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Old 06-15-06, 01:26 AM
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Default How I will manage to be the first player eliminated form the 2006 WSOP...

Mathematical Proof to follow...

Assumptions:
1. 8000 players
2. Forget all this Day 1a, Day 1b, etc. crap. Assume the first hand dealt is the first hand, regardless of the number of starting days.
3. 10 players per table to start.
4. 52 card deck, Texas Hold'em, etc, etc, etc - all the normal crap.

Proof:
-8000 players / 10 players per table = 800 tables in action.
-The odds of ONE player being dealt AA on the first hand are 220:1; therefore the odds of AA being dealt at a given table are approximately 22:1. That means that about 36 tables will have AA dealt on the first hand.
-Of those 36 tables, how many will have AA dealt TWICE??? This is where I usually like to call in the MathBabe, but she's not here (and she was WRONG about that last probability thing - hehe), so I'll take a stab at it:

With one player holding AA that leaves 50 cards and only 2 Aces left. So the odds of any ONE player (at that table) being dealt both of the remaining two Aces are: 2/50*1/49 = .08% * 9 other players at the table = .73%. Is that right? I have no idea, but let's just assume it is.

Given that, we can now deduce that the odds of AA being dealt to two players at the same table in the first hand are 36 * .73% = 26%.

(Oddly enough, if my admittedly suspect math is correct, this means we can EXPECT this to actually happen within the first 4 hands dealt at the tourney. Does this really happen every 3200 table hands dealt? I guess that sounds reasonable)

Now, back to fantasyland: With two players being dealt AA, there is a 95.65% chance of a chop, and a 2.17% that one of the players will win (each player has this chance, so it's really a 4.35% total chance).

That means that the chances of two players at the same table getting it all in prelfop (assuming they keep raising each other until they get all in), each holding AA on the first hand of the WSOP and one of them actually 4 flushing the other guy are: .26 * .0435 = 1.15%.

And we all know that is PLENTY to all but guarantee that it will happen to me. So there you have it, mathematical proof.

The good news is, ESPN will interview me, and I'm semi-confident I can (on camera) talk Harrahs into buying me into the 2007 WSOP as a pity prize.
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