Pot Odds Question

[Crooked Paul]

What other posters have said is correct, but I think it might help you more if someone pointed out specifically where you seem to be confused. Might as well be me. =) You wrote:

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Here's the thing though. Let's use your example of getting 10:1 on your money and let's say you have like 4 outs on the river - about 9:1 probability. The one time you hit it has to make up for the other 8 times you miss.

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This is exactly correct. In fact, this is the very definition of pot odds, albeit reworded. The entire reason we calculate pot odds at all is to find the answer to the question "How often do I have to win this pot so that I won't be losing money in the long run?" So for example, if you have to call $10 for a chance to win $100, as long as you win that pot 10% or 1/10th of the time, you won't lose money.

Now, with that number in mind, we then look at our mathematical odds to make a killer hand which we assume will bring us the pot. If our chance to make the killer hand is greater than our pot odds (>10% from the example above), then we say this is a "correct" play, because probability being what it is, if you made that same decision 1000 or a zillion times, you would make money.

It seems like you might be confused because we are not, in fact, faced with an identical decision 1000 or a zillion times. But you see, that doesn't matter. As long as you use this formula on each hand, plugging in the current values each time, you're sitting pretty.

Look at it this way: Suppose you have to call $10 to win $100 and you have great hand odds, like 20%. Probability is fickle in the short term, but over time the results even out, right? So if you made this decision 1000 times, you would expect to lose 80% (800 times at $10 a pop = $8,000) and win 20% (200 times at $100 each = $20,000). That's a net gain of $16,000. Now, if you have any faith in mathematics and probability at all, I hope you will grant that if you could do that same thing over and over, that 16k gain would actually be the result. Okay, but you're not doing it 1000 times, but just once. So divide your expected winnings by 1000. $16,000/1000 = $16. That means, for each of those $10 calls you make, you can expect $16 back, on average. And that is what we mean by expected value or EV.

Not to beat a dead horse, but let me quote you again to make sure you've got it.

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If in your successful attempt you put in 2$ and won $20 but the other 8 times you missed you put in an average of 3$ you lost 24$ for a net loss of 4$. So even though you had good pot odds each time and you matched the statistical expectation, this was not a profitable play.

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Dude, you did a bait-and-switch here. You changed the value of the bet on the "missed" attempts. But changing the value of the bet changes the pot odds, so you're comparing apples and oranges. This part of your post should have read: (italics indicates corrections)

"If in your successful attempt you put in $2 and won $20, and the other 8 times you missed you put in $2 and lost $16, for a net gain of $4."

Okay, I'm sorry to have blathered on. But I don't see how you could possibly still be unclear on the idea after all that, so it was time well spent.


Crooked

[Medibart]

I wasn't trying to do a bait and switch but I can see how that could be construed.

I said "If in your successful attempt you put in 2$ and won $20 but the other 8 times you missed you put in an average of 3$ you lost 24$ for a net loss of 4$. So even though you had good pot odds each time and you matched the statistical expectation, this was not a profitable play".

What I was suggesting was that for each of the unsuccessful attempts you had the same 10:1 pot odds. The pot odds didn't change just the size of the wager and the size of the pot. So I don't think it's a case of apples and oranges but of different sized apples.

But if I understand what you are suggesting - if you have the same 4 outer on the river - let's say for the nut inside straight draw (about a 9:1 probability) and one time you have to bet 2$ to win $20 and another time you have to bet $10 to win $100, those two events have to be considered completely independent of each other. So even though you might win a few of the $20 pots and you might lose more of the $10 wagers, thus creating a net loss on the same play that isn't a consideration here because they are entirely independent events.

Is that correct?

[ThisHo]

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But if I understand what you are suggesting - if you have the same 4 outer on the river - let's say for the nut inside straight draw (about a 9:1 probability) and one time you have to bet 2$ to win $20 and another time you have to bet $10 to win $100, those two events have to be considered completely independent of each other. So even though you might win a few of the $20 pots and you might lose more of the $10 wagers, thus creating a net loss on the same play that isn't a consideration here because they are entirely independent events.


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First of all, a gut-shot ("inside") straight draw is closer to 11:1 than it is to 9:1. I use 10:1 because its easy to remember, but the actual math is 10.5:1. I just wanted to point this out.

Second:
stop thinking in terms of $ and think in terms of units bet. It doesn't matter if you bounce between limits. If you play for the rest of your life then the law of large numbers takes over and things return to the average (i.e. 1 time in 11.5 tries you'll hit your gut-shot no matter what level you play).
If you have to call 1BigBet when there are 10BigBets in the pot then you are getting 10:1 no matter what that BigBet is ($1, $6, $10, $100, $1000, $10,000 -- doesn't matter).

The reason that its so important to play within your bankroll is that its not unlikely that you could go 15 or 20 gut shots in a row and miss them all. If you are getting correct pot odds to make the call you are "winning" in the long run, but short term variance is a cold hearted biyatch!

Pot Odds = $in pot / $ to call current bet
Cards Odds = (cards left - outs) / outs

If Pot Odds are larger then card odds you need to call no matter what size the apple is.

ThisHo

[Paul2432]

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I said "If in your successful attempt you put in 2$ and won $20 but the other 8 times you missed you put in an average of 3$ you lost 24$ for a net loss of 4$. So even though you had good pot odds each time and you matched the statistical expectation, this was not a profitable play".

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I think you are getting the hang of it. I'd just like to point out that whether or not a play is profitable is not determined by the short term results. If you have favorable pot odds to make a play, then that play is profitable. Period. In fact it is quite likely that from time to time a large number of profitable plays will yield no actual profit.

Paul

[Medibart]

First of all, a gut-shot ("inside") straight draw is closer to 11:1 than it is to 9:1. I use 10:1 because its easy to remember, but the actual math is 10.5:1. I just wanted to point this out.

<font color="green"> </font> [img]/images/graemlins/diamond.gif[/img] [img]/images/graemlins/diamond.gif[/img]I was just using the handy dandy 2+2 method. 4*2 + 2 = 10% = 9:1. [img]/images/graemlins/diamond.gif[/img] [img]/images/graemlins/diamond.gif[/img]

Second:
stop thinking in terms of $ and think in terms of units bet. It doesn't matter if you bounce between limits. If you play for the rest of your life then the law of large numbers takes over and things return to the average (i.e. 1 time in 11.5 tries you'll hit your gut-shot no matter what level you play).
If you have to call 1BigBet when there are 10BigBets in the pot then you are getting 10:1 no matter what that BigBet is ($1, $6, $10, $100, $1000, $10,000 -- doesn't matter).

The reason that its so important to play within your bankroll is that its not unlikely that you could go 15 or 20 gut shots in a row and miss them all. If you are getting correct pot odds to make the call you are "winning" in the long run, but short term variance is a cold hearted biyatch!

Pot Odds = $in pot / $ to call current bet
Cards Odds = (cards left - outs) / outs

If Pot Odds are larger then card odds you need to call no matter what size the apple is.

<font color="green"> </font> [img]/images/graemlins/diamond.gif[/img] [img]/images/graemlins/diamond.gif[/img]This is exactly what I said in my original post. Was just looking for logical or mathematical confirmation. So you agree that as your number of hands increase the bets average out i.e. become a constant. [img]/images/graemlins/diamond.gif[/img] [img]/images/graemlins/diamond.gif[/img] [img]/images/graemlins/cool.gif[/img]

[ThisHo]

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I was just using the handy dandy 2+2 method. 4*2 + 2 = 10% = 9:1.

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I've never seen the "outs*2 + 2" calc. I just use "outs *2" as a rough estimate. Using that, 2*4 = 8 .. 92/8 = 11+... anyhow... its close and in most situations being close is enough, just understand that in situations where the decision is close, 9 isn't quite right.

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This is exactly what I said in my original post. Was just looking for logical or mathematical confirmation. So you agree that as your number of hands increase the bets average out i.e. become a constant.

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In your original post your example got messed up, but yeah - as your number of hands approaches infinity the variation from the odds approaches zero (the more hands you get the closer to the actual odds you get). Keep in mind that the cards have NO memory, so you can't say "I've missed my last 9 gut-shots, I'm Due! ALL-IN!!!!!!!!!!!!!" That is most assuredly -EV. [img]/images/graemlins/grin.gif[/img]

You're on the right track... keep working at it!
ThisHo

[Crooked Paul]

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What I was suggesting was that for each of the unsuccessful attempts you had the same 10:1 pot odds. The pot odds didn't change just the size of the wager and the size of the pot. So I don't think it's a case of apples and oranges but of different sized apples.

But if I understand what you are suggesting - if you have the same 4 outer on the river - let's say for the nut inside straight draw (about a 9:1 probability) and one time you have to bet 2$ to win $20 and another time you have to bet $10 to win $100, those two events have to be considered completely independent of each other. So even though you might win a few of the $20 pots and you might lose more of the $10 wagers, thus creating a net loss on the same play that isn't a consideration here because they are entirely independent events.

Is that correct?

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Yes, that's all correct. I think you see what we're saying, but this might make it even clearer:

What your example is essentially saying, what with its very low sample size and the contrivance that you win the small pot but lose the big ones (with the same odds), is really nothing more than "Wouldn't it suck if you made the same correct play 10 times, and you only won the one time the pot was the smallest?"

Yes, that would suck. But it's not statistically/mathematically relevant. It should not factor into your decision-making. Which I think you see now, but this seems a concise way to sum up.


Crooked