STOP OVERPLAYING POCKET PAIRS AND WHINING WHEN YOU LOSE WITH THEM!1!!1

[Sir P]

I had a vision last night while masterbating, which brought me to this theory.

ok, lets say you have one 6 versus one ace. the ace will hit a straight or flush about 60% of the time while the 6 will only hit around 40, clearly because the ace is higher in value and can go from 10-a and/or a-5.

now, if we add another card to the 6, since hold'em deals with 2 cards,66 versus one ace will have about half the chance of getting a straight or flush because the 6 counterfeits it and lessens the percentage.

this makes the new percentages around 20% for 66 hitting a straight/flush to 80% for just one ace. if I added in the other possibilities, since the ace can win by hitting a pair, trips, full house, quads, or royal flush, then the ace would be about a 89% favorite overall to win. also, since the ace can win by quads coming on the board and having an ace kicker, this bumps it up to 92% to win, and we didnt even add another card to it yet!

I hope you are still following. ok, so mathematically currently the ace is 92% versus 66 who is 8%. we now add a King to the ace, and this doubles the percentage and also halves 66's winning %

the new percentages are approx. 92*2= 214% to 8/2= 4%.

You may be asking, "wait, 214+4 is only 218%? what accounts for the other 782%?"

I'm not finished.

If you think of the numbers as a triangle, there would be an ace on one corner, a king on the other, and a 6 on the final one (since you cannot have two 6s in it). the Ace and King represent 2/3 of the triangle, so now we must multiply 214*2/3= about 200 and add it to the original number.
also, 4*1/3 is 2 so we subtract that and it gives us 2.

now we have AK 414% to win VS 66 2%
this part is frequently overlooked. Most people dont realize they are playing with two cards and view their hole cards as a whole. this makes people forget to multiply the percentage by 2 or divide, if necessary.

we are at 828% and 1% now. almost finished.

back to jman's theory... you are not completely wrong, your ideas do eventually tie in to the intricate problem solving of this situation. since AK has 6 outs while 66 has only 2, we multiply this by the amount of cards in a deck. and this gives us 312 and 104. we simply add/subtract these numbers to our percentiles, giving us 1140% and -103%. If I worked out all these numbers in decimals the -103 would actually be -104, but I thought it was +EV to do it this way.

ok, finally, we must switch the 0 and 4 in -104% for 66's winning chance clearly because you are playing no limit and not pot/fixed limit (if it was pot limit, for example, you would switch the 4 and 1.)

thus, we have 1140% and -140%, added together gives us 1,000% or 1.

I hope this clears things up. AK vs 66 is a 1140% to -140% favorite.

I spent much thinking, time, and money on this project. I have done several experiments that back up my theory.

NOTE: this was not tested with pinochle decks, or when you are playing wrap arounds. also, I am not liable for any losses from applying this strategy to games using blackjack decks. (which decreases ak's winnning percentage drastically)
I know that I do not even have 20 posts, which decreases reliability by 120%, but if you think about it mathematically we all have less that 20 posts.


Thank you for listening and if you need any tips or advice on how to improve your poker game feel free to give me a PM.