[quote]
Quote:
Let me show you an example, backed by a real life event.
Sorry.
Quote:
My brother and I like to play heads up poker against one another...
Quote:
Do you ever play HU not against each other...like on the same side as each other?
No. We are very competitive with each other. It runs in the family, on the weekends we'll sit around and play HU against each other, sometimes our Dad or my cousin James will play too.
Quote:
So let's examine the final hand:
Quote:
The *final* hand. No more poker?
Yeah, I was extremely angry that he had called me with Jack high and I went on a cussing rampage. Haha. I asked him why he called me with J*4 when I pushed on the flop, and he simply responded "Because I didn't think you had sh*t". Good answer.
Quote:
Why did he win? Here is the reason (and the basis of my hypothesis). I did not have numerical advantage. If you view the cards simply as numbers, then you will understand what I mean:
2-10 represent the numbers they are.
Jack = 11
Queen = 12
King = 13
Ace = 14.
To understand numerical advantage multiply the cards by each other.
Thus; A*J = 154, 7*6 = 42, J*4 = 44, Q*K= 156.
In that hand, I would've never won with AJ, despite the fact that Sklansky would consider hand to be dominant over QK. However, if I had QK, I would've won. AJ would've certainly beat J4, and 76 never had a chance against any of the other 3 hands described. This is because of numerical advantage.
Quote:
So, AQ (168) would have beaten KQ? Because here is where your theory may need some refinement. KQ (156) < AQ (168), yet, counter-intuitively, in the final hand, KQ would have prevailed despite the fact that AQ clearly enjoys the numerical advantage.
Yeah, I forgot to address this when I was writing last night. AQ does enjoy the numerical advantage, and has math on its side. KQ enjoys more synchronicity (they are not gapped, and closer together) which would mean it has more luck on its side. Of course, any two cards can beat any other two. Despite A*K having a distinct NA over 2*4, 2*4 always has the completely random possibility of winning over A*K, but according to NA it is less likely.
Also, the only way to employ this theory is knowing and having a very good read on your opponent. It is only useful for helping to calculate your luck odds. If you have a good read that you have a numerical advantage against your opponent, that your hand is "luckier", then this could help to aid you in winning the hand.
Thanks for asking this, and thanks for the constructive criticism. My theory definitely needs refinement, I agree.
Quote:
Perhaps the sample size is too large.
Look for a dollar from me shortly.
Yay! Thank you!
If you or anyone has any doubts that I'm broke as a joke, I can provide screenshots of my Poker Stars cashier as proof.