What would you rather be against....
 

OK, ill set the scene.

Its a tourney, everyone has the same amount of chips, blinds and stacks dont matter.

This question is about correct pot odds working out.

The flop is Q T 7 rainbow.

You have Q T

WHich of the above hands would you rather be against from a % to win preference, basically, which of the hands listed would give you the most chance of winning the pot if you went all in and the ther player called??



I want people to work this out themselves, not vote after Zybomb has worked out all the odds and runner runner factors and made his vote!

PLease read this, then vote, then read other peoples reasoning on their decisions.

AA - 28.622%
AQ - 16.395%
AT - 15.604%
A7 - 21.698%
KJ - 32.590%
J9 - 30.404%
89 - 30.014%
Q7 - 9.192%

So, Q7.

Grr, use your mind not an odds calculator

I just looked at each hand (glanced, really) and picked the one with the least outs. Seems simple enough.

:confused:

BTW - I voted before I even read the FIRST post in this thread.

yea no odds calc necessary -- by just looking you can tell that Q7 has only 2 outers, the two remaining 7s...everythingelse has at least 3

IM just thinking if people could figure it out for themselves:thumbsup:

I shouldnt have added the obvious one. is there anyway to delete it? if you can do that then it would be more on a challenege

if u eliminate Q7, it's still fairly basic

First lets eliminate 89 J9 and KJ, all of which are roughly 32% (8 outs)

Next lets take out AA. It has 2 outs with Aces 3 outs with 7s and 3 outs whatever the turn comes

That leaves us with AQ AT and A7.

A7 is out with 3 aces and 2 7s.

which leaves AT and AQ -- I could care less which one they had

Q7 any day, you are going to win more and there is a lot less of a chance that you will get sucked out on