New question on Pot Odds: Sklansky vs. Harrington

[wuwei]

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Could be. I'll re-read both and find out. I've probably just stirred up a whole lot of foam over a simple mis-understanding. Again, I apologize.

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No problem, Bill. That's why the forums are here.

[Paul2432]

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My apologies to all.

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No need to apoligize. The purpose of this forum is to discuss poker.

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You're heads-up. The pot starts out at $200. First-to-act bets $100. There is now $300 in the pot. It's on you, $100 to call.

If you call, there will be $400 in the pot. Do you stand to win $400, or do you stand to win $300? Is the money you will bet money you will win, or does it not count?

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In this situation your pot odds are 3:1. Both Sklansky & Harrington would agree to this as would any experienced player in this forum.

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Let's try a thought experiment... same situation: heads-up. It's $100 to you, and there is $300 in the pot. You raise it to $200. There is now $500 in the pot.

Your opponent re-raises $200. There is now $700 in the pot, and it's $200 to call. Are the pot odds 7-2, 9-2, or are they only 5-2 because "you can't win back your original raise of $200?"

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None of the above. Pot odds are now 8:2, because there is $800 in the pot not $700. The pot started at 200, opponent bet 100, you raised 100 to 200, and your opponent raised 200 to 400. The pot contains the original 200, 400 from your opponent and 200 from you. You need to put in 200 more to call.

I would like to point out one other problem from another of your posts. You wrote:

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"If there is $300 in the pot, and it will cost you $100 to call, you stand to gain a $300 reward for a $100 bet, so you pot odds are 3-1." Note that he is talking about a player considering making a $100 bet into a pot that already has $300 into it. I interpret this as a $300 pot that will have $400 in it AFTER the call; hence, 4-1, as I have always understood it, as as Sklansky explains it.

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In this quote you have been loose with your terminology. There is a difference between making a bet and calling a bet. I think this may be the source of some of your confusion. If you make a bet where you want your opponent to fold your odds on the bet are (size of pot)size of bet). Harrington talks a lot about continuation bets where you might bet half the pot giving yourself 2:1 on the bet. This is a completely different situation from your opponent betting half the pot, in which case you are getting odds of 3:1.

Paul

[Bill Kolter]

Thanks. Great response! This helps a lot. I think I get it now.

[jwombles]

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My apologies to all. At the risk of complicating what I thought was a fairly simple question, let me rephrase the situation as I understand it.

You're heads-up. The pot starts out at $200. First-to-act bets $100. There is now $300 in the pot. It's on you, $100 to call.

If you call, there will be $400 in the pot. Do you stand to win $400, or do you stand to win $300? Is the money you will bet money you will win, or does it not count?

Let's try a thought experiment... same situation: heads-up. It's $100 to you, and there is $300 in the pot. You raise it to $200. There is now $500 in the pot. Your opponent re-raises $200. There is now $700 in the pot, and it's $200 to call. Are the pot odds 7-2, 9-2, or are they only 5-2 because "you can't win back your original raise of $200?"

In other words, does your own bet affect pot odds? I have to conclude that it does, thus your bet should be included in the odds, whether you're about to make it or you have made it. Pot odds don't apply if you're not in the hand....correct? Thus, if you are considering what the odds are IF YOU PLAY, then it seems to me that the bet you are ABOUT to make must count. It will certainly count on the next betting round. The question, it seems to me, is not "What are the pot odds at this point in time before I bet," but "What are the pot odds vs. my chances of winning IF I BET?"

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Whenever you are calculating your odds before you make a call, you don't include your bet. The reason is because your money or your call is not in the pot yet, ie, you can still fold and hold on to your 100.

For example, $300 in the pot and it's $100 to call means you WILL be getting 3 to 1 on your money. You don't include your $100 b/c it's NOT in the pot yet.

It's really that simple.

Much success,
Wombles

[starvs]

when using percentages to determine if you should call or not you DO include the money you will be putting into the pot to call in calculating what percent of the pot you are calling, correct?

[EarlCat]

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when using percentages to determine if you should call or not you DO include the money you will be putting into the pot to call in calculating what percent of the pot you are calling, correct?

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I only think that matters if you are raising the pot...then you would (in your head) first call (adding to the pot) then raise the total of the pot after you called.

For instance, if there's $30 in the pot after the flop and your opponent bets $5, raising the pot would cost $45 ($5 call plus $40 raise).

[BugsBunny]

That's correct, because with percentages you're actually calculating equity. And your bet is part of the equity calculation.

Pot contains $400. Your hand has a 20% chance of winning. It's $100 to call.

This is a breakeven situation:
20% * (400 + 100) = 100

20% is the same as 4 to 1 ( (1 - .2)/.2 = 4 ) amd 4 to 1 is obviously break-even here.

[btwnfdngs]

For any newbies like me who have been misapplying this concept (what a leak!), I'd like to sort of belatedly restate what I think Bugs and others have said (thanks), because I'm not sure it's really been made explicit where the confusion may come from. I can't answer the original question of how to correctly define pot odds, and in fact I'm sure I'll bungle the terminology, but I want to point out how the distinction discussed in this thread affects the overall purpose of calculating pot odds, which of course is to be able to compare the pot odds "ratio" (I'll call it) to the chances of winning "ratio" to determine the correct play in the long run.

The most important thing is not to use the commonly accepted definition, but rather to use whatever definition you choose in conjunction with the appropriate corresponding method of expressing your chances of winning. To use Bugs' example, say the bet's to you, the pot is 400 before you bet, it's 100 to call, and your chances of winning the hand are 20 %. (Again, I know the terminology is wrong, but I hope the meaning is clear.) If you express pot odds as 4:1 (size of pot before you bet : your bet), then you must express your 20% chances of winning as 1:4 (number of times you'll win per : number of times you'll lose).

But if you do include your unmade bet the in pot odds ratio, so that your pot odds under this scenario would be expressed as 5 to 1, then the corresponding correct way to express your 20% chances of winning is, 1 in 5.

Obviously, if you were express your pot odds as 4:1 and your chances of winning as 1 in 5 (or 5 to 1 and 1:4), you wouldn't realize this is a break even bet (unless you've learned to make the translation in your head, which I'm sure many veterans do).

If, like me, you forget, think of a coin flip. If the prize for calling it correctly is 1 (the pot), and the bet is 1, then your thinking is either: 1 to 1 odds, 1:1 chances of winning, so it's break even; or 2 to 1 odds, 1 in 2 chances, so it's break even.

If any of this is wrong, please correct it and I'll never post again. If it's right, hopefully there's another newbie moron like myself out there who'll find it helpful.

[andyfox]

How can you win $400 if there's $300 in the pot? If the bet to you was $1,000,000, would you stand to win $1,000,300 if you call?

[Kevin K.]

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But if you do include your unmade bet the in pot odds ratio, so that your pot odds under this scenario would be expressed as 5 to 1, then the corresponding correct way to express your 20% chances of winning is, 1 in 5.

Obviously, if you were express your pot odds as 4:1 and your chances of winning as 1 in 5 (or 5 to 1 and 1:4), you wouldn't realize this is a break even bet (unless you've learned to make the translation in your head, which I'm sure many veterans do).


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You're making this more complicated than it is. The problem is you can't include the unmade bet. Just don't do it. Anybody who participates in any form of gambling won't know what you're talking about when you call 5 to 1 and 1 in 5 the same thing, because they aren't. 5 to 1 = 1 in 6, period. You're speaking a different language altogether.

It seems that some of the confusion comes from attempting to convert the pot odds into a percentage. If that's the problem, just skip the conversion. It's not necessary. Just compare the pot odds to your odds to complete your hand, win the pot, etc.

Ex:
$90 in the pot, $10 to go = 9 to 1.
8 outs to hit your straight on the river = 38 to 8 = 4.75 to 1. Call.

You don't HAVE to know that you're 1 in 5.75 or 17% to make the call. The percentages, translating 9 to 1 to 1 in 10 or 10%, etc. will become second nature before too long.