3 card probability question

[jjp]

At a home game, we play a version of "guts" where three cards are dealt, and you make a two card "high" hand. How can I calculate what the odds are of having various hands such as (and the odds of anyone at the 6 handed table having):
66 or better
Any pair
AJ or better

We then deal two more cards (total of 5) where we play a 3 card low (straights and flushes ignored). What are the odds of a:
Rough 5 or better
Rough 6 or better
Rough 7 or better?

Thanks.

[UtzChips]

There are 52*51*50/3*2*1 = 22,100 possible hands.

There are 4*3*48/6 = 96 ways to be dealt a pocket pair, but not Trips.

66-AA = 8*96 = 768 possible ways to be dealt 66-AA

1-[(22,100-768/22,100]^6 = 19.12% chance that at least one person will have a pair of 6s or higher, but not 3 of a kind.

Add in the 3 of kind possibility:
4*3*2/6 = 4*8 = 32

1-[(22,100-800)/22,100)^6 = 19.85% chance at least one player will have a pair of 6s+ or three 6s+

[jjp]

So 22 or better would be
96 * 13 = 1,248 +32(trips) = 1,280
Odds of me getting pair or better = 1,280/22,100 = 5.6%
Odds of anyone at the table with pair or better
1 - ((22100-1280)/22100)^6 = 69.9%

So if I'm in first position with AK, with 6 people to follow there is a 70% chance I have the best hand. Correct?

[UtzChips]

Not quite.
There are 52*51*50 = 132,600 variations of how you can be dealt 3 cards. This includes the permutated variations, such as getting As as your 1st card, Ad for the 2nd and Ah for the 3rd, or Ad first and on and on and on.

So, we have to remove the permutations. So you multiply, starting with the number of cards being dealt (3) by (2) by (1) of course the one really isn't necessary.

If someone asked you how many different Boards there are in Hold Em after your dealt AA; there's 50 cards left, so it would be 50*49*48*47*46 and then divide that total by the sum of 5*4*3*2*1. That would give you the total number of possible Boards, without counting the different order in which 5 specific cards cards could appear on the Board.

Now in your example you dealt yourself AKx. Now there are only 49 cards left in the deck to be dealt to the other players. So, it's 49*48*47/6 = 18,424 different hands available to be dealt to your opponents.

Now for a pair, it's 4*3*(50-the 2 other cards of that rank)/6
4*3*48/6 = 96 ways to be dealt a pocker pair.

NOTE: You do not remove the 3 cards in your hand, when determining how many different ways there are to be dealt a random pair. Your three cards only affect the number of variations in which your opponents can be dealt a pair in the ranks you hold in your hand.

And now.........I've come to the realization that there is a good chance I do not know what I'm talking about when determing the chances of someone being dealt the hands you're concerned about.

I'm not sure if you do remove the cards that were dealt to you or not. However, that's the way I've been doing it since I saw this post, and I think I may have been doing it the wrong way.

I am going to refer you to this post:

http://tinylink.com/?qGOnTK9wV4

You may want to PM one of those guys.