Odds

[T'ai Chi]

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.

Stuff in the complex plane came much later. It was introduced in working with natural numbers.

I'm thinking that

e^-t*t^x ~ t^x = 1/t^-x
since e^-t ~ 1 when t ~ 0.

And the p-test for convergence implies that it converges when -x < 1, or equivalently when x > -1.

 

[Yllanes]

Stuff in the complex plane came much later. It was introduced in working with natural numbers.

But since the 19th century we know about complex analysis and lots of other things. Euler introduced the function, but not the Gamma notation. That comes from the 19th century. Then they could have established Gauss's Pi notation, but they didn't because of historical reasons.

I'm just saying that there is no profound motivation behind the usual definition.

 

[69dodge]

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

[qimg]http://www.randi.org/latexrender/latex.php?%5Cdisplaystyle%3Cbr%20/%3E%0A%5CGamma%28z%29=%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7Bn%21%20n%5Ez%7D%7Bz%28z+1%29%28z+2%29%5Ccdots%28z+n%29%7D,%5Cquad%20z%5Cneq0,-1,-2,%5Cldots%3Cbr%20/%3E[/qimg]

(This definition is equivalent to the one I gave my previous post).

Ha. That formula looks better the way it is (instead of having n^z+1 on top, and starting from z+1 on the bottom).

Oh well. It's one or the other, I guess.

More on notation: check out Dijkstra's take on the standard definition of 'n choose k'.

 

[T'ai Chi]

More on notation: check out Dijkstra's take on the standard definition of 'n choose k'.

I've seen it written as nCk, so 10 choose 3 is written 10C3, for example.

No taking up multiple lines, and no taking up parentheses.

 

[Art Vandelay]

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.

Actually, I think that you could apply it to any field, and some rings. There would the issue of convergence, though. For instance, if D is the differentiation operator, then gamma(D)=integral(0, infinity, F(t), dt) where
F(t)=(t^(D-1))(e^-t)=
(e^((D-1) (ln t )))(e^-t)=
(e^(D ln t))(e^-t)/t

gamma(D)f(x)=integral(zero, infinity, f(x+ln t)/(te^t), dt)
= integral(-infinity, infinity, f(x+s)(e^-(e^s)), ds)

Unless I’ve screwed up somewhere, which is quite possible.

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.

That was a bit unclear. At first, I though that you were saying that it provides an answer to the question “is there a way to explain” rather than “what is 3.7!”.

How about this: suppose that you were to consider the number .010409162536496481
This number is equal to (1^2)*100^-1+(2^2)*10^-2+(3^2)*100-3...
So if we were to write this in base 100, it would be a sequence of squares. We now need to make three modifications:
1. Make it base e, instead of base 100
2. Normally, when we express something in a base, we have integral powers of that base; in base 10, for instance, we have a tens (10^1) place, a hundreds (10^2) place, etc., but we have no 10^1.5 place. But we're going to have every power of e, not just the integral ones.
3. In the example I gave, I squared the numbers. That is, I raised them to the second power. But we're going to allow any power, and the way we're going to decide which power to use is to subtract one from z. For instance, if z=4, we'll raise everything to the third power.

If we raise everything to the power of 3.7, then we get 3.7!.

 

[T'ai Chi]

Re: 3.7!, and how to get the answer, one could use a Taylor series for the gamma function. It is somewhat complicated for me to post.

 

[joe87]

52! = 80658175170943878571660636856403766975289505440883277824000000000000 ~ 8.065817517094388· 10⁶⁷

I used Excel to calculate it directly and got 8.06581751709439000000E+67, (that was easier than looking up the factorial approximation equation). This value is also an approximation but is good for about 14 decimal places, which is better than I had thought Excel would be capable of. I rounded it to 8E+67 which seemed to be good enough for arguing purposes, although I now see that it wasn't.

If the cards are randomly shuffled between each hand, it will take considerably longer than 8E+67 deals for all possible sequences to appear because there will be many duplicate sequences turning up before all possible ones appear.

 

[Art Vandelay]

I used Excel to calculate it directly and got 8.06581751709439000000E+67, (that was easier than looking up the factorial approximation equation). This value is also an approximation but is good for about 14 decimal places, which is better than I had thought Excel would be capable of.

I also used excel, but I used the e^((ln 1)+(ln 2)+(ln 3)...) method.

If the cards are randomly shuffled between each hand, it will take considerably longer than 8E+67 deals for all possible sequences to appear because there will be many duplicate sequences turning up before all possible ones appear.

Depends on what you mean by "considerably" longer. An extra factor of 200 would be way more than enough to make it overwhelmingly likely that every sequence will appear.

 

[joe87]

I also used excel, but I used the e^((ln 1)+(ln 2)+(ln 3)...) method.

Depends on what you mean by "considerably" longer. An extra factor of 200 would be way more than enough to make it overwhelmingly likely that every sequence will appear.

I meant 10E+68 or 10E+69 more hands. My reasoning is, after the first 10E+67 hands are dealt at random, there will probably be 10E+65 possible hands that haven't been dealt yet, and each of these will probably take 10E+67 more hands to appear. I'll have to think about this some more, though.

 

[joe87]

Well, I did some more thinking about this and also some calculations to get a feel for how long it takes to get all possible permutations by randomly shuffling cards. I set up an Excel macro to do a monte carlo calculation for a 4-card deck. Since 4! is 24, there are 24 possible hands that can be dealt using 4 cards. Excel randomly picked numbers between 1 and 24, and when all 24 numbers had appeared, that sequence was terminated and the number of hands dealt was logged. The sequence was also terminated if 200 hands had been dealt and all permutations had not yet been observed. The macro then repeated this process 200 times, for a total of 40,000 shuffles & deals. The results are as follows:

The average number of hands necessary to get all 24 possible hands = 89.6
Minimum number = 42
Maximum number >200 during 4 of the 200 sequences.
Standard deviation of number of hands necessary = 31.6

So while it is possible to get all 24 possible hands in just 24 shuffles, this did not happen in 200 sequences of 200 hands. On average, it took almost 90 hands to get all 24 possible permutations. To have 95 percent confidence that all 24 possible hands would occur it would be necessary to deal 148 hands.

The distribution of the number of hands is skewed to the right with a long tail, as can be seen from the max, min and mean values, somewhat like a poisson distribution, but probably not exactly since the standard deviation is quite different from the mean minus the minimum possible number.

So how does this translate to a 52-card deck? I think that it shows that “considerably” more than 8x10E+67 hands need to be dealt before all 8x10E+67 possible hands will occur. I suspect that increasing the number of cards in the deck increases the ratio of the average to the minimum possible value, but I can’t prove it with one calculation. I considered taking the next step and looking at a 5-card deck with 120 possible hands, but that would probably be beyond the capability of an Excel calculation on my computer (2.20 gigahertz AMD Athlon 64 chip), since the 40,000 calculations with a 4-card deck took about 10 minutes.

 

[Art Vandelay]

The exact formula is discussed in this thread:
http://www.internationalskeptics.com/forums/showthread.php?t=53430

For such a large number, it would quite impossible to calculate exactly.

 

[joe87]

Yes, that thread has some good info on closed-form solutions to the problem. Using Alkatran’s formula for the mean number of hands that it would take to deal all possible hands, I came up with the following:

Number of-------------Number of------------------Mean Number of Deals
Cards in Deck---------Possible Hands-------------to Get All Possible Hands
3---------------------------6---------------------------------14.7
4--------------------------24---------------------------------90.6
5-------------------------120-------------------------------644.3
6-------------------------720------------------------------5153.2
7------------------------5040----------------------------45,876
8----------------------40,320---------------------------450,851

The mean value for a 4-card deck is reasonably close to the 89.6 that I got using the monte carlo calculation mentioned earlier in this thread. I stopped doing the above calculations at an 8-card deck because Excel only goes up to 65,536 rows, so it would probably take a dedicated program to go beyond 8 cards.

I then fitted a power equation to the ratio of mean time to deal all possible hands divided by the number of possible hands and got a good fit (r=.99992) over the above range. Then taking a wild leap into the unknown, if we extrapolate that to 52 cards, the mean number of deals to get all 8.0657E+67 hands is 1.7E+70, which is a bit larger than the 10E+68 or 10E+69 that I had earlier guestimated. Not that the new value is very accurate, but it gives some indication of the magnitude of the number of hands that it would take to deal all possible hands with a 52-card deck.

 

[Walter Wayne]

As a back of the envelope type calculation ...

If we approximate that when N is very large (52!) then after N trials that the (expected) fraction of values that haven't appeared yet is 1/e we can start to get a first order approximation of how long it will take to see all possible hands.

So after dealing out 52! hands the odds a given hand hasn't appeared is about 1 : e. Complete a second shuffling of 52!, and the odd becomes 1:e², and so on. So using this approximation the odds that a given sequence hasn't occured after M shuffles, is (1/e)^M/52! and the expected number of sequences that haven't occured is 52!*(1/e)^M/52!.

Applying this formula to Joes values above I should get more accurate numbers as the deck gets larger, since my first assumption holds as N approaches infinity. Do the math I get expected number of hands remaining undealt to be:

Cards--Possible Hands--Number of hands dealt--Exp. number of undealt sequences
3-----------6-----------------14.7--------------0.518
4----------24-----------------90.6--------------0.550
5---------120----------------644.3--------------0.558
6---------720---------------5153.2--------------0.561
7--------5040--------------45786----------------0.561
8-------40320-------------450851----------------0.561

If I solve my equation to find Number of hands dealt to get an expectation of 0.561 I find

0.561 = 52!*(1/e)^M/52! I find that 156.94 = M/52!, or that we will need to deal out 1.266x10⁷⁰ on average to get every possible hand.

An approximation, but wouldn't shock me if I had it correct to within 25%,it depends on how good my assumption that my formula and Alkatran's formula from the other thread are roughly equal if I set my expectation to 0.561 for large numbers.

Walt

P.S. Why I am printing that many sig-figs when I am using so many approximations I don't know.

 

[Art Vandelay]

Yeah, I used basically the same reasoning to conclude we would need, at most, a factor of 200 (note that an extra factor of 200 gives us 1.6 x 10^70, rather close to joe87's figure of 1.7 x 10^70).