Reply to thread

Message

<blockquote data-quote="jerichothebard" data-source="post: 1694392" data-attributes="member: 4705">I think you are all making it too complicated.If we are assuming that:Monty will always remove one of the doors, andthat he will never eliminate the door with the car, and that he will always give you a second guess, Then there is actually only one round here.&nbsp; What happens with the first guess is entirely irrelevant.&nbsp; It's just showmanship.Here's why:It is inevitable [we assume] that Monty will eliminate a goat door.&nbsp; No matter whether you initially choose the door with the car, or one of the goat doors, Monty will ALWAYS eliminate a goat door.&nbsp; That's part of the game, and what makes it exciting.&nbsp; But, since it's inevitable, that door doesn't matter.No matter what happens, the first guess always has exactly the same result:&nbsp; Monty removes one (wrong) choice and gives you another chance.The game doesn't start until that first door is out of the picture.&nbsp; Which means that the game isn't about guessing one out of three doors, because you are always assured that one of those three possibilities will be removed and you will get to guess again.Which means you are always guessing between 1 of two doors.&nbsp; A 50/50 proposition.&nbsp; Whether you &quot;switch&quot; or &quot;stay&quot;, you really are just guessing between two doors.so it doesn't matter.&nbsp; Stay, switch.&nbsp; It's irrelevant.&nbsp; You'll win half the time. as for the one-in-a-million guesses, that's a totally different problem.&nbsp; The difference is in the base assumptions - in the first case, 1/3 of the options are payoff.&nbsp; In the second, 1/1000000 options is a payoff. In this case, your switch analysis is correct - because he is narrowing the field by so many more than just 1/3 of the options.In order for it to be the same problem, 333333 out of the original million would have to be payoff doors, and he would only get to eliminate 333333 of them - then you have a 333333/666667 chance - or just about 1/2 again.jtb</blockquote>

[QUOTE="jerichothebard, post: 1694392, member: 4705"] I think you are all making it too complicated. If we are assuming that: Monty will always remove one of the doors, and that he will never eliminate the door with the car, and that he will always give you a second guess, Then there is actually only one round here. What happens with the first guess is entirely irrelevant. It's just showmanship. Here's why: It is inevitable [we assume] that Monty will eliminate a goat door. No matter whether you initially choose the door with the car, or one of the goat doors, Monty will [I][B]ALWAYS[/B][/I] eliminate a goat door. That's part of the game, and what makes it exciting. But, since it's inevitable, [I]that door doesn't matter[/I]. No matter what happens, the first guess always has exactly the same result: Monty removes one (wrong) choice and gives you another chance. The game doesn't start until that first door is out of the picture. [I]Which means that the game isn't about guessing one out of three doors, because you are always assured that one of those three possibilities will be removed and you will get to guess again.[/I] Which means you are always guessing between 1 of two doors. A 50/50 proposition. Whether you "switch" or "stay", you really are just guessing between two doors. so it doesn't matter. Stay, switch. It's irrelevant. You'll win half the time. as for the one-in-a-million guesses, that's a totally different problem. The difference is in the base assumptions - in the first case, 1/3 of the options are payoff. In the second, 1/1000000 options is a payoff. In this case, your switch analysis is correct - because he is narrowing the field by so many more than just 1/3 of the options. In order for it to be the same problem, 333333 out of the original million would have to be payoff doors, and he would only get to eliminate 333333 of them - then you have a 333333/666667 chance - or just about 1/2 again. jtb [/QUOTE]

Verification