brain teaser see if you can get this holdem problem

 You are dealt the 2- 7 off suit in a 9 player game.

You are allowed to put any cards in any of the other 8 hands. Both hold cards must be higher than your 7. Thus you might give player A an 8-9 ,player B a 10- J  and so forth.

 Also you  select any 3 cards on the flop but are only allowed to put up  1 deuce and no 7's . The last 2 cards come up at random.

  The board is only allowed 1 pair.

 Set up hands in this 9 player game that will allow you to win the whole pot.

 

The idea is to be able to get the whole pot every time so it is impossible for anybody to beat or tie you after the flop. Remember the fourth and five card are turned at random. In your case an A-thru 5 straight is still possible and you must split.

I know I posted this in the other forum, but I was not sure which one you plan to post in... 

I notice you did not put the stipulation in this forum that a card on the flop, turn or river could pair with any hole card in another player's hand....


Here is my solution and explanation.... looking forward to the correct solution...

Since you must deal 16 cards to the other 8 players.  Those 16 cards must be made of 4 quads.  You can choose any 4 sets of quads (other than deuces and sevens since you have one of each) but they MUST be dealt in such a way that no player has either a pair or a suited hand.  For example, let's deal everyone all the aces, kings, queens and jacks (We will make them think they all have the best hand, so they can go all in) 


Players cards might look like this
(AsKh, AdQh, AcJs, KcQd, KdJc, QsJh, AhQc, KsJd)

Doing this eliminates the possibility of the last two random cards matching one (or more) hole cards of one (or more) of the other players - i.e. there are no pairs, trips or quads.  If you eliminate trips, you eliminate the full house as well.

 
The flop will go something like 2-3-4 offsuited (since you are allowed to give yourself a deuce here). 

This does two things: 1) eliminates the possibility of a high end straight draw by the other players (no A-K-Q-J-T or K-Q-J-T-9 or Q-J-T-9-8), and 2) eliminates the possibility of a flush draw, since  the other players will never have more than two cards of the same suit after the flop. 

Now flip over any two remaining cards at random. 

Three possibilities will happen.

1)  If the cards come up 5-6, everyone has a straight, but your 7-high straight is higher, than everyone else's 6-high straight.  Result...

2)  If the cards come up paired, there are four scenarios...

   a)  If they come up dueces, you get quads...
   b)  If they come up sevens, you get a boat...
   c)  If they match one of the other two cards on the flop, you get a boat...
   d)  If they do not match one of the other two cards on the flop, you get at most two pair to everyone else's pair. 

In any situation, result... 

3)  If the cards come up unpaired (suited or not), other than 5-6, you always have at least a pair (trips if one is a deuce) and everyone else's crap hand, A-high, K-high, etc.  Result...


You still leave yourself the possibility of a full boat, quad deuces or even a straight or flush draw.

How's that for a solution???  Did I understand it correctly to get it right??? 

kangy has it almost right and the godfather makes a valid point...but the way to guarantee a win everytime is to select 2-5-6 rainbow flop.  That eliminates any chance of a flush and cancels out any wheel draw with an opponents ace (i.e. turn and river come 3, 4, you still have top straight with the 7 and no other combination of cards can come to beat you or cause a split pot.)

Where's the professor?