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  #1  
Old 03-22-2004, 03:54 PM
Steve S Steve S is offline
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Default Lou Krieger\'s pot odds article

In the March 26, 2004 issue of Cardplayer - Pot Odds Made Easy

Could someone read this article and tell me if this is correct? In the article he mentions the odds for a flush draw on the turn as 37 to 9. I always thought the odds were calculated by the number of outs from the total number of cards remaining, in this case 46 to 9.

"For example, whenever you hold four cards to a nut flush on the turn in a hold'em game, there are 46 unknown cards (52 minus your two cards and the four on the board). Of those 46 cards, 37 won't help you, but the other nine cards are the same suit as your flush draw, and any one of them will give you the nut flush."

"The odds are 37-to-9, or 4.1-to-1, against making your draw. Percentage poker players will call a bet in this situation only if the pot is four times the size of the bet. In a $20-$40 game, the pot would need to contain at least $160 — or else you'd have to be able to count on winning at least a total of $160 from future calls (this is called "implied odds," and is a guesstimate of sorts) — to satisfy this requirement."

Thanks,

Steve

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  #2  
Old 03-22-2004, 04:48 PM
uuDevil uuDevil is offline
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Default Re: Lou Krieger\'s pot odds article

[ QUOTE ]
I always thought the odds were calculated by the number of outs from the total number of cards remaining, in this case 46 to 9.


[/ QUOTE ]
Steve,

Krieger is correct.

When stating odds we are saying that there N cases where some event of interest occurs and comparing that number to the number of cases (say M) where that event does not occur. We state the odds as being M:N against the event occuring.

In this instance, the event of interest (a flush) involves N=9 out of 46 remaining cards. The flush will not occur when one of M=46-9=37 cards falls. So the odds against a flush are M:N or 37:9.

You may be confusing odds with probabilities. The information is the same, just stated differently. The probability of a flush is 9/46. This is the same as odds of 37:9.

Hope that helps. [img]/images/graemlins/tongue.gif[/img]
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  #3  
Old 03-22-2004, 06:28 PM
Steve S Steve S is offline
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Default Re: Lou Krieger\'s pot odds article

[ QUOTE ]
You may be confusing odds with probabilities. The information is the same, just stated differently. The probability of a flush is 9/46. This is the same as odds of 37:9.

[/ QUOTE ]

"The odds are 37-to-9, or 4.1-to-1, against making your draw."

I may just be confusing odds and probabilities. If the probability of hitting the flush is 9/46, wouldn't the pot odds need to be 5.11 to justify a call and not 4.1.

Thanks for your help!

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  #4  
Old 03-22-2004, 06:52 PM
RocketManJames RocketManJames is offline
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Default Re: Lou Krieger\'s pot odds article

[ QUOTE ]
I may just be confusing odds and probabilities. If the probability of hitting the flush is 9/46, wouldn't the pot odds need to be 5.11 to justify a call and not 4.1.

[/ QUOTE ]

If the probability of hitting the flush is 9/46, then that means that 9 times you will hit out of 46 and 37 times you will miss out of 46 (over the long-term). Thus, you need 37:9 pot-odds to justifiy a "break-even" call. Any more pot odds, and your call is positive in expectation (assuming you know you will win if you hit your flush).

Examples...

EV Calculation where pot is laying you 37/9 pot odds:
37/46 * (-1 unit) + 9/46 * (+37/9 units) = 0

EV Calculation where pot is laying you 46/9 pot odds:
37/46 * (-1 unit) + 9/46 * (+46/9 units) = +9/46

Therefore, as you can see with the above, if you had 5.11 to 1 pot odds (46/9), you are clearly +EV, meaning you don't need quite that much to justify a break-even call. At exactly the 37/9 pot odds, do you see that your call is 0EV.

Hope this helps.

-RMJ
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  #5  
Old 03-22-2004, 07:03 PM
BradleyT BradleyT is offline
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Default Re: Lou Krieger\'s pot odds article

9 times out of 46 you're going to win.

1 out of 5 times you're going to win. 20%
4 out of 5 times you're going to lose. 80%

80% loss : 20% win

Now can you see why it's 4:1 to call and not 5:1?
if it were ~16% win it would have to be 5:1 to call
If it were ~10% win it would have to be 9:1 to call

What would it have to be if you were 33% to win, 66% to lose?
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  #6  
Old 03-22-2004, 08:46 PM
uuDevil uuDevil is offline
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Default Re: Lou Krieger\'s pot odds article

[ QUOTE ]
If the probability of hitting the flush is 9/46, wouldn't the pot odds need to be 5.11 to justify a call and not 4.1.

[/ QUOTE ]

Steve,

No. Expressing your chances of making the flush in terms of odds is useful because you can directly compare these odds with the pot odds you are getting to make a call. Assuming there would be no further betting and that you win if you make your hand and lose if you don't, the pot would have to be 4.1 times the size of the bet you are calling for you to break even. If you are getting any amount more than 4.1:1, then calling makes money, and you should do it.

The number 46/9=5.1 isn't relevant in this problem. The probability 9/46 is the same as odds of 4.1:1 against. To convert this probability to odds, you would do the following calculation:

odds against=(1-probability)/probability

So odds against flush=
(1-9/46)/(9/46)= (37/46)*(46/9)= 37/9= 4.1

Next you should look at implied odds, which can sometimes be used to justify a call when you are getting less than the required pot odds.

[I saw the other responses, but I thought it would be better not to introduce new terminology until these ideas are clear.]
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  #7  
Old 03-23-2004, 12:26 PM
mosta mosta is offline
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Default Re: Lou Krieger\'s pot odds article

There is a terminology difference. If you bet red on a roulette wheel, the payout can be described as: 'one TO one', or 'two FOR one'. ie, the first thing says for each chip you bet, if you win, you get one chip. the second says, you pay one chip for the chance of winning a two-chip pot. '37 to 9' ('37:9') = '46 for 9'. same thing said different ways. suppose the pot is fair to draw to the flush on the river (no implied odds--your opponents are all-in) you could say, the return on my $9, IF I bet it, is $37 if I win. Or you could say, it would cost me $9 to win a $46 pot. 4.1 to 1 is the same as 5.1 for 1. if something has a 33 & 1/3rd% chance of occuring, fair odds are 2 to 1 or 3 for 1. 40% chance is 4 to 6 (= 2 to 3) or 4 for 10 (2 for 5).
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  #8  
Old 03-23-2004, 06:58 PM
Steve S Steve S is offline
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Default Re: Lou Krieger\'s pot odds article

I think I was getting caught up w/ the terminology and ended up confusing myself with an easy (yet so important) topic.

Thanks to everyone for your responses.

Steve
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  #9  
Old 03-23-2004, 11:41 PM
uuDevil uuDevil is offline
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Default Re: Lou Krieger\'s pot odds article

[ QUOTE ]
ended up confusing myself with an easy (yet so important) topic.

[/ QUOTE ]
Steve,

You have plenty of company. In this thread:

http://forumserver.twoplustwo.com/sh...mp;o=&vc=1

David Sklansky, in an interesting response to a comment by MajorKong, estimates that only 25% of "good players" can do relatively simple counting probability problems. So it seems that basic statistical skills, though they may be useful, aren't essential to a "good player." [img]/images/graemlins/tongue.gif[/img]
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