Pot odds on the flop re flush draw

Hi guys

Im a bit confused and was hoping someone can iron something out for me.

I have been reading up quite a bit on pot odds, implied odds, expected value etc.

What i am struggling to get my head around is this....

Lets say I have A A. My opponent has two suited cards.

I raise x amount and he calls.

The flop comes down with two of his suit, but im still ahead.

Lets say the pot is $10 at this stage.

How much do i need to bet to make give him the wrong odds to call?

As far as I can see at the moment, even if I bet the pot $10 then he will have to bet $10 to return $30, thus giving him a break even percentage of 33%. Therefore, he is still right in calling this bet, correct?

It seems to me that unless you bet like twice the pot he will pretty much have decent odds, especially implied odds to call....

Am i right?

You don't actually count his call in the size of the pot - so after you've bet $10, the pot is $20, and he needs to put up $10 to win that $20. The pot's offering him 2:1, which is a losing proposition in the long run for a flush draw.

No, a $10 bet into a $10 pot would be ($10+$10)/$10 = $20/10 = 2:1. (edit: what oz said )

Speaking strictly in terms of pot odds (omitting implied odds), to give him bad odds in your example you need to bet more than x...

4.11 = (10+x)/x
4.11x = 10+x
3.11x = 10
x = 10/3.11
x = $3.2

Considering implied odds, we usually want to give them worse than 3:1 though, so we'd bet like 2/3 (giving ~2.5:1) or 3/4 the pot (giving ~2.3:1).

Ok then.

So switching roles, if i have the flush draw and there is a $10 pot, the most i should call to see the turn is $3.20?

I was thinking that my call counted towards the pot i would win and therefore had to take that into consideration...

Nope, your call never comes into consideration with pot odds. It's:

[amount in middle]+[any bets ahead of you]/[amount to call] = x/y => x:y

Thanks very much. That helps alot.

Just read this though, which I though was interesting in terms of when you should call/fold etc....

ESPN - Forget pot odds - Poker

All Phil's doing in that article is demonstrating a way of converting pot odds to percentages to show what % of the time we need to be ahead to call.

He uses an example where we're getting 4:1 pot odds, and goes on to calculate with his equation that we need to win the hand at least %20 of the time. This is exactly the same as saying we're a 4:1 dog, so we need to win the hand 1 in 5 times (or %20).

Don't let it confuse you - if that method's easier for you, then use it, but he's talking about the same thing there.

My brain hurts now....

TY Chuck for the good answers

If it is an all-in bet, yes. If your opponent still has money left though, you likely will still be able to make some money if you hit a flush, so you can call slightly more than $3.20. It all depends on what kind of hand you think your opponent has and whether he will bet or call a river bet.

Just quickly re visiting this to make sure my head is in the right place.

Take this example:

I have J 10 and I have put my opponent on AA or KK through my amazing read (bare with me hear).

The flop comes 9 8 2 (suits don't matter in the e.g.)

Now according to the odds calculator i have about a 34% of winning vs his 66%.

So lets say the pot is $60. My opponent bets $25 into the pot and is all in.

That means I have a 34% to win $85....correct? So because 34% of $85 is $28.9 this is a profitable move as I will return more than i put in in the long run and make a $3.90 profit?

I am correct here? or do i take the $85 pot plus my $25 call to make it $110 and then times that by 34% to get $37.40 and show a $12.40 profit in the long run?

I know this may sound jibberish but please i need to get this right in my head hahaha

The theory's correct - you count your opponent's bet, but not the amount of your call, in the pot size.

The percentage is slightly out though - assuming all your outs are clean, an open-ended straight draw actually gives you eight outs (four queens, and four sevens).

With eight outs, you're approximately 31.5% to hit on either the turn or river. 31.5% of $85 is $26.35, so you're still going to show a profit, just not quite as big a one as you mentioned.

It's possible that you actually are 34% on the flop, BTW - if you used an odds calculator, it's probably including the possibility of runner-runner trips, two pair or maybe a backdoor flush. But the straight draw on its own is only 31.5%.

Aleeki, it's much easier to just use ratios instead of %, makes the math much easier when you are playing. Here's a chart to check out you should try to memorize some of the ones like 8 outs (open ended str8 draw), 9 outs (flush draw), 4 outs (gutshot).

Your browser is too old - Texas Holdem Poker

Now using your last example where you have J 10. Pot is $60 and he's all in for $25, so you're pot odds are 3.4:1. You have 8 outs for the straight plus two overcards. Jack or a ten might be an out but they might not if he has a set or overpair. But if he has Ace 9 or King 9 then they are outs. So this is where it gets tricky because you have to make an educated guess, I'd count them as another 3 outs. So 8 + 3 is 11 outs total. If you look at that chart, odds against you hitting one of your outs on both the turn and river (since you get to see both by calling because he's all in) is 1.4:1, so you have to be getting better than 1.4:1 pot odds so easy call.

What are you doing calling a $30 bet preflop with 10 J when your opponent only has another $25 left? You're not getting the implied odds you need to call. Fold pre-flop.

j/k looks like your getting it, but I too prefer to use odds rather the percenages.