Two Plus Two Older Archives Stuff that's tough to believe
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#1
07-23-2003, 09:32 PM
 Rushmore Senior Member Join Date: Apr 2003 Location: Tampa, FL Posts: 868
Stuff that\'s tough to believe

This is, at the very least, an interesting article:

http://www.pokerpages.com/articles/archives/hench33.htm

C'mon. That's good stuff, right? [img]/forums/images/icons/confused.gif[/img]
#2
07-23-2003, 09:52 PM
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636
Re: Stuff that\'s tough to believe

What's interesting to me is the kind of idiots that write poker columns. A whole article about whether there will be 52! shuffles in the life of the universe? Of course there won't be, and it isn't even close. The guy had to research this? [img]/forums/images/icons/grin.gif[/img]

Non-random shuffling can cause arrangements to repeat more often though.
#3
07-24-2003, 10:29 AM
 Nottom Senior Member Join Date: Feb 2003 Location: Hokie Country Posts: 4,030
Re: Stuff that\'s tough to believe

OK so we know the number of shuffles possible, so how many shuffles are needed before there is a 50% chance of a repeat. I expect this number is much smaller than we would expect much like the "two people in a room sharing a birthday" situation.
#4
07-24-2003, 06:36 PM
 redsamurai Junior Member Join Date: Jun 2003 Location: Seattle Posts: 29
Re: Stuff that\'s tough to believe

No longer having access to a top Mathematics software package I can't do the calculation, but here is the formula. Like in the "birthday" problem we take 1 minus the odds all of the shuffles of a deck are different ( http://www.cs.jcu.edu.au/~david/Tours/birthday.html ). For N shuffles this comes is:
P(@ least on match in N shuffles)
= 1 - ((52!-1)/52!)*((52!-2)/52!)*...*((52!-N)/52!)
= 1 - Product of ((52! - n)/52!) for n=1 to N

Using MatLab or Mathematic someone can likely solve this for P=50%. I estimated this using a different inexact probalistic technique and it came out to approx 1*10^33 (this could be an order or two of ten off). In other words if you shuffled a billion decks each second it would over 10,000,000,000,000,000 (10^16) years to be 50% likely to repeat an exact random shuffling.

All that said hold'em players only care about the position of 25 cards in the deck (10 players with 5 common cards). Anyone care to do that calculation?
#5
07-25-2003, 04:54 PM
 irchans Senior Member Join Date: Sep 2002 Posts: 157
Re: Stuff that\'s tough to believe

I had some fun with the article's claims. It claims that

"In the history of the universe or in it's foreseeable future, no two decks of cards have ever been or will ever be in the same sequence (unless artificially so arranged )."

and it goes on to claim

"In other words, you would run out of atoms in the galaxy before you run out of different sequences in a deck of cards."

The first claim is likely to be true, but it is possible to imagine a situation where it is false. The truth of the second claim is hard to determine.

Here is my reasoning,

If you have an experiment that has n possible results, each of which are equally likely and you repeat the experiment Sqrt[n] times then there is approximately a 40% chance that there will be a match. (For n large, the probability is about 1-e^(-0.5).) For n = 52! (approx 8*10^67), the probability of a match is 50% after about 1.05*10^34 deals. (To get this number I used the approximation

prob of getting a match ~= 1 - E^(-(m^2*(m + n))/(2*n^2))

for m experiments with n possible results.)

So, to get a match we need about 10^34 deals. This is unlikely, but if humans continue to play poker and colonize 10,000 galaxies, then we could get a match. Assuming each person shuffles 1 deck per year on average, then we get

10000 galaxies
* 10^10 planets per galaxy
* 10^10 humans per planet
* 10^10 years

= 10^34 deals.

As for the number of deals exceeding the number of atoms in the galaxy. The number of atoms is about

(mass of galaxy in grams)*( Avogadro's Number)

which is

6*10^11 * 2*10^30 * 1000. * 6 10^23
= 7.2*10^68.

This is about 10 times more than the number of deals.

I think we can conclude that if you have a poker probability result that you are going to put in an article, it would be best to check your result first by posting it to the 2+2 fourum. (2 missspellings intended [img]/images/graemlins/tongue.gif[/img] )

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