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#1
02-10-2007, 02:51 PM
 kevin017 Senior Member Join Date: Mar 2005 Posts: 624
guaranteed aa?

this doesn't necessarily have any practical purpose, but i saw the same kinda thing argued somewhere else and couldn't figure out the answer.

Let's say you play infiinite hands of poker. Is it 100% guaranteed that you will be dealt AA? Ive taken some math before, and i vaguely remember some formulas that would probably tell me that this converges to 100%. but, just thinking about it theoretically, i see no reason why you would ever HAVE to get aa, you would just get infinitely close to a 100% chance.
#2
02-10-2007, 04:30 PM
 ispiked Senior Member Join Date: Dec 2006 Location: whizzing by Posts: 437
Re: guaranteed aa?

If you play infinite hands, then you will have been dealt every single hand (of varying suites, too) at least one or more times.
#3
02-10-2007, 05:04 PM
 QuattroFour4 Senior Member Join Date: Mar 2006 Location: Evanston, Illinois Posts: 195
Re: guaranteed aa?

If you are dealt an infinite amount of hands, you will actually get every single variation of 2 cards (suits and rank) an infinite amount of times. Kolmogorov's zero-one law states that in an infinite series (in this case, an infinite number of hands), the probability of an event occuring must be either zero or one. Since in our case the probability of getting aces over an infinite amount of hands is definitely not zero, we can say that it must be one.
#4
02-10-2007, 06:17 PM
 SpaceAce Senior Member Join Date: Dec 2003 Posts: 3,017
Re: guaranteed aa?

QuattroFour4 has it covered, but any question of this type involving infinity is silly, anyway. Whenever you have an infinite number of trials, all possible outcomes will happen an infinite number of times. If you sit at a table and get dealt an infinite number of hands, not only will you get AA an infinite number of times, but an infinite number of cards will also fly off the table and hit you in the eye and an infinite number of dealers dealing the cards to you will have an infinite number of heart attacks and an infinite number of cocktail waitresses will spill an infinite number of drinks on your head. Obviously, those events are outside the scope of your question, but I am illustrating a point. Infinity is usually useless in this type of question.

A better question might be, "If I get dealt a million two card hands..." in which case you have to choose your own threshold for "guarantee" because the chance of you NOT getting AA in a million hands is very, very tiny but not quite zero.

SpaceAce
#5
02-10-2007, 06:40 PM
 kevin017 Senior Member Join Date: Mar 2005 Posts: 624
Re: guaranteed aa?

That is the basic answer that half of the other place arrived at.

but, why must there be an infinite amount of AA's dealt? Is it not POSSIBLE that you could be dealt 88 every single hand for 2 million, 100 million, infiite hands?

I'm not familiar with kolmogrov's law, but based on wikipedia it seems as though the probability of AA occuring is "almost surely". An example given on wikipedia in the definition of almost surely says exactly what I'm thinking.

"Suppose that a coin is flipped again and again. The infinite sequence of all heads, "HHHHHH....," ad infinitum, is possible in some sense -- it does not violate any physical or mathematical laws to suppose that tails never appears -- but it is very, very improbable. In fact, such a sequence has probability zero. Thus, there will almost surely be at least a single tails flip in an infinite sequence of flips."
#6
02-10-2007, 07:38 PM
 SpaceAce Senior Member Join Date: Dec 2003 Posts: 3,017
Re: guaranteed aa?

[ QUOTE ]

but, why must there be an infinite amount of AA's dealt? Is it not POSSIBLE that you could be dealt 88 every single hand for 2 million, 100 million, infiite hands?

[/ QUOTE ]

No, that's what makes it infinity. I'm not a statistician, so I don't know how best to explain this, but I will try. You know that .1 is 10% and .01 is 1%, so .0-&gt;infinity (with an infinite number of deals, you end up with an infinite number of 0s) is 0. You will never get to the "1" after the decimal place because it comes behind an infinite number of 0s, and "infinite" means never ending. If the 0s never end, the 1 never shows up. I am sure someone else can phrase it better.

People tend to confuse any really big number with infinity, but infinity has its own properties that no other number, no matter how large, has. In an infinite test, every possible outcome happens an infinite number of times.

[ QUOTE ]

I'm not familiar with kolmogrov's law, but based on wikipedia it seems as though the probability of AA occuring is "almost surely". An example given on wikipedia in the definition of almost surely says exactly what I'm thinking.

[/ QUOTE ]

I believe the "almost" part of "almost surely" only exists because you can never actually reach infinity. If you have a truly infinite number of flips, you WILL get all possible outcomes. The "almost surely" comes from the fact that if you stop at infinity-1 (obviously not possible since infinity-1 is still infinity, but you know what I mean), there is some non-zero probability that you never flipped a tail. True infinity doesn't leave any room for anything besides 0 or 1 and since tails is a possible outcome, it is 1 and will happen in an infinite number of flips. So will a series of 100 tails in a row, 1,000 tails in a row and 1,000,000 tails in a row IF you actually flip to infinity, which you cannot do, and therefore you are stuck with "almost".

&lt;Paging the math and stat supergeeks to come in here and set us straight&gt;

SpaceAce

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