. :-)

Let's try this again.

It's easier to start with why a permutation works. It's the same equation except without the k! In the denominator. So it only has (n-k)!

When I pick the first card there are 52 total possible outcomes. Then I am down to 51 cards left, and therefore 51 possible outcomes for that pick. Etc. Just to shorten it, I'm only going to pick 5 cards. So I have 52(51)(50)(49)(48) possible ways to pick these five cards. (It's a really big number too!)

That's not such a terrible thing to do on your calculator, but if you don't feel like it or it's a lot of numbers (13 would be a hassle), we have a shortcut:

\frac{({52}\cdot{51}\cdot{50}\cdot{49}\cdot{48})({ 47}\cdot{46}\cdot{45}\cdot{44}....{4}\cdot{3}\cdot {2}\cdot{1})}{({47}\cdot{46}\cdot{45}\cdot{44}.... {4}\cdot{3}\cdot{2}\cdot{1})}

Which is:
\frac{52!}{(52-5)!}\ =\ \frac{52!}{47!}
And you said understand the factorial thing.

So if we were to cancel out in both numerator and denominator the part from 47 on down, then we're left with the original of just multiplying 52 through 48. There wouldn't be much point in doing that manually. But if you use that permutation equation, it will multiply all the way down to the 1, and then by dividing by 47! You cancel out the ones in the numerator you didn't really want. In other words, I only want to multiply the first 5 cause I'm only picking 5 cards. Dividing by the 47! Cancels out the 47 we don't want and leaves us with only the first five.

Since you can use the factorial key on your calculator this is a shortcut way to multiply a bunch of numbers when it becomes a hassle to do it manually.

The problem with that is that if we're picking cards, order doesn't matter. That is, if I pick a Queen of hearts and a 2 of clubs, does it matter which one I picked first? No, it's still the same two cards. In other words, ABC and ACB are the same thing cause it didn't matter what order I picked them in. So for something where order doesn't matter, it's not a permutation. Cause a permutation is for when order matters. For a permutation ABC and ACB are two different things. But for a combination they're not. For picking cards I need a combination cause order doesn't matter. If I use a permutation, I'll also get BAC, BCA, CAB and CBA, and all six of those are the same thing if the order doesn't matter! That's 5 extras.

So the combination adds the 5! Into the denominator:

\frac{52!}{5!(47!)}

That 5! In there divides back out all those duplicates you would get if you did a permutation equation. That is, like the 5 extras in my example above.

Why it's 5! Is a little trickier to understand, let alone explain. I spent an hour one day pondering over that cause I insisted I would understand why, and I did figure it out. This probably won't make much sense, but if I were to add another letter and start doing ABCD, ABDC, ACBD, ACDB, etc. there are actually less duplicates as you work your way down the line. So it's like the 5 is the first set of duplicates, then you move one letter over and it becomes 4 sets of duplicates, etc. Hence the 5!

And after all that, I just hope that is what you were actually asking, since you asked why the equation works. :p

BTW, probability and odds are actually two different things. If your probability is .01, that is, one out of 100, then odds is 1:99. That is, the probability is 1 "for" out of 100 possible, whereas odds is 1 "for" and 99 "against."

Speaking of terminology, your very first question makes no sense. You asked what the probability was of picking 13 cards out of 52. Well, if you're picking 13, then the probably is 100% that you'll pick 13 of them. :) What you are meaning to say is how many ways are there to pick 13 cards, not what is the probability. How many ways there are to do it isn't the same as the probability of what might happen.

To get into a probability, you need to state what you expect of these 13 cards. Like, the probability that all 13 will be spades, or something like that. A probability is the number of ways to do something that results in some particular end (they're all spades), divided by the number of results that could conceivably happen (spades and non-spades both). The answer to your question is actually the denominator of that equation, because you haven't chosen any particular result for the 13 cards.

(And the probability that they're all spades is 1/635,013,559,600, so I hope you didn't bet on that. :p)

WOW Thank you again Galactus and Ultra Member Morgaine300.
That was really kind of you to get into such explanatary detail. Are you both math instructors? I kind of felt myself back in high school class, but I was actually paying attention this time. Funny how that is when you really want or need an answer or how to do it.
This is one fine community at Askmehelpdesk. So glad I found such great people.
Thank you again - most heartily. Blurij

Galactus I believe is either a high school or college math instructor. My degree's in accounting and I'm like a professional tutor I guess you could say, tutoring college level accounting and math. Most of the math I do, though, is business-related stuff, so basic algebra as applied to some business stuff and some stats. (I hated trig and calculus and that sort of thing.)

And as anyone who knows me at all knows, I hate plug n chug, so I like people who want to know why something works and am all too happy to explain it. Besides, next time someone asks, I can just link to this and never have to do it again. :p