Outs & Odds

[ark123]

I am trying to get a grip on outs & pot odds stuff & what I dont understand is this:
If after the flop I have say 5 outs,I understand that with 2 cards to come my odds of making my hand are about 4 to 1 against.However,if I don't make it on my 1st card my odds of making it with 1 card to come are about 8 to 1.If the
size of the pot does not justify calling at 8 to 1,should I fold?If so,whats the point of using the odds for 2 cards in the 1st place?
I don't know if I explained that vey well.The point is why
use the odds for 2 cards to come when you have to recalculate when you dont hit it on the 1st card.

[AKQJ10]

This is actually a very good question. As I understand it, the assumption is that with strong draws such as most flush draws, you're assuming that you will in fact have odds to stay in for both cards. Therefore, when appying pot odds to the decision whether to raise for value, you're happy to raise and get two callers because you're 2:1 to hit your flush with two cards to come. The implicit assumption is, you'll certainly be getting 4:1 to stay to the river if you don't make your flush on the turn.

For most weak draws (gutshots, middle pair), as I understand it you would be more interested in hitting your draw on the turn because you're not likely to stay past the turn anyway.

There's also a more complicated way to look at it which goes under the name effective odds (http://poker.wikicities.com/wiki/Effective_odds). This was discussed on this forum a week or so ago, so if you search on +effective +odds you might find that. But in short my understanding is that they're pretty difficult to figure accurately, and if you look at it as separate call-or-fold decisions on the flop and turn you won't go too far wrong.

I'm eager to hear if others agree with this, though, because this is a topic that's confused me a bit too.

[MrStretchie]

Basically what AK says.
I find an easier way to look at it is this:
Odds with 2 cards to come are only valid when you will be seeing two cards. If you don't have odds to call the flop just based on the next card (with implied odds), you'll just about never have odds on the turn, when the bet doubles. (The only times you would would be when the flop gets raised, and then you'd wish you hadn't called anyway.)

So if you don't have odds to call the flop based only on the turn card, you can't use 2-to-come-odds to justify it ever. When that takes effect is when you *do* have odds to call, and moreover, your odds to complete after two cards are better than your number of opponents.

Say there's 9SB in the pot, it's 5-way, and you have a bet and 3 callers to you on the button after the flop. You should raise.
Obviously you have odds to call, with the pot laying 9:1 even without implied odds. But by the river, your odds are about 1:2. In other words, you'll win about 33% of the time. Since you're putting in less than 33% of the money when you raise (there are more than 2 callers), the raise is for value.

That's about the only time to use 2 cards to come odds, unless your flop call will put you all-in. Then you know you get to see two cards without calling the turn, so you can use the 2 card odds.

[TaoTe]

outs & odds, that sounds like it should be the name of a poker book.

[TaoTe]

[ QUOTE ]


Say there's 9SB in the pot, it's 5-way, and you have a bet and 3 callers to you on the button after the flop. You should raise.
Obviously you have odds to call, with the pot laying 9:1 even without implied odds. But by the river, your odds are about 1:2. In other words, you'll win about 33% of the time. Since you're putting in less than 33% of the money when you raise (there are more than 2 callers), the raise is for value.


[/ QUOTE ]

Are you referring to a flush draw? Sorry if I missed that part. So what are the odds of making the flush on the turn and river seperately? I need to read SSHE again. It's 3:1 with two cards to come right? Also, you're raise may buy a free card on the turn.

[MrStretchie]

[ QUOTE ]
Are you referring to a flush draw? Sorry if I missed that part. So what are the odds of making the flush on the turn and river seperately? I need to read SSHE again. It's 3:1 with two cards to come right? Also, you're raise may buy a free card on the turn.

[/ QUOTE ]

Hmm.. yes I missed saying that didn't I? Yes, flush draw.
With two cards to come it's actually 1.86:1 against or about 2:1. So if you have 2 or more callers, with a bare flush draw, you can bet/raise for value on the flop. (Although with 2 callers the value's pretty thin, so you might wait if you thought it would help you get more later, or if you didn't have the nut draw (so your outs are worth somewhat less).)

The odds for the next card are really easy. You have 9 outs to your flush. On the flop there are 47 cards unseen, so your chances of hitting are 9/47 = 19%. In terms of odds, you have a 9:47-9 = 9:38 = 1:4.2 chance of hitting.

Same calculations on the turn, except only 46 unseen cards (works out about the same - 1:4.1)

Oh, and .. sorry but.. "your raise", meaning the raise belonging to you, not "you're raise", meaning "you are raise". I see that one so much, I've decided to crack down /images/graemlins/wink.gif

[AKQJ10]

[ QUOTE ]
[A flush is] 3:1 with two cards to come right?

[/ QUOTE ]

As the other respondent said, it's 2:1. However, the probability is 1/3, so your memory might be calling that number up instead.

As you probably know, odds of n:1 correspond to a probability of 1/(n+1). Or to look at it another way, odds are the ratio of p(success) to p(failure), p standing for probability.

[TaoTe]

Your, for you're? You're kidding! I wrote that? I'm a writer and have as many books on grammar as I do for poker, but I tend to be lax when posting on a message board. Good job, though.

[MrStretchie]

[ QUOTE ]
Your, for you're? You're kidding! I wrote that? I'm a writer and have as many books on grammar as I do for poker, but I tend to be lax when posting on a message board. Good job, though.

[/ QUOTE ]

lol, ok, good. /images/graemlins/smile.gif I just read that one so much.. I think I'm starting to have an aneurysm each time.