How bad can a loose call be? Plenty

[binions]

If I have 4 hearts in Holdem. 9/46 chance to hit the
heart on the last card - 19.5% chance or odds of approx 4:1
against.

Assume there is $100 in the pot and it costs me $50 to call and see the last card, then the pot is only giving me 2:1 odds on a 4:1 shot. 4 times I will lose $50, and one time I will win $100. Over 5 tries, I will lose an average of $20 per hand.

How good a situation do I need to find to offset this loose call? Will getting 8:1 on a 4:1 shot offset the time I took 2:1 on a 4:1 shot? NOT EVEN CLOSE. How about 20:1? Will that balance out the loose call? NO WAY.

Assume there is $104 in the pot, and it only costs me $1
to see the last card, the pot is giving me 104:1 odds
on a gamble where I am 4:1 to win. 4 times I lose a dollar. One time I make $104. Expected value is $100/5 or $20 per hand.

As you can see, for every loose call where you are only getting 2:1 on a 4:1 shot, you need to find a 104:1 situation to break even on your flush draws.

Sorry folks, but 104:1 situations don't arise very often. 2:1 situations arise all the time in PL and NL poker.

I don't think the ordinary player appreciates this phenomenon.

PS This analysis does not account for implied odds. You can at times make up for a loose call by winning an extra bet on the end.

If you call $50 to win a pot of $100 with one card to come on a flush draw, then you have to be able to win another $100 on the end if you hit to make the pot odds 4:1 on that original $50 call (ie 50 to win 200). Lots of times, you can't win extra money when the flush draw hits, but you can win extra money when the straight draw hits since it is more hidden.

[phish]

You ought to write a book. And I hope every poker player buys it, studies it, and cherishes it. You'll be rich and I'll be rich.

[PairTheBoard]

Not sure what you mean by "offset". But if half the time you took 2:1 on a 4-1 shot and half the time you took 6:1 on the 4-1 shot you would break even in the long run.

eg. For every 10 times you win twice, getting paid 2 and 6. For a total of 8 in winnings when you win. The other 8 times you lose 1 each for a total of 8 lost when you lose. 8-8=0. You break even.

Of course that still would not justify accepting 2:1 on a 4-1 shot.

PairTheBoard

[TheShootah]

It's almost as if he has read TOP...but maybe I'm wrong.... [img]/images/graemlins/grin.gif[/img]

[Ace_Ren]

Big problem in your logic:

When you are taking the losing end of the draw, you are paying $50 and losing an average of $20.

When you are on the winning end of the draws, you are reducing the bet to $1. A really really bad thing to do.

What you should do is keep the bet at $50, or normalize everything back down to $1. So, with 104:1 pot odds on a 4:1 draw, a bet of $50 into $5200 on 4:1 odds would be an accurate comparison. And of course, you would make a huge profit on this. In 5 hands, you would win $5200 once and lose $50 4 times. Total profit: $5000 or $1000 per hand.

The actual balance is 6:1 pot odds to even out a 2:1 pot odds call on a 4:1 draw. With 6:1 pot odds normalizing to +0.4 per hand and 2:1 pot odds normalizing to -0.4 per hand.

[theRealMacoy]

Nice work Ren,

Thanks for clearing that up.

Cheers and welcome to the forums.

the Real Macoy

[binions]

[img]/images/graemlins/ooo.gif[/img]

[Kevin K.]

Hehe. Was wondering how someone could have reged over a year ago and still believe this.



[img]/images/graemlins/smile.gif[/img]