Two Plus Two Newer Archives  

Go Back   Two Plus Two Newer Archives > General Poker Discussion > Beginners Questions
FAQ Community Calendar Today's Posts Search

Reply
 
Thread Tools Display Modes
  #1  
Old 05-19-2007, 08:28 PM
Beauregard Beauregard is offline
Junior Member
 
Join Date: Apr 2007
Posts: 1
Default Question About Hand 2-1 in Harrington Book

I have recently begun reading the Dan Harrington/Bill Robertie book “Harrington on Hold’em, Volume 1 – Expert Strategy for No-Limit Tournaments”. On page 64 there is a discussion of the play of a hand, titled “Hand 2-1”, in which you, the reader, are player E, the last player before the dealer, with a total of eight players. Blinds are $50/$100. Player E’s hand is 8c6c.

The action begins as follows: Players A and B fold. Player C calls $100. Player D folds. Pot is now $250.

Harrington states that player E should fold; however, if player E decides to play, the action proceeds: Player E calls $100. The dealer, player F, raises to $200. The blinds fold. Player C calls for another $100. The pot is now $650, and it costs player E $100 to call.

At this point Harrington states, “The pot is offering you 6.5-to-1 on your money, and your flush and straight possibilities make this an easy call.”, and consequently player E calls.

However, the equity – the probability of winning or splitting the pot - for a 86s starting hand in a heads-up match is 46.2%, according to Stephen H. Landrum’s table, shown in the following link:

http://gocee.com/poker/he_ev_hand.html

With four remaining players before player E’s pre-flop bet, by my calculations the probability of player E’s winning (or splitting the pot) is .462^4 = .046; that is, the odds against a favorable outcome are about 21:1. The fact that two of the remaining players have folded after player E bets does not alter these odds; it simply means that both the remaining two players probably had better hands than player E from the start.

Therefore, the pot odds of 6.5:1 are not nearly good enough to call the $100 raise.

As it turns out, although player E makes the flush on the showdown (after going all-in on fourth street), the dealer, player F, who raised to $200, makes the winning flush with QcTc. This should not come as too great a shock to player E.

Is my analysis correct? If not, why not?
Reply With Quote
  #2  
Old 05-19-2007, 08:58 PM
ShaneP ShaneP is offline
Member
 
Join Date: Aug 2006
Posts: 80
Default Re: Question About Hand 2-1 in Harrington Book

First, I think as you mentioned the correct odds probably aren't 46.2, since we now have a bit more information about the hands.

The error in your calculations are that you've assumed independent hands--that is, the calculation above is for 8c6c against one hand, spread the board, then redeal an entire new hand against a random hand. That's not what's going on here. We've got our 8c6c, and we've got *one* board against the other hands. Totally different than the assumptions that make the calculation correct.

Seen another way, the % chance of 8c6c beating a random hand--what does that mean the 8c6c hit? Sometimes one pair, but often a flush, trips or something else (probably what's needed against the hands that we're up against in this example). So, if we hit trips, or a boat, to beat one of the hands, it's incredibly likely we've beat all the other hands too.
Reply With Quote
  #3  
Old 05-19-2007, 08:59 PM
Niediam Niediam is offline
Senior Member
 
Join Date: Jan 2005
Location: Michigan
Posts: 4,269
Default Re: Question About Hand 2-1 in Harrington Book

You are not considering the fact that you are only going to play past the flop when you make a hand/have a good draw...
Reply With Quote
  #4  
Old 05-19-2007, 10:27 PM
Kal With A K Kal With A K is offline
Member
 
Join Date: Apr 2007
Posts: 38
Default Re: Question About Hand 2-1 in Harrington Book

Basically, you have to consider implied odds. Let's say you're in the big blind position with 8[img]/images/graemlins/diamond.gif[/img] 6[img]/images/graemlins/diamond.gif[/img] and you're sitting with morons who would never fold AA, KK, QQ, JJ, or TT even to terrible flops.

Let's set the blinds at $1-$2, for simplicity. Players all fold to 5th position, who raises it to $10. Players 6 and 7 call, and everyone fold to you. So we have 3 players, and we're quite sure that they either have decent late position cards or great cards like high pocket pairs or over cards.

Now, if you weren't last to act, you should fold. But because your decision dictates whether or not you get to see a flop, I would call easily. There is $32 in the pot and it's $8 to call, so you have 4-1 pot odds. Not very great for your suited connectors, honestly.

But what if they hit? Say the flop comes up 457 (preferably of diamonds, which gives you the nuts no matter what, but rainbow is fine) and you check. Now, the player left of you thinks you're on a draw, because you checked. So he'll bet. The players next to him will call because they refuse to fold pocket pairs, though there may be one fold of overcards.

If the board pairs, you can be sure that you're safe, because they're playing high pocket pairs, so they probably have two pair, not a full house. So unless they hit runner-runner for a full house, you're safe.

Let's assume they're fairly deep stacked, at $200, and that they're not afraid to push all in with their pocket pairs. Pre-flop, it cost you $8 to win $32. But you'll win far more if you hit your flush or your straight. In fact, you'll win roughly $380 more, costing you $8 (the remaining $190 being money you won't be risking, since you'll be sure you've won) to win $420, giving you about 52-1 odds. HUGE odds.
Reply With Quote
  #5  
Old 05-19-2007, 11:49 PM
rakemeplz rakemeplz is offline
Senior Member
 
Join Date: Feb 2006
Location: +ev grimmstar bux vs everyone
Posts: 1,803
Default Re: Question About Hand 2-1 in Harrington Book

".462^4 = .046"

Either your math is terrible or you're trolling the forum (I'll assume your math is bad). Calling 100 to win 650 immediately with more implied odds is a simple decision, something you can easily learn from watching poker on tv (no offence intended to the OP, I'm no poker genius). My advice is to simplify your thinking rather than complicating easy decisions with faulty and/or complex mathematics (which in this case is not only wrong, but the assumptions making it are wrong).

Apologies if I've partially repeated someone else's post...

Basicly your pot equity against the types of hands you could be against is pretty good (You are NOT up heads up against a random hand i.e. 26 offsuit, you're up against the range of hands your opponent would play in their "position", i.e. a better than random hand), your position is good, and you have decent implied odds as well.
Reply With Quote
  #6  
Old 05-21-2007, 05:43 PM
gedanken gedanken is offline
Senior Member
 
Join Date: Feb 2007
Posts: 261
Default Re: Question About Hand 2-1 in Harrington Book

hmm, My hand is .462 to win, there are 4 hands about as good as mine, so each is .462^4?

but if they're all about as good as mine, we're each .25 to win.

get pokerstove and play with that. That will give you some idea of what 86s is worth 4-handed. you certainly are not .046!

don't forget implied odds. In NL holdem, you don't need anything like actual pot odds to call.
Reply With Quote
Reply


Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump


All times are GMT -4. The time now is 05:54 AM.


Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2024, vBulletin Solutions Inc.