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?