Poll: 30 outer against an overbet?

[c9h13no3]

A thread riley makes degenerates into a discussion about how retarded riley is.... Somehow I'm not surprised.

 

[dj11]

Seems to me that GDRiley is making the better points. We do tend to play and encourage others to play 'by the book'.

While the usage of an "outs' number here differs, I will go with the 15 (14) number, and on the flop think I get 15 twice, so that over 2 cards I might hold an edge. Fact is there are only 15 (14?) unique cards that will fill our draws. Being as we have lower cards in the suit, I would have a lesser confidence level and maybe treat it as if I had 10 outs. If one chooses to use a '30' number here it has to be 30 outs out of 93 cards once. figure 30% ish.


In ring, if I thought that fast, I would probably stack either myself or the villain, in tourney, I don't wanna go broke chasing a draw, especially since villain got the vig in first. I might push or shove if I had first shot at opening post flop, but calling is different.

Perhaps how one perceives, and then acts on 'outs' is the real point here. And to that end, GD offers some nice insights.

And stop calling people names.....

 

[Cdub512]

If you had 30 outs then i would call, but you might not have 30 b/c he could have a pair that would make a 6 and 7 no good.

I would fold this hand you cant bank on a draw coming up everytime. If this was in a tournament i may call depending on the situation.

 

[Stu_Ungar]

I'm actually leaning towards a fold now.

The pot is 24.5BB. If we shove, then we are risking 40BB to win 44.5BB, meaning we need a draw of 1.12:1 or better.

15 outs (0.85:1) makes this OK, but if we discount to 12 (1.22:1) then its a very slight losing play.


From my analysis before, I was assuming that the slight mistake made by calling the flop was cancelled out by the increase in odds offered on the turn, but when both streets are considered together, it becomes a slight losing play.

We need to be accurate with a hand like this as there is no fold equity and unimproved the hand has no showdown value.

 

[jdeliverer]

While the usage of an "outs' number here differs, I will go with the 15 (14) number, and on the flop think I get 15 twice, so that over 2 cards I might hold an edge. Fact is there are only 15 (14?) unique cards that will fill our draws. Being as we have lower cards in the suit, I would have a lesser confidence level and maybe treat it as if I had 10 outs. If one chooses to use a '30' number here it has to be 30 outs out of 93 cards once. figure 30% ish.

This shows the problem with the system. The estimate you get from doing this is not even close. Even if villain has a set of kings you have a 40% chance to win, and if you make it top 2 pair you are a favorite to win.

15 outs twice =/= 30 outs once.

 

[GDRileyx]

What happens if you had 30 outs on the turn and 30 on the river? 60 outs? What if nlhe was a 7-street game so 4 more instead of 2 more cards to come? 15*4 = 60?

Going the non-poker route if it's 75% to rain on Saturday and 75% to rain on Sunday, is it 150% to rain for the weekend? Simply put even ignoring the fact that you're just using poker terminology completely wrong, you're also making an error in probability.

Can you name a hand where there would be 30 outs for one draw? I don't think there is one. Therefore your argument is flawed.

The second example is also flawed. There is a 1/4 chance it won't rain the first day, and a 1/4 chance it won't rain the second day. For it to not rain either day, both events have to occurr. It has to not rain the first day and not rain the second day, so you multiply the probabilities, and get a 1/16 chance it won't rain either day. This is an entirely different situation than in this poker example, where only one event has to happen, and the probabilities are added together.

 

[GDRileyx]

This poll was really just a chance for you to plagiarise something from that 'play poker like a pigeon' book that your really like.

I don't see how you come to this conclusion. This was a poll, not an essay. I was curious to see how people would answer.

 

[GDRileyx]

This was a reallyinteresting discussion to me and I learned some things I didnt kniow before but my question to you would have to be is how sick would you feel with that many out in a hand to push and then just flat miss it all?

It's the same as any other bad beat. In fact I think it's almost the same situation as discussed in the thread about cowboys and stray bullets. KK is enough of a favorite over Ax that you want to get all-in, but sometimes you lose. But in the long run, you win more than you lose, so you go all-in anyway.

 

[GDRileyx]

ROFL
Roflstiltskin..and citing your own book, nice touch. Also if you have 15 outs on the turn you'll hit ~1/3 and won't hit ~2/3 so you would have to multiply that times the 15 on the river, you don't get bonuses for hitting your outs twice (and 2 hearts more likely hurt you obv).

I don't know what makes you think I am citing my own book. I am not.

But you are right about the two hearts. I would have to subtract 4% chance of that happening from my equation, to be more accurate. On the other hand, poker odds are always approximations anyway, since the burned and mucked cards affect the percentages, but there is no way of knowing how to figure their impact. So that 4% is actually less than the average margain of error inherent in approximating poker odds, and becomes essentially irrelevent.

 

[GDRileyx]

Seems to me that GDRiley is making the better points...

Ahhh, the rare open-minded person I am actually writing for. They do exist

 

[Stu_Ungar]

I don't know what makes you think I am citing my own book. I am not.

But you are right about the two hearts. I would have to subtract 4% chance of that happening from my equation, to be more accurate. On the other hand, poker odds are always approximations anyway, since the burned and mucked cards affect the percentages, but there is no way of knowing how to figure their impact. So that 4% is actually less than the average margain of error inherent in approximating poker odds, and becomes essentially irrelevent.

They dont effect the calculation.

The 4% in this situation is very relevent because you are on pure draw and its very close .

 

[blankoblanco]

if you agree with me, you are open-minded! if you disagree with me, you are close-minded!

makes sense

 

[GDRileyx]

Stu, the mucked and burned cards DO effect the odds, but there is no way of including them in the equation.

If you have AA, and both the other aces were mucked, you have 0% chance of making a set. But you have no way of knowing they were mucked, so you can't account for that factor in determining your approximate odds.

Suppose one of the mucked aces was accidentally exposed. Would you still argue that you have the same 12% chance of hitting your set?

 

[Stu_Ungar]

Stu, the mucked and burned cards DO effect the odds, but there is no way of including them in the equation.

If you have AA, and both the other aces were mucked, you have 0% chance of making a set. But you have no way of knowing they were mucked, so you can't account for that factor in determining your approximate odds.

Suppose one of the mucked aces was accidentally exposed. Would you still argue that you have the same 12% chance of hitting your set?

If you dont see the card, then it does not effect the calculation. What if one of the cards you needed was on the bottom of the deck? - you would have no chance of it being dealt. It is exactly the same thing.

If you see the card, or someone tells you they have the card then yes you can include it, but otherwise it has no bearing on the odds given to you.

 

[blankoblanco]

but gdriley, you're wrong in saying the 4% is essentially irrelevant. it's quite relevant. we're looking at the % under the conditions that all other cards are unknown, which they are. if you somehow know what one of the mucked hands is, or a card was exposed, then yes, that changes the odds. but if all cards are unknown, that is what the odds are. if you played the hand out a million times, that 4% would average out to 4% impact. that's the fundamental basis of probability

suppose i shuffled a deck of cards and say "what is the chance that the top card is the J of diamonds?" there's a 1 in 52 chance. it doesn't make sense to say "well what if the J of diamonds is actually on the bottom? then it's 0%!" the chance that the J of diamonds is on the bottom is included in the odds. just like the possibility that every single person at the table mucked one of your outs is included in the odds. all cards are unknown so your outs are evenly distributed throughout the deck from a probability standpoint

 

[UFCcantstopme]

push

You have the straight draw, flush draw and if he has AQ or trying to pull a bluff, you will have even more outs. I would instantly call, and in that situation i would say you have a 60/40 advantage depending on his cards. Best bet is to just push allin.

 

[Stu_Ungar]

I'm actually leaning towards a fold now.

The pot is 24.5BB. If we shove, then we are risking 40BB to win 44.5BB, meaning we need a draw of 1.12:1 or better.

15 outs (0.85:1) makes this OK, but if we discount to 12 (1.22:1) then its a very slight losing play.


From my analysis before, I was assuming that the slight mistake made by calling the flop was cancelled out by the increase in odds offered on the turn, but when both streets are considered together, it becomes a slight losing play.

We need to be accurate with a hand like this as there is no fold equity and unimproved the hand has no showdown value.

This is why the 4% is very relevent in this particular hand.

 

[Dwilius]

You'd have to be pretty openminded to abandon the constructs of math. First it was basic algebra 2 x 15 = 30, now its close enough. Plus, its not a 4% difference, your way 60/93 = 64.5%, the correct way is 54%,

...and Rofl once again over Zach's example being flawed, not yours. Zach's example is right on, taking your method a little further to show it can result in more than 100% (proof its wrong) with enough streets or outs.

The more wrong you are, the more you post so I'm looking forward to the next round of responses. I would like to be dismissed as a sucker beneath your advanced play...maybe throw in an example of a pro who doesn't rely on math...and a little 2nd life for old times, please.

(I'm not ready for the "I'm done with this thread" post yet. Let's get a little more nonsense in before lock)

 

[Stu_Ungar]

suppose i shuffled a deck of cards and say "what is the chance that the top card is the J of diamonds?" there's a 1 in 52 chance. it doesn't make sense to say "well what if the J of diamonds is actually on the bottom? then it's 0%!" the chance that the J of diamonds is on the bottom is included in the odds. just like the possibility that every single person at the table mucked one of your outs is included in the odds. all cards are unknown so your outs are evenly distributed throughout the deck from a probability standpoint

Thats very nicely explained. I often find myself struggling to explain this to people and usually end up getting out a pen and paper and drawing pictures of decks of cards inside little clouds LOL

 

[smd173]

Shit is this why everyone always hits gutshot straights on me because they think they have 8 outs?

 

[Stu_Ungar]

Shit is this why everyone always hits gutshot straights on me because they think they have 8 outs?

and a 4% margion of error, which allows you to treat those 8 outs as though they were 9

 

[NineLions]

(9 hearts + 3 threes + 3 eights) x 2 draws = 30 outs. You're 60/93 to hit a straight or flush.

I dunno why I started reading this thread, but,

If you're going to count 15 helping cards against two chances, you get 30. If you count all the cards unknown in those two draws, you get 52 - 2 in you hand - 3 on the flop = 47, plus one less on the river for 46 = 93. Then, assuming you are going to get to see both draws, you have 30/93, not 60/93. You can't double the helping cards twice.

Edit: just saw that dj already corrected that part.

But that's not how probabilities work.

If there's a 50% chance of rain one day and a 50% chance the next and you want it to rain one day, that doesn't give you a 100% chance of rain. If it rains day one, you're good. If it doesn't, then on day 2 you still have a 50% chance of rain. That means the 50% that it doesn't rain the first day you have 50% of 50%, so another 25% overall. Before the start of the two days, you have a total chance of 50% + 25% of the dry 50% = 75%.

You only have to count your chances on the second try against the times you missed the first try.

 

[GDRileyx]

You'd have to be pretty openminded to abandon the constructs of math. First it was basic algebra 2 x 15 = 30, now its close enough. Plus, its not a 4% difference, your way 60/93 = 64.5%, the correct way is 54%,

Plug this hand into a poker calculator against a random hand, and it comes out between 63 and 65 percent to win. So it seems to me, my way comes closer to the mark.

 

[Dwilius]

Um, vs. a random hand pairing your 6 or 7 will often be an out also, we weren't counting those. Generally you don't get too much action from random hands postflop. Thanks for switching the basis of the problem and obliging with the nonsense post sir, may I have another?

 

[Snowmobiler]

I am amazed how far 1 person will go to try to convince people, that their

flawed thinking somehow equates to fact.



Snow