Betting to Disrupt the Pot Odds

[dabluebery]

**Also posted in Probablity, for you pot odds junkies**

A situation arose in a NL holdem tournament that has me asking about how to bet with the lead to disrupt people drawing.....

With even stacks (2000), 9 handed, Jason raises to t-200 with AKo and the blinds at 25-50. Chris, on the button calls, and so does the small blind.

The flop was AJ8, two diamonds. Jason bets t200. Button calls, small blind folds. Turn is a 4, not a diamond. Jason bets another 200, button calls. The turn is a 6, and not a diamond. Jason checks his top pair top kicker, button bets 400, Jason calls.

Button turns over 56 of diamonds, and was drawing to a flush, but backed into a straight. Jason knows he underbet the pot twice, giving the button odds to call. But we're not sure of how much is *optimal* to bet here.

You can calculate the break even point where the pot odds give you an EV of zero on the drawing players call. But how much MORE than that does Jason have to bet, to force the drawing player to fold? (I realize that betting more than that break-even point is +EV for Jason, because then the caller will be taking the worst of it from a pot odds position.)

Theory of poker doesn't talk much about an *optimal* betting strategy when betting from the lead. Can anyone fill us in?

Rob

[Milky]

When I have TPTK on a board with a flush draw, I always bet the pot at least. You don't want anyone drawing to the flush, and (especially in this case) you'd be happy to pick up the pot right then without having to see another card.

These are just the thoughts from a regular $5 + $.50 SnG player so take from them what you will /images/graemlins/smile.gif

[blendedsuit]

<font color="blue"> I do the same with TPTK, and make a bet atleast the size of the pot on the flop with draws available. Sometimes, when the blinds and pot are large enough, you just push all-in at the flop, and people on those draws will still call even when they clearly don't have the correct odds to call. I find this occurring often to me in the lower $$$ sng's and MMT's.</font>

[naphand]

Flush draws are drawing at 1.9:1 with 2 cards to come, OESD 2.2:1. By the Turn this is 4.1:1 and 4.8:1 respectively.

Of course they will make some other hands too (like runner trips or 2-pair etc.), but equally the board may go 4-suited or pair and they still lose to a better suit or FH.

The odds you need to consider then are approx. 2:1 on the flop and 4:1 and 5:1 on the Turn.

One thing to consider is that when they call the flop bet, they are drawing for 2 cards at these odds, but in reality they are likely to have to pay again on the Turn. I think HFAP or TOP deals with this in some detail, but I cannot remember it off-hand.

Betting the pot on the flop, means the caller is getting odds of 2:1, he has to call your pot-sized bet, but has that plus the pot, so 2X his bet to call. Is betting the pot enough? From this calculation any flush still has enough odds to draw even HU. If any other player is likely to call then they easily have the odds to call.

My personal feeling is you need to over-bet the pot against a draw HU. A pot bet is nice and easy, what with the "bet the pot" button, but is it enough?

If you think someone else is likely to call, then it is unlikely you can force the flush draw to draw incorrectly. You cannot let him draw free though, so a bet must be made. If the Turn is friendly to you, then a pot-sized bet is certainly correct and makes it impossible for him to draw correctly, he must fold.

However, if you consider each street as a single-card draw, (ie. you definitely bet the Turn as well) then the draws are only 4:1 and 5:1 respectively, and a pot sized bet is correct in both cases.

Can anyone explain the mathematical concept behind this? It seems to me that the odds to draw to a flush (at 2:1 with 2 cards to come) is much more relevant to limit poker, where your opponent never faces more than a few BB to draw. Can we ever consider a flush as drawing to TWO cards? where another bet is cewrtain on the Turn? Or is it more correct to only consider each as single card draws?

[BrettK]

Thanks for bringing this up, Naphand. I've always calculated for one card at a time, figuring that if I'm making a mistake on the flop, it's not by much and not often.

[ QUOTE ]
Betting the pot on the flop, means the caller is getting odds of 2:1, he has to call your pot-sized bet, but has that plus the pot, so 2X his bet to call. Is betting the pot enough? From this calculation any flush still has enough odds to draw even HU. If any other player is likely to call then they easily have the odds to call.

[/ QUOTE ]

Like you said, the calculation isn't this simple unless one of you is all-in, because the turn bet has to be taken into account. Using a HU situation as an example, if you plan to bet the pot again on the turn when any card other than a third card of that suit falls, you'll be betting three times what the pot was on the flop(the original pot plus your flop bet and his call), so while it's true that over two cards after the flop the odds are 2:1 against him making his hand, it's also true that his pot odds are only 5:4. Stack sizes are important here, especially if you'll pay him off when he hits, because implied odds may make it a correct call.

I recently read in either TOP or TPFAP that there are even situations in which it's correct *not* to bet the best hand, because there are so many scare cards in the deck that as long as your opponent is a good player and knows that you'll fold if he bets a scare card, he actually has a better chance of winning the hand. This is not very relevant, but I thought it was interesting to note.

[ QUOTE ]
However, if you consider each street as a single-card draw, (ie. you definitely bet the Turn as well) then the draws are only 4:1 and 5:1 respectively, and a pot sized bet is correct in both cases.

Can anyone explain the mathematical concept behind this? It seems to me that the odds to draw to a flush (at 2:1 with 2 cards to come) is much more relevant to limit poker, where your opponent never faces more than a few BB to draw. Can we ever consider a flush as drawing to TWO cards? where another bet is cewrtain on the Turn? Or is it more correct to only consider each as single card draws?

[/ QUOTE ]

I agree. Being able to create your opponent's odds for him makes things much more complicated. It would be great to hear from one of the math gurus.

Brett

[poolshark]

Just my 2c:
In my mind for NL - especially true for NL cash game than tourneys - the correct odds are dictated not only by what's in the pot 'Pot odds' , but also by the stack sizes of the people who might be likely to call all in if your draw hits 'implied odds'. I know this is a very fundamental concept but is often ignored by people betting against the person drawing. In tournament - another factor that comes into play is the blind level. The reason for that is because while drawing, you also need to protect your stack from severe damage if the draw misses. In lower levels - it is much easier to call a pot sized bet than in higher levels. example - 10/15 - 4 ppl call a $30 raised pot - pot size 120 - person flopping top pair (or better - even better odds of calling all in!) bets out pot - 120 - and has a stack size of 560 left (assuming first hand) - person on flush draw has to call 120 for the 240 in the pot (assuming no one is calling) and also the 560 potential win - or atleast half of that - depending on how you play it. In this case though - the important factor is that you CANNOT count on two cards if you are the one drawing. You will have to hit your flush on turn for this play to be profitable - because most likely you are not going to be able to see the river without calling all in if you missed on turn. Ofcourse all of this is hypothetical - but the point is - you will have to bet MORE than the pot for the drawers to get out.
In higher levels - it is much tougher to call a t400 bet for instance in the same situation.

Thoughts?

[TheDrone]

[ QUOTE ]
Theory of poker doesn't talk much about an *optimal* betting strategy when betting from the lead.

[/ QUOTE ]
How do you know that you have the lead with TPTK? For example, let's say you raise 3BB preflop in MP with A /images/graemlins/heart.gif, Q /images/graemlins/club.gif, the button calls, and the flop comes Q /images/graemlins/diamond.gif,T /images/graemlins/diamond.gif,6 /images/graemlins/club.gif. There's 7.5BB in the pot and you bet 8 BB. Button calls. Now what? At this point you have 11BB invested (could be half of your stack) and you don't know if your opponent is on a draw, has a weaker kicker, or is slowplaying a set.

I have found that making pot-sized or larger bets with TPTK can be very costly in a tournament except in certain situations, therefore I do not employ a standard bet with TPTK. For example, I prefer to be more aggressive with TPTK when I am last to act and there are only 1 or 2 other in the pot with me. There are many scenarios where I will tend to bet less than the pot and shut it down on the turn.

Being able to judge when you are behind with TPTK is a skill that comes more from experience in assessing all variables of the situation at hand. I don't think you can boil it down to a formulaic strategy in NL tournament games. Or maybe someone can, but the formula would be quite complex given the number of variables.

[ddubois]

As mentioned, there are a number of factors. It's not just pot odds, but also: "Will this opponent give me his whole stack if my draw comes in?"

It's probably possible to make a mathematical formula that emcompasses the size of the pot, the odds involved, and the stacks size of both players. I think it would be complicated though.

Let's look at one example:
Stacks: Hero 700, villan 900
Blinds: 15/30
Pre-flop: Hero bets 100, one caller form the BB
Flop: Pot 215, hero with TPTK bets pot 215. At this point it costs villan 215 to get a shot at somewhere at least 430 and maybe as much as 815. Paying this fee only lets him see one card, and he will catch his draw roughly 1/4th of the time (again, varies on number of outs). So far, it looks like villan would be making a small mistake to call.
Turn: Draw missed, pot is 645, hero has 385 left. Hero pushes, now villian is getting charged 385 for a chance at 1030. With only a 1:4 chance of hitting, again, he's making a mistake to call.

The sucky part about this is hero has to commit his whole stack if he wants to charge the draws enough. So if he's behind a set or two-pair, he's screwed - there's no way to bet small enough to get away without opening a hole for flush draws to slip through. If someone has more than 9 outs, like an open-ended straight-flush draw, you are definately SOL. There's no way you can charge them enough, at least not on the flop. (On the turn it might be possible, depending how much stacks are left.)

Change hero's stack to 2000, and and two things change: 1) villan's implied odds on the flop change to a range of 430-1015, which means 215 might not be charging enough if hero will pay off, and 2) it's much less painful for hero to bet the requisite amounts to force out draws, without worry of being eliminated by 2-pair/sets.

[ddubois]

It seems to me the point of 0 EV in terms of pot odds (ignoring implied odds for now) for the drawer is this:

(pot + bet) * outs / (47-outs) = bet

This is like saying reward*odds = cost.

So far so good, yes? When I plug in numbers, like pot 300 and and bet 100, odds 1/4, the formula seems correct and intuitive.

I do some algebra and solve the equation for bet, I get this:

bet = (pot * outs) / (47 - 2 * outs)

and that seems good when I plug in numbers. (For the t300 sized pot, charging the flush draw requires a t93 bet, assuming no implied odds). So let's go back and consider implied odds now:

(pot + bet + (rest of stack)) * outs / (47-outs) = bet
(pot + bet + (stack - bet)) * outs / (47-outs) = bet
(pot + stack) * outs / (47-outs) = bet

So for my example from the previous post of the t600 stack (was t700 at start of hand, its t600 after pre-flop raise) and the t215 pot, hero needs to bet: t253. This is the break even point, if and only if hero will lose his whole stack when the draw hits.

So if hero can play perfectly (according to FTOP), he only needs to bet t94+, but if he can't let go of his AA under any circumstances, then he needs to bet over t253 to protect his hand.

Comments welcome.

[ddubois]

Just for giggles, I extrapolated this into pre-flop raise sizes. Like, if you have no self-control and can't fold your QQ when the flop comes Axx or 9c8c7c, how much do you have to bet to kill the implied odds of hands you are 2:1 against (AXs, KXs), 7:3 against (AXo, KXo), or 4:1 against (98s, 55), etc? The answer seems to be:

bet = (pot + max_stack) * heads_up_odds

(800 + 50) * 1/2 = 426
(800 + 50) * 3/7 = 364
(800 + 50) * 1/4 = 213

Betting 60% of your stack would be rather silly obviously.