Stud hi: What's your plan when 3-bet on 3rd by an overpair?

[PoorLawyer]

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One is an "absolute or constant" with Monty, with a poker table the range is 0 to 3 (or 8 if you like). Its Random, not constant.

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I see. There is definitely a prize in the Monte puzzle, and we are only concerned with finding it. In the poker game there may or may not be a good hand out there.

I shall retire to the study to ponder.

Have I mentioned how much fun it is to read y'all? It's great fun. Thank you.

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Steve, you have now shifted this thread from Stud (something I know precious little about) to the Monty Hall problem (something I actually know quite a bit about). Thank you.

Re: Monty. The way the problem is presented is of utmost importance. In order for you to have an advantage by switching doors, Monty must have shown you a door with a goat INTENTIONALLY and not RANDOMLY when he revealed one. Why? Because he knows where the prize is, and by showing you a goat he is essentially giving you a "tell". If he chooses the door at random, and happens to show a goat, it's still 33% that your door had the prize and 33% that the other door had the prize. What happened to the last 33%? The goat ate it! That is, the other 33% of the time Monty will have accidentally revealed the prize and that makes your choose moot.

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Ok, so I have to ask for a clarification here. There is one prize and 2 goats behind 3 doors, correct? 33% chance by just picking a door that you'll win your washer/dryer combo and a year's supply of tide. Monte opens a door and there is a goat behind it. Whether Monte picked his door intentionally or not, there are 2 doors left and one of them has a goat and one has a prize.

What makes your choice of door not 50-50 to win? Why would switching doors change those odds? how would this change if you would rather win the goat because your mom washes your clothes anyway?

[SGspecial]

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Ok, so I have to ask for a clarification here. There is one prize and 2 goats behind 3 doors, correct? 33% chance by just picking a door that you'll win your washer/dryer combo and a year's supply of tide. Monte opens a door and there is a goat behind it. Whether Monte picked his door intentionally or not, there are 2 doors left and one of them has a goat and one has a prize.

What makes your choice of door not 50-50 to win? Why would switching doors change those odds? how would this change if you would rather win the goat because your mom washes your clothes anyway?

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Mutlipart answer:
1. This is not pocketfives, so assume you do your own laundry and would rather have the washer/dryer combo than the goat.
2. If you want more explanations than a goat could eat in an hour, google "monty hall problem". It was originally unleashed on the world by Marilyn Vos Savant in her "Ask Marilyn" column and the academic world's reaction to it was LOL.
3. I will give you my personal take on the answer which I did not find in any other explanation.

Think of the problem in terms of "equity" rather than "probability". Like if you have a small flush draw and are drawing against someone you read for Jacks over Tens (in stud). Your "probability" of winning the hand is 20% if your read is correct, but say he also has a third pair or a 4 flush to go with it? So your "equity" in the pot has to take into account the odds that your read is right, and the odds you make your hand and it holds up.

So in the Monty Hall problem, with all three doors closed you have a 33% probability that the prize is behind the door you chose, and also a 33% "equity" in the prize. If we conduct 100 trials, then when Monty opens one door, your average "equity" will go up to 67% (since if he opened two doors it would be 100%).

If he opened the goat door KNOWING it was a goat, then every trial will be the same... you see a goat. Your original "equity" was 33%, thus the "equity" in the other two doors was 67%. Only one of these other doors is still closed, so it's 67% certain the prize is behind it.

If Monty opens a door at RANDOM, you will be shown a goat 2/3 of the time, but the prize 1/3 of the time! If you're shown the prize, you have 100% "equity" if you're allowed to switch to that door. So your "equity" when you're shown a goat must be 50% to balance the equation:
(Shown prize)*100% + (Shown goat)*X = Avg "equity"

Where X is your "equity" for the "Shown goat" case.

(1/3)*100% + (2/3)* X = 67%
X = 50%

So the if the opened door has a goat behind it in this case, the "equity" of each remaining door must be 50%.

[PoorLawyer]

Yes, I do my own laundry; however goat is tasty and I live in an apartment with a free washer/dryer in the unit [img]/images/graemlins/grin.gif[/img]

Very interesting puzzle though. For those that are visual learners, this has pictures of goats and clears it up nicely:

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

[SGspecial]

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Yes, I do my own laundry; however goat is tasty and I live in an apartment with a free washer/dryer in the unit [img]/images/graemlins/grin.gif[/img]

Very interesting puzzle though. For those that are visual learners, this has pictures of goats and clears it up nicely:

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

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Yes, it's a tough choice between the goat diagrams and the 37 line Bayesian probability proof. The funny thing is that when the problem was first posed, it drew an angry response from Monty himself since that's not the way the Let's Make a Deal program worked in real life.

[Spladle]

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So you want exactly two, but not three, knowing full well that your hand suffers from reverse implied odds. In addition, calling down is correct here, but the raise just makes it more correct.

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The reason you want two bets to go in on third street is because you are a favorite over the ace's range in that situation. If three bets go in then you are a dog.

Because your hand suffers from reverse implied odds, you will not win this pot as often as equity calculations indicate, and you'll win less money when you win the pot than you'll lose when you lose it. In situations like these, if it is correct to play, then it is generally best to inflate the pot while you're guaranteed to have an edge rather than wait - unless by keeping the pot small you allow yourself to fold on a later street when you would have been forced to play on had the pot been only slightly bigger. I don't think there are many ways the later streets can develop where it will be correct to fold if one bet went in on third but call if two bets went in. If I am wrong about this then calling is a lot more attractive.

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So it comes down to "Correct" vs "More Correct", in your opinion.

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Again, I am basing this judgment on the assumption that there are not many situations where inflating the pot by two small bets will force you to call as a dog on the later streets when you could have otherwise folded. If this assumption is accurate then I am right - inflating the pot on third with a guaranteed edge is correct.

[Spladle]

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The free card? It could be available on any street, not just 4th, it depends on how we play it. The Ace knows we are not calling with crap and that we are in call down mode. By betting 4th we're just extending it to 5th.

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I think most of the time that the ace checks to us we are going to have an equity edge and should bet. The only time I can imagine wanting to check behind is if the ace's board gets so strong that the check is obviously a slow-play, and in that case we are probably drawing dead and will not benefit from the extra card at all.

[Poker CPA]

So Howard it appears we st8 on the free card business. I didn't say "on 4th Street". This dance between the Ace and 8 produces a lot of check/bet, check/check and CRs on 4th, 5th and 6th. I hope you're happy with my reponses concerning this.

"The reason you want two bets to go in on third street is because you are a favorite over the ace's range in that situation. If three bets go in then you are a dog."

You are really cutting it close here, especially when the 8 is in "call-down" mode AND has reverse implied odds; and the ACE knows it. The Ace can put the 8 on a much tighter range of hands and can fold anytime. So when he wins he gets the extra bet and when he loses he saves the extra bet. This gives him control because the 8 just never knows, as the hand progesses, whether he is ahead or behind. I believe you have said as much with this comment

"you'll win less money when you win the pot than you'll lose when you lose it."

You're basing your argument on this

"Again, I am basing this judgment on the assumption that there are not many situations where inflating the pot by two small bets will force you to call as a dog on the later streets when you could have otherwise folded. If this assumption is accurate then I am right - inflating the pot on third with a guaranteed edge is correct."

while at the same time hoping for no 3 bet AND a 5.

I base my argument on the fact we HAVE to play this type of marginal hand because it makes a difference in winning and losing over the long run. You have to play these hands well, at the BEST PRICE. If you don't play, then you get no action when you do play.

The "guts" of the hand is how we play cards 9 thru 14 and how we handle the 3 1/2 BBs still to come. Your guaranteed edge on 3rd means nothing on the river.

I call and then play stud poker.

[electrical]

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Re: Monty. The way the problem is presented is of utmost importance. In order for you to have an advantage by switching doors, Monty must have shown you a door with a goat INTENTIONALLY and not RANDOMLY when he revealed one. Why? Because he knows where the prize is, and by showing you a goat he is essentially giving you a "tell". If he chooses the door at random, and happens to show a goat, it's still 33% that your door had the prize and 33% that the other door had the prize. What happened to the last 33%? The goat ate it! That is, the other 33% of the time Monty will have accidentally revealed the prize and that makes your choose moot.

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We know for a fact that Monte showed us a goat. There is no possibility that he shows us a prize instead. He shows us a goat. Whether he got there by accident or by choice, he has shown us a goat and we should switch.

Imagine the puzzle expanded to 1000 doors. You pick a door at random, and it has 0.1 percent chance of having the prize. Monte gets the other 999 doors. One by one, he exposes goats until there is only one door left. Do you think it is equally likely that you picked the prize initially, or that the prize is behind the final door under Monte's control?

I should point out that Monte's intention has nothing to do with it, only the 99.9 percent likelyhood that he had the prize under his control in the beginning of the game, and he has eliminated all but one door from consideration.

In the original puzzle it isn't the case that Monte could have shown you the prize one-third of the time. We know for a fact that he showed you a goat. He showed you (a) goat(s). You should switch.

[Spladle]

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So Howard it appears we st8 on the free card business. I didn't say "on 4th Street". This dance between the Ace and 8 produces a lot of check/bet, check/check and CRs on 4th, 5th and 6th. I hope you're happy with my reponses concerning this.

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I'm not. If we are checked to on a later street then it will be because our equity against the ace's distribution is either > 50% or < 1% and it should be pretty easy to decide which based on his board. And in case this needs to be said again, when we are ahead of our opponent's range, we benefit from putting money in the pot.

Please provide an example of a situation where you think we are likely to both want and receive a free card.

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You are really cutting it close here, especially when the 8 is in "call-down" mode AND has reverse implied odds; and the ACE knows it.

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What am I cutting close here? You never clarified.

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The Ace can put the 8 on a much tighter range of hands and can fold anytime. So when he wins he gets the extra bet and when he loses he saves the extra bet.

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I know. That's why I said this:

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you'll win less money when you win the pot than you'll lose when you lose it.

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What I should have said is that you'll win less money on the later streets when you win the pot than you'll lose on the later streets when you lose it.

By the way, do you know why this is? It's because the ace has a wider distribution than we do. Any time your range is narrower than your opponent's, you are at a disadvantage for all future betting where that disparity continues to exist.

Your problem may stem from this misunderstanding - do you think we're going to lose this pot more often than we're going to win it? That is not the case. We're going to win what's in the middle more than half the time, so it's in our best interests to increase the amount in the middle. This is really basic stuff.

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This gives him control because the 8 just never knows, as the hand progesses, whether he is ahead or behind.

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This is obviously true but irrelevant. The ace just never knows, as the hand progresses, whether he is ahead or behind, either. All you can ever know at any point is how your hand compares to your opponent's distribution.

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You're basing your argument on this

"Again, I am basing this judgment on the assumption that there are not many situations where inflating the pot by two small bets will force you to call as a dog on the later streets when you could have otherwise folded. If this assumption is accurate then I am right - inflating the pot on third with a guaranteed edge is correct."

while at the same time hoping for no 3 bet AND a 5.

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Okay. What's your point? Unless you can show that there are a significant number of situations where it will be correct to fold a later street when only one bet went in on third but you will be forced to call down when two bets went in, then raising is preferable to calling. You have not done this.

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I base my argument on the fact we HAVE to play this type of marginal hand because it makes a difference in winning and losing over the long run. You have to play these hands well, at the BEST PRICE. If you don't play, then you get no action when you do play.

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In this paragraph you have not shown that calling is preferable to raising.

I would like to take this opportunity to again point out how important it is to knock out the bring-in if he has a hand that he is correct to play for one bet but not two.

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Your guaranteed edge on 3rd means nothing on the river.

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This could not be more wrong. The more equity you have on third, the more likely you are to have the winning hand at the river. At all times, on all streets, our hand's equity against our opponents' distribution should be the primary factor in our decision of whether to fold, call, or raise.

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I call and then play stud poker.

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Awesome. HU 4 rollz?

[Poker CPA]

"Please provide an example of a situation where you think we are likely to both want and receive a free card"

No improvement on 5th after betting 4th.

Cutting close - That you want EXACTLY two bets and if you go to the river you have to improve via a 5 or 8 (5 cards)

You agree the Ace has control but not important.

"Your problem may stem from this misunderstanding - do you think we're going to lose this pot more often than we're going to win it? That is not the case. We're going to win what's in the middle more than half the time, so it's in our best interests to increase the amount in the middle. This is really basic stuff."

My point is the Ace is not in call down mode, the 8 is, and the Ace knows it. The Ace wins more of the 3 1/2 BBs.

"Okay. What's your point?"

The 8 needs to improve AND hope for no 3 bet. And this is the bottom line Howard. IMPROVEMENT. Without it he's losing the contest for the 3 1/2 BBs. And based on your "tone" I assume you're calling 7th with unimproved 8s. Good Luck