Poker Video: SNG by Luceboy (Mid Stakes)

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In this opening episode Luceboy and md261 discuss conditional probability and ICM adjustments based on Future EV theory.

Luceboy takes theory of SnG's into a whole new level with this series of theoretical discussion as well as practical applications.

- Game: sng
- Stakes: Mid Stakes
- 42 minutes long
- Posted about 2 years ago

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Time Link to 00:07:22

Could you bet here and if he folds, show the Jack? this way we get an image of betting when we hit?

Could you bet here and if he folds, show the Jack? this way we get an image of betting when we hit?

You could bet here, in fact I could go as far as to say you should bet here. It looks fairly apparent that you have your opponent beaten so betting for value is obviously a good option. I will admit that I wish I had bet in this particular situation but sometimes when you're playing a lot of tables you don't play as perfectly as you would like!

A few thoughts of my own, the best thing you can do to look at this, is to use the holdemresources beta. Take that chipstacks and ranges for all players in the possible end stacks. It will tell you expectation for the chip setup and ranges.

My take on the utg shove problem is that if we live, we are taking a similar -EV setup with 5bb in the big next hand too. So in effect, we are still losing about the same equity next hand with 5bb vs 3.5bb, it hurts a little less psychologically because we are shoving 3.5bb instead of calling 3.5bb. Im not too sure about 6m, but I have seen lots of spots for 9m and its usually a small difference in edge, like .1%.

Hope this helps.

A few thoughts of my own, the best thing you can do to look at this, is to use the holdemresources beta. Take that chipstacks and ranges for all players in the possible end stacks. It will tell you expectation for the chip setup and ranges.

My take on the utg shove problem is that if we live, we are taking a similar -EV setup with 5bb in the big next hand too. So in effect, we are still losing about the same equity next hand with 5bb vs 3.5bb, it hurts a little less psychologically because we are shoving 3.5bb instead of calling 3.5bb. Im not too sure about 6m, but I have seen lots of spots for 9m and its usually a small difference in edge, like .1%.

Hope this helps.

Thanks for the heads-up regarding this resource - I didn't know there was a beta!

Deciding whether to shove UTG or fold and consider calling all-in when in the BB is a real tough one. One side of me wants to always side with Collin Moshmann. He has written an excellent 6max SnG book and is of the opinion that once you lose fold equity you might as well consider yourself out of the tournament. However there are obviously a lot of factors that you cannot overlook (such as good regs behind who may realise you are shoving wide and call wide themselves).

If by what you said you mean that the difference in 9mans for this situation (between shoving UTG or calling all-in when in the BB) is only very small, then I would almost always advocate shoving if you think its marginal. It all goes back to Al and my SnG philosophy: if you're debating a call, fold, and if you're debating a raise, raise.

I think this problem will be solved with a pretty high degree of accuracy within the next few years. So far, I just have never seen any better explanation for it than 'a big blind is worth xyz equity, so next hand is worth abc less.' This is of course true whether we have 3.5 or 5 bb. At anyrate, I think its a tiny mistake.

I understand I am in the minority on this one.

I think this problem will be solved with a pretty high degree of accuracy within the next few years. So far, I just have never seen any better explanation for it than 'a big blind is worth xyz equity, so next hand is worth abc less.' This is of course true whether we have 3.5 or 5 bb. At anyrate, I think its a tiny mistake.

I understand I am in the minority on this one.

It seems like you have thought about this a lot and just because you are in the minority it does not mean that you are wrong. Thinking for yourself and avoiding conventional wisdom is an important habit if you want to maximise the amount of money you make playing poker.

I believe that you are right that there is a good chance that this will be solved to a much greater degree of accuracy in the next few years. Until then I'm going to shove my trash UTG when crippled and pray to be a 45% underdog when called!

You actually put the Monty Hall problem in the video, and have it wrong.

THE KEY TO THE PROBLEM IS THE HOST _KNOWS_ WHICH DOOR HAS A GOAT.

Therefore, due to the host ALWAYS being able to open a goat, it's no longer a "2/3" shot.

If you really believe you have a case for changing doors, i think you should try it with your friend..

You pick a door, he shows you a goat (he knows where it is), and you change.. Run that scenario 100+ times and you will see changing door will give you exactly as many cars as your first pick.

Jesus christ, it's embarassing to look at you making this an example to imply to poker.

Welcome to DC, Nice first post XioXKraD

You pick a door, he shows you a goat (he knows where it is), and you change.. Run that scenario 100+ times and you will see changing door will give you exactly as many cars as your first pick.

If you do this you should get twice as many cars if you change doors, and this is the point.

I do hope you are not just trolling.

My first post too (having watched all of 4 minutes into my first video!)....

Welcome to DC, Nice first post XioXKraD

If you do this you should get twice as many cars if you change doors, and this is the point.

But you don't. I know this because basic logic told me it was not true so I wrote a computer program to simulate it and find out for absolutely sure.

I ran the simulation 10 times over a sample of 1000 choices each time and the range varied by no more than about 10% (i.e. 600 / 400) either way.

You have a choice between door A or (door B or C). Since one of (door B or C) is automatically eliminated as the wrong choice you are effectively choosing between door A (50%) or whichever door isn't opened by the host, which is the other 50%.

I'll explain the logic of this by extending the game, it's exactly the same except that instead of 3 start doors there are 100 doors, still one car prize but 99 goats, you pick one door initially and the the host has to open 98 doors showing a goat and give you the option to change your initial choice to the last unopened door.

When you pick the first one there is a 1 in a 100 chance of picking the right door.

So there is a 99% chance of the host having the prize in this other 99 door selection.

So at this point the host with 99 doors is 99 times as likely to have the prize as you.

According to the game the host *has to* show you 98 'goat' doors, most of the time the host has no choice he has to show you all of the 98 'goat' doors left in his selection, leaving closed the only one with the prize. It is only once in 100 times when you have picked the prize initially that he can choose any of all the 99 doors to show you any 98.

At the end you are still left seeing two unopened doors but it should seem that the initial choice is a 1 in 100, and the offered option is now a 99 in a 100 chance. It is not a 50 50.

In this altered game it is 99 times more likely that you will win if you swap your 1st pick for the hosts only unopened door.

The logic is the same with just 3 doors, you pick and the host has the prize 2 out of 3 times, if you don't change you simply keep a 1 in 3 chance but when the host opens the 'goat' door still 2 out of three times the prize is behind this host unopened door so you should change.

If you simulation works the initial pick is a 1/3rd chance of a win but by swapping to the unopened there is a 2/3rds chance of a win.

http://en.wikipedia.org/wiki/Monty_Hall_problem

The explanations of the solution will be much better on this link.

Welcome to DC, I hope you liked the video and I do hope you are not just trolling

I would have put a decent amount of cash on the probability being 50:50 given the choice between my door and one other door, regardless of how many were possible in the first place. I'm happy to admit I was wrong though I am still thoroughly confused - it doesn't take much:-)

We are in good company:

quote from the wiki page:*Accounts of the Hungarian mathematician Paul Erdős's first encounter of the problem can be found in The Man Who Loved Only Numbers and Vazsonyi 1999; like so many others, Erdős initially got it wrong.*

Paul Erdős has written more math papers than just about anybody but was still initially convinced that changing your choice didn't matter.

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