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View Full Version : Best way to think of odds when drawing to straight


thwang99
08-19-2003, 03:37 PM
Hi all,

Say the odds of hitting a straight on the turn are 4 to 1, and of getting your straight by the river is 2 to 1, and if you hit you'll win, if you miss you lose.

Say someone bets X on the flop, do you need the eventual pot to be 4X to break even? Or 2X? Or something inbetween? Say you go with the 4X figure, and call if the eventual pot will be greater than 4X. That's of course profitable. But you can call with a smaller pot too. But it's not 2X. Since if you miss the turn, assume the other player will bet the turn again.

Assume the other player has a set and will bet every round, and when you hit you will raise and the other player will call a raise.
- Tony

Copernicus
08-19-2003, 05:43 PM
You will be surprised by the answer, although your scenario is "perfect world", and there is a bit of a paradox, so maybe my math is wrong.

Assume the current pot (before his flop bet) is C big bets. Under your simplified assumptions that none of his cards can win but that he bets out on the turn and calls any raise of yours, if you hit on the turn the betting will go his bet, your raise, his call on the turn, and his check your bet his call on the river. So the pot (not counting your future bets) is C+3.5 big bets. If you hit on the river, the turn betting will be his bet your call, the river his bet, your raise, his call, so again its C+3.5 chips in the pot. If you miss on both you have committed only 1.5 additional chips (.5 big bets on the flop, one of the turn, fold on the river).

The math is 39/47*38/46*(-1.5) = (1-39/47*38/46)*(C+3.5)
.685*-1.5 = .315C + .315*3.5
1.028=.315C+1.103
C= -.237!!

In other words there could be nothing in the pot and calling is +EV.

You can check the reasonableness of the answer by saying that you will hit your straight a little less than 1/3 of the time, and miss a little more than 2/3 of the time, so you need a little better than 2/1 odds. On future bets only, you are getting 3.5 of his when you hit, but committing only 1.5 of yours when you miss, so you are getting 2.333/1 in future bets alone....you need nothing in the pot.

The paradox is this: You must commit to the whole scenario, because neither the flop bet (turn card) nor turn bet (river card) are plus EV. All of the value is in the river bet.

On the flop bet (turn card) you are only getting even money (his .5 to your .5), but you are a 39/8 dog and on the turn bet you are getting 2/1 but are a 38/8 dog.

It is only the 2 bets that you have committed him to on the river (that carry no risk to you) that make it a +EV situation.

OK..where is my math wrong, because it must be! Implied odds cant make an open ended straight draw +EV with nothing in the pot!...or can they. /images/graemlins/laugh.gif

thwang99
08-19-2003, 06:47 PM
Wow, I don't think implied odds can make a -$.23 pot even money! That's just "crazy talk" as my friend says.

The reason I asked is that I lost a $78 pot last night when I raised KK and someone called in the BB with an open-ended straight at a 3/6 game. Small pot on the flop, but he made his straight on the turn and raised me, and I called him down. (he had 97 and flopped T85).

It seems that in situations like this, with a straight draw versus someone who won't fold his hand to a raise, a straight draw has HUGE +EV. Mostly due to the ability to raise if hit. Let's say he misses, I'm sure the pot would have been way less than $78, and he wouldn't have lost much (just $9 to go for $78). My KK isn't looking as good as I thought intuitively! There was another player in the hand, 3 to the flop but the other player folded on the flop or turn, I forgot.
- Tony

BruceZ
08-19-2003, 08:48 PM
OK..where is my math wrong, because it must be! Implied odds cant make an open ended straight draw +EV with nothing in the pot!...or can they.

In this case they do. You will make 3 big bets after you hit on the turn, so of course you should pay 1 small bet to see the turn since you'd be getting at least 6-1 on a 5-1 shot, even if there were nothing in the pot.

Even if the opponent would not call a raise, you would win 2 big bets plus at least the 1 small bet you called on the flop, so you'd still be getting the required 5-1. In real life, the pot will always be at least 3 small bets on the flop, so you'd be getting at least 7-1.

You need nothing in the pot to call on the flop if he will always call a raise, and you only need 1 small bet in the pot even if he won't call a raise, and of course there will always be 1 small bet.

I disagree that you have to commit to the river to have +EV. You have +EV even if you bail on the turn when you miss. You're still getting at least 6-1 on a 5-1 shot. Your EV will be better if you call after missing on the turn as long as there are at least 3 big bets in the pot (always in a real game), assuming you will win 2 more on the river. If your opponent won't call the raise, then you need the pot to be at least 4 big bets to call on the turn.

BTW, odds of hitting a straight on one card are not 4-1, they are almost 5-1.

Copernicus
08-20-2003, 10:00 AM
[ QUOTE ]
OK..where is my math wrong, because it must be! Implied odds cant make an open ended straight draw +EV with nothing in the pot!...or can they.

In this case they do.

actually the "or do they" was a weak attempt at irony on my part...the math is straightforward enough

I disagree that you have to commit to the river to have +EV. You have +EV even if you bail on the turn when you miss. You're still getting at least 6-1 on a 5-1 shot. Your EV will be better if you call after missing on the turn as long as there are at least 3 big bets in the pot (always in a real game), assuming you will win 2 more on the river. If your opponent won't call the raise, then you need the pot to be at least 4 big bets to call on the turn.

I was only talking about the need to commit in the specific "nothing in the pot" scenario with thwangs betting/calling "rules". In a real game with blinds you can certainly have good current as well as implied odds.


[/ QUOTE ]

Bruce, in an earlier thread I "challenged" the textbook determination of "implied odds", which, as I read it, calls for projecting the ultimate pot (excluding your own money) and dividing by the current bet. If it may take 2 cards to hit, or if there is some risk that your hand isnt the nuts when you do hit, I believe the above calculation over-states your position in both the numerator (which should be weighted for the likelihood that you win if you do hit), AND, more importantly, in the denominator, since you have to commit additional money which is at risk to achieve the ultimate pot (and just excluding your bets from the numerator doesnt reflect that risk).

Any thoughts?

BruceZ
08-20-2003, 11:19 AM
I was only talking about the need to commit in the specific "nothing in the pot" scenario with thwangs betting/calling "rules". In a real game with blinds you can certainly have good current as well as implied odds.

That's what I mean, even when the pot only has only your opponent's flop bet (impossible) you can bail on the turn when you miss and have implied odds of 6-1 under his rules. In fact, in this case you should always bail on the turn since you'd only be getting 3.5-1 to hit on the river.


Bruce, in an earlier thread I "challenged" the textbook determination of "implied odds", which, as I read it, calls for projecting the ultimate pot (excluding your own money) and dividing by the current bet. If it may take 2 cards to hit, or if there is some risk that your hand isnt the nuts when you do hit, I believe the above calculation over-states your position in both the numerator (which should be weighted for the likelihood that you win if you do hit), AND, more importantly, in the denominator, since you have to commit additional money which is at risk to achieve the ultimate pot (and just excluding your bets from the numerator doesnt reflect that risk).

Of course you have to take all these things into account, implied odds are only part of the calculation. The extra money you have to put in to draw two cards comes under "effective odds". Getting drawn out on is "reverse implied odds". If all of these apply, then they must all be combined, or at least considered separately. None of these alone can always give you the complete picture. Note that if you bail on the turn when you miss, you don't have to take into account effective odds (except for a raise behind you). If you can't be outdrawn, then you don't have reverse implied odds.

Here are some examples that take into account the possibility of getting drawn out on (reverse implied odds). There are no effective odds in these examples since I'm not considering calling on the turn when I miss.

With a set, you can never be sure that your straight will win, because the board can pair giving your opponent a full house. In the above example, the implied odds were enough to overcome the fact that there was no money in the pot. Letís see what happens when we consider that the set will have 9 outs to pair the board on the river. He will fill up about 1 time in 5 after we make the straight on the turn. So if we are getting implied odds of 7-1 on a 5-1 shot by calling 1 SB on the flop, that means out of 6 hands we lose 5 SBs and win 7 SBs, for a net gain of 2 SBs. After 24 hands, we will be up 8 SBs. Now we will play 6 more hands, again losing 5 of them so now we are only up 3 SBs, and again we hit our straight on the 6th hand, but this time we get drawn out on by the full house on the river. Instead of winning 7 SBs, we lose 3 SBs (flop and turn, and we fold to a bet on the river). Now we are dead even. Our implied odds were wiped out by our opponent's reverse implied odds.

In the real world, there would be at least 3 SBs in the pot before we call the flop bet, so we would be getting 9-1 instead of 7-1. Then out of 30 hands we would be ahead 8 SBs. Note that these extra bets come from money already in the pot, not from implied odds.

Take the case where our opponent does not raise. This would cost us 1 big bet, so we would be getting only 7-1 odds, assuming the minimum of 3 SBs in the pot before we call the flop bet. These are the same implied odds as in the case where our opponent did raise, but we assumed no money in the pot, so the results would be the same. Again the implied odds are cancelled by our opponent's reverse implied odds.