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

[Poker CPA]

This free card business Howard is pretty silly. All I said was "Can produce a free card", whether we take it or not depends on 9 thru 14.

[Poker CPA]

On the matter of raising to knock-out the 2; I believe this works against the 8. His involvement actually helps the 8 in dealing with the Ace. Our profit potential increases 100% while our EV, if ahead of the Ace, goes down only 25% (60 to about 45). So I don’t mind a call from the 2. What generally happens on 4th is the Ace will check and the 8 will bet, hoping the 2 calls. Now its decision time for the Ace; fold, call or raise. If its split Aces, then the Ace will raise to thin the field and bet out on 5th. A call will be a sign of weakness and probably a 5th St check, allowing us a potential free card or a bet. So the 2 allows us to get a better handle on the Ace and will make his 4th st decision to fold easier. My EV goes way up on this fold.

I understand your point on the knock-out, but in this case, he helps more than he hurts IMO. More money too.

[Michael Davis]

Not playing this hand strikes me as incredibly weak. Unfortunately, once you commit to playing it, you're in it to win it unless your opponent pairs on board or catches a three flush on fifth. There's already too much money in the pot once it goes three bets on third.

I frequently fold when I complete and am reraised by a higher upcard showing, but that's only the second bet, your kicker makes your hand slightly better, your opponent is going to be less scared by you pairing your doorcard 8 than if it were a Q, and you're still in a steal situation so there's a greater than normal chance that he's on BS (although this is unlikely for most).

-Michael

[mosta]

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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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I'll give my understanding of the game show situation just for comparison.

Imagine repeating 2 different games indefinitely. Game 1, you pick a door from three that have 1 prize and 2 losers. Your expectation is 1/3. You do this forever.

Game 2, you pick a door, a loser is removed (every time, always a loser), and you are allowed to switch. By always switching, as you play forever, you get the inverse of the above game. Your losers (on the initial choice) are always switched to winners, and your winners are always switched to losers (since a loser is always removed and a switch therefore always inverts your initial choice). Therefore your expectation is now 2/3.

You see it in the repetition. You intitially expect to pick a winner only 1/3 times, but you get to completely invert that process when you _know_ you will get to switch _all_ initial losers (the majority) to winners and _all_ initial winners to losers.

Now as far as requiring the loser to be removed intentionally, I go back and forth and need to review. If a meteor reveals a loser after you have picked and you can switch now--Is that a new game that is 50-50, or are you in the same situation where you know you probably picked a loser but now get to invert your original selection process. I'm not sure. And not sure how or whether repeatability fits in here.

[Micturition Man]

Wikipedia has a great article on the Monte Hall problem. There is a Venn diagram that explains it all in a glance.

The difference between the scenario where Monte randomly opens a door and happens to reveal a goat and the scenario where Monte deliberately reveals a goat is subtle.

In the first case the probability that one of the 2 doors you did not pick holds the prize does not change. It is still 2/3rds, only now only one of those doors remains to be opened.

However if Monte opened one of those two doors randomly and happened to reveal a goat, we can use Bayesian probability to infer that there is an increased likelihood that neither of those two doors held the prize in the first place.

Before he opens a door at random the possible combinations for the two doors you did not choose are goat-prize, prize-goat, and goat-goat.

But once he randomly opens a door and reveals a goat, that eliminates one of the prize-goat possibilities. So it's now 50/50 between goat-goat and goat-prize. Since a goat has already been shown that means the remaining door is 50/50 between goat and prize.

It's similar to if you know someone in poker has either AK or AA (and would play both 100% of the time let's say).

If he decides for no reason to show you an A, that has no bearing on the probability of whether he has AA or AK, although it does make it more likely his other card is a K.

However if some drunk guy came up and snatched one of his cards and turned it over and it happened to be an A, that now makes it less likely that he holds AK.