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Odds

Walter Wayne said:
As a result, the standard deck of cards can be ordered in 52 x 51 x 50 x ... X 3 X 2 X 1 ways, or simply 52!.
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In case anyone is interested, 52! = 8 x 10^67, which is a very large number. So large that no matter how many card games people play with a 52 card deck, and no matter how many times the decks are shuffled, all possible sequences of 52 cards will never occur.
 
Really? I got (8.065817517...)*10^67. (Which is basically the same as what joe87 said, except that I think that he should have made it more clear that it is an approximation). What did you get? 156?
 
Art Vandelay said:
Really? I got (8.065817517...)*10^67. (Which is basically the same as what joe87 said, except that I think that he should have made it more clear that it is an approximation). What did you get? 156?
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Well, yeah! It's larger by at least 6*10^65, which is an incredibly HUGE number!

- Timothy
 
joe87 said:
In case anyone is interested, 52! = 8 x 10^67, which is a very large number. So large that no matter how many card games people play with a 52 card deck, and no matter how many times the decks are shuffled, all possible sequences of 52 cards will never occur.
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Obviously wrong. It will repeat ever 10^67 times.

:)
 
Walter Wayne said:
Exclaimation marks signify "factorial". Basically, n-factorial is equal to the product of the integers from 1 to n. So ...
3! = 3 x 2 x 1 = 6
6! = 6 x 5 x 4 x 3 x 2 x1 = 720
The reason factorial occurs so often in counting things is the nature of the beast. How many ways can a standard deck be ordered. Well, the first card can be any one of 52 cards. After selecting the first card, the second can be any one of the remaining 51 cards, the third any one of the remaing 50, the fourth ...

As a result, the standard deck of cards can be ordered in 52 x 51 x 50 x ... X 3 X 2 X 1 ways, or simply 52!.

So in the original problem, you have 5 cards. So if you put them in random order, you would have 5! possible shuffled decks.
5! = 5x4x3x2x1=120
When you wrote down as you mentioned in your very first post that you wrote down all the combinations, how many do you get? 120 I expect.

Walt
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Well I was only interested in the colour, not the value, so it was many less, just 10.

factorial.jpg



I then counted how many columns had 2 reds in the first 3 rows. There are three, so that makes 3 from 10, or 30%.

Thanks for the info on the bang symbol, makes sense now cheers.
 
Rob Lister said:
Obviously wrong. It will repeat ever 10^67 times.

:)
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Actually, it doesn't take that long. He said "all possible sequences of 52 cards will never occur" but I think that he meant "not all possible sequences of 52 cards will ever occur".
 
Just thinking said:
Correct .. but then how does one get 0! = 1? ;)
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If you have zero objects, in how many different ways can they be ordered? One way: the order consisting of no objects.

The same way that zero is a perfectly good number, a sequence of length zero is a perfectly good sequence.

There are six ways to order three distinct objects:
(1, 2, 3)
(1, 3, 2)
(2, 1, 3)
(2, 3, 1)
(3, 1, 2)
(3, 2, 1)

There are two ways to order two distinct objects:
(1, 2)
(2, 1)

There is one way to order one object:
(1)

There is one way to order zero objects:
()

Makes perfect sense. Is () not an ordering of zero objects? (If not, what's wrong with it, exactly? It has the correct number of items---zero, in this case---and no repetitions of any item, just like all the other orderings above. What else does it need?)
 
69dodge said:
Makes perfect sense. Is () not an ordering of zero objects? (If not, what's wrong with it, exactly? It has the correct number of items---zero, in this case---and no repetitions of any item, just like all the other orderings above. What else does it need?)
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Yes, that's how I learned it and understood what a factorial meant as opposed to how it is calculated.

BTW ... is there a way to explain (in everyday English) how a value is derived for 3.7! ?? If you Google it, you do get an answer.
 
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Just thinking said:
Yes, that's how I learned it and understood what a factorial meant as opposed to how it is calculated.

BTW ... is there a way to explain (in everyday English) how a value is derived for 3.7! ?? If you Google it, you do get an answer.
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Through the already mentioned Gamma function, an extension of the factorial to real and complex numbers

latex.php


This is defined for all complex numbers, except negative integers, where it goes to infinity. With this definition, Gamma(n)=(n-1)! I don't like to use z! for non integers, but some people do. What Google gave you is 3.7! = Gamma(4.7).

This function appears everywhere in physics and mathematics. It is related to all kinds of special functions, it is very common in statistical physics, etc.
 
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Why is Gamma(z) defined with tz-1 under the integral sign, instead of tz? The latter would make Gamma(n) be n! rather than (n-1)!, which seems simpler.
 
69dodge said:
Why is Gamma(z) defined with tz-1 under the integral sign, instead of tz? The latter would make Gamma(n) be n! rather than (n-1)!, which seems simpler.
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For unfortunate historical reasons. The modern notation (using the Greek letter Gamma) is due to Legendre. Gauss used Pi(x) = x!, a simpler choice, but today Legendre's notation is universal. The first one to work with this function, by the way, was Euler. He used an infinite product which, in modern notation, would look like this

latex.php


(This definition is equivalent to the one I gave my previous post). By the way, I said

Yllanes said:
This is defined for all complex numbers, except negative integers, where it goes to infinity.
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I should have added that 0 is also a singularity (a simple pole in this case).
 
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I think maybe for notation purposes. too.

We see a lot of gamma(x) pop up in applications. If gamma(x) is defined as integral tx e-t dt, t=0 to oo, then we'd see a lot of gamma(x-1)'s, which isn't as 'clean'.
 
T'ai Chi said:
I think maybe for notation purposes. too.

We see a lot of gamma(x) pop up in applications. If gamma(x) is defined as integral tx e-t dt, t=0 to oo, then we'd see a lot of gamma(x-1)'s, which isn't as 'clean'.
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Not really. The applications you are thinking about were developed after the notation had been established, so they couldn't influence it.

Remember Gamma(x+1)=x Gamma(x), you can always take out integer summands. So Gamma(x-1+1) = Gamma(x)= Gamma(x-1) (x-1) => Gamma(x-1) = Gamma(x)/(x-1). What I mean with this is that people, once they get a formula with Gamma(something · x + something), usually move things around so they get, if possible (something)·Gamma(x). It's easy to manipulate this function to get an easier formula, check for example Mathworld for some of the possible identities.
 
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Yllanes said:
Not really. The applications you are thinking about were developed after the notation had been established, so they couldn't influence it.
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I'm not sure how you could possibly know the applications I am thinking about. Maybe you should apply for the million dollars?

I was actually speaking about Euler and others' work with integrals (beta function, for example). But of course there are tons of more recent applications.

I was thinking about this for some time this morning, and the answer to 69dodge's question is that it is due to the domain. Gamma as he defined it would be on (-1, oo). To get the domain on (0, oo) which is more natural use gamma as it is typically defined.
 
T'ai Chi said:
I'm not sure how you could possibly know the applications I am thinking about. Maybe you should apply for the million dollars?
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I'm sorry, I meant 'the applications I can think about'. Anyway, as I said, you can always take the 1s out in the formulae, so that's not the reason.

I was thinking about this for some time this morning, and the answer to 69dodge's question is that it is due to the domain. Gamma as he defined it would be on (-1, oo). To get the domain on (0, oo) which is more natural use gamma as it is typically defined.
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The domain is the whole complex plane, except for simple poles at 0 and the negative integers. With Gauss's definition we would get simple poles at the negative integers only. An easier domain is not the reason.

Legendre found the now usual notation more convenient and he used it. But nowadays there is not a real mathematical motivation behind this definition. Hence my answer: 'historical reasons'.
 
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