Hi, I have a deck of 52 cards.  Taking the cards 2 at a time gives us
52*51/2 = 1326 different 2 card combinations.  I then rank those 1326
combinations in descending order starting with pairs and then
combinations starting with higher value cards and suited cards ranked
above their offsuit combinations.  In the below table cards postfixed
with an s represents suited.  Absence of an s means the cards are
offsuit or of differing suits.

Here is an example of the ranking

Cards	# of occurances	Odds Against	% occurance
AA	6	         220:1         	0.45%
KK	6	         220:1	        0.45%
QQ	6	         220:1	        0.45%
TT	6	         220:1	        0.45%
99	6	         220:1	        0.45%
88	6	         220:1	        0.45%
77	6	         220:1	        0.45%
66	6	         220:1	        0.45%
55	6	         220:1	        0.45%
44	6	         220:1	        0.45%
33	6	         220:1	        0.45%
22	6	         220:1	        0.45%
AKs	4	         331:1	        0.30%
AK 	12	         110:1	        0.90%
AQs	4	         331:1	        0.30%
AQ 	12	         110:1	        0.90%
AJs	4	         331:1	        0.30%

If starting with a deck of 52 cards I deal 2 piles of 2 cards each for
4 cards total, and I look at the first pile and see that it is AJs (or
Ace of Spaces and Jack of Spades), what percent chance is there that
this two card combination is ranked higher then the unknown two card
combination in the second pile?  Is it simply 92% or 100 - (the
combination of the probabilites of those items ranked above the AJs) ?

If I were to increase the number of piles next to 3 then 4, 5, 6 ....
10 (without knowing what 2 cards are in these piles), what would the
probablity be that the AJs in the first pile would be the highest
ranked hand?

I believe the answers to be as follows:

Against 1 Unknown Pile - 92%
Against 2 Unknown Piles - 85%
Against 3 Unknown Piles - 78%
Against 4 Unknown Piles - 72%
Against 5 Unknown Piles - 66%
Against 6 Unknown Piles - 61%
Against 7 Unknown Piles - 56%
Against 8 unknown Piles - 52%
Against 9 unknown Piles - 48%

Are these calculations correct?  If they are not can you show me what
the correct method and answers would be?

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Request for Question Clarification by mathtalk-ga on 20 Jul 2004 06:19 PDT

Hi, tonyvu-ga:

Were you satisfied with the information provided in the Comments, or
did you still wish to have an Answer posted on this one?

regards, mathtalk-ga

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Clarification of Question by tonyvu-ga on 21 Jul 2004 09:21 PDT

I do have one question that I would like answered based on the comments.
VballGuy-Ga States:
"50*49/2=1225 possible hands left for the other players. So the formula should be: 
1225 - 96(hands that can beat us) -3(the matching hands) -1(our hand) 
we arrive at 1125 hands that loose to us.... Depending on how you
factor the ties, you can go from there...."

The point I am interested in is arriving at 96 hands that can beat / tie us.

It is obvious to use the method by creating a table and manually
listing the higher combination and summing them as in the AJ example. 
We can use our knowldege of the cards to describe them in another way.
 The cards 2 to Ace could be ranked in cardinal order from 1 to 13 and
offsuit or suited can be denoted by an integer value of 1 or 0, where
1 is equal to suited and 0 is equal to not suited.

Knowing the ordinal rank of each card in the 2 card hand the and the
integer value representing matching suits, in the 2 card hand, can we
devise an algorithm or formula for calculating the number of hands
that can beat us?

In the Ace jack Suited example we would denote that as (13,10,1) or
generally as (a,b,x) where a is the ordinal rank of card 1 and b is
the ordinal rank of card 2 and x is the integer represented the suited
property.




Thanks

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Clarification of Question by tonyvu-ga on 21 Jul 2004 09:22 PDT

In my above clarification please ignore my interchaning of the words
cardinal and ordinal.  I believe cardinal order would be correct since
the rank has a meaning that corresponds to the fact that the higher
the rank, the more valuable the hand.

Thanks

1) You forgot JJ
2) The second person really only has a deck of 50 cards to choose from
   A) AA has 3 combos (and JJ) loose 3 combos each
   B) AKs, AQs each loose one combo each
   C) AK, AQ loose 3 comos each
I end up with the following:
AA	3
KK	6
QQ	6
JJ	3
TT	6
99	6
88	6
77	6
66	6
55	6
44	6
33	6
22	6
AKs	3
AK	9
AQs	3
AQ 	9
AJs	96 total
(assuming AJs V AJs is a push)

96/1225 = .078367....

or you have a 92.21633 % chance of beating another single player
you have a 92.21633 % of the 92.21633 % chance to beat a second person
 or about 85.0% chance of beating two oppontants or 92.21633 ^ n-1
power
where you and two opponants make n-1 = 2  or just make it to the N
power where n is the number of opponants.
<Intersting question - and I am not a researcher...>

I was reading an old book on 5 card draw from an IBM scientist from
the 1970's and he had stated in his book if your cards have a 50%
chance or greater of being high (The game in my example is texas
hold'em where each player starts with 2 cards) then you should open
the betting.

With something like AJ suited you would probably always open the
betting since you probably have the best hand and it would be unlikely
to face all 10 opponents at once (thus only having a 48% chance of
being high for instance).

ok...
U r right. 
(1) There are 96 piles combinations above (of higher order) AJs.
(2) There are 3 pile combinations (AJs) of same order as the pile we picked (AJs).
(3) Remaining piles combinations (lower order) = 1326 - 96 - 3 - 1 = 1226.

So, total number of piles which can be ranked higher than the pile we picked are 

So, The chance that this two card combination is ranked HIGHER  than
the unknown two card combination in the second pile would be
= 1 - (99/1326)
= 1 - 0.0746606
= 0.92534
= 92.534%

Similarly we can calculate for the rest ... i.e. for 2 piles, 3 piles, ... etc.

In Response to pamya:
"Remaining piles combinations (lower order) = 1326 - 96 - 3 - 1 = 1226."

After we know what two of the cards are, we are cetain that there are
only 50 cards left in the deck so there are no longer 1326 hands left
for the other player in stead there are:

50*49/2=1225 possible hands left for the other players. So the formula should be: 
1225 - 96(hands that can beat us) -3(the matching hands) -1(our hand) 
we arrive at 1125 hands that loose to us.... Depending on how you
factor the ties, you can go from there....

in response to vballguy :

Nice explanation! Thanx for pointing out my error.