uhhmm, i dunno if i got the additional reply above in off-red correct or not. it makes sense to me now, but my head is spinning, just thought i'd run it past yah, lol.As for your second paragraph, that is my point. People are making subjective (not objective) value based decisions on how to reallocate the value of the original probability of the exposed door (having a 1/3 probability before being exposed by the game show host).

well maybe, here is the thing though, people are at least doing something, at least really exerting some mental prowess in some regard over and on real phenomenon that exist, beyond just applying math with no justifiable reason.

Okkkkkay?

As for 1 above, how? How do we know we "screwed up" when we made our initial selection? We don't and can't unless we can read the future (See, Nicholas Cage as Chris Johnson in Next) or the answer to the problem was disclosed to us (or we stole the information).

just playing the game is a screw up. gambling is a screw up. mindlessly following monty’s orders is a screw up. ok, it’s not a screw up because it’s a free roll, right? well, let’s imagine the monty devil thing again, where behind two of the doors is DEATH. is it a screw up to play the game then? hypothesis proved: it’s a screw up.

So what, continuing to play the game might be a screw up. How do you know the devil (if such a thing/entity really exists) isn't just Monty Hall playing more mind games with you, and that there is really no deadly option behind 1 of the 2 remaining doors, or perhaps death lurks behind both doors. Booyah!

As for 2 above, why? I can see where this would be true if our initial selection was made on a "known" bad choice. there yah go!!! our initial selection was made on a “known” bad choice! But since our initial selection was made in a truly random fashion, and not based on bad facts, it was done merely through gambling (cannot do more, but could choose not to play the game) on the odds 1:2 no matter which door we selected. So, how would our initial selection of Door 1, Door 2, or Door 3 merit "not deserv[ing] to have any more value added to it."?

merely gambling!? SoFlaLaw you didn’t just say that did you and you a card counter?

Okay, I was sloppy. You got me there. Yes I am a CC. I can appreciate your characterization of the decision to play the game in the first place as being a bad choice, thereby rendering the selection of Door 1 as a bad decision, assuming that you are placing your life, health, loved ones or significant financial interests at risk. But that doesn't change the bad status had we selected either of the other 2 doors. ALL ARE BAD CHOICES, SINCE THEY ARE SELECTED WITHOUT AN ADVANTAGE AGAINST THE HOUSE.

As for 3 above, what "facts" do we "know" (or more appropriately, what reasonable assumptions can be made)?facts? reasonable assumptions? there yah go, now we are getting somewhere! and somewhere is where we want to get, not just spinning our wheels. hey it’s not a crystal clear perfect world out there in advantage play land but we do the best we can with what we’ve got, when we can. may seem like grasping at straws, but no out and out gambling, just gotta be thinking on your feet, savvy. i mean hey, you can go with your original losing gamble if you want, or you can use what information you have and refine your action, it’s up to you. science, my friend, but not rocket science.

Science and math is what I have been trying to get to the underlying fundamentals of, and what I have been utilizing (proving or disproving a hypothesis). I have no way of prejudging whether my initial selection was "your original losing gamble" or if Door 3 would be a "losing gamble".

Fact 6 - We do not know if the game show host "knows" behind which door the prize has been placed (if relevant at all); incorrect, we know & it’s relevant but I contend it is not necessary info as long as the prize is not exposed by monty. but we know, if nothing other than this problem is on the internet, lol. and if it’s on the internet the switch solution must be true. Bonjur?

But he said he was a French model! Even if Monty Hall knows which door the prize is behind, provided that he exposes 1 of the 2 doors behind which the prize is not located, at best his knowledge is irrelevant.It could be behind the door I selected (Door 1) or the remaining door (Door 3 after he exposes Door 2).

exactly !!!! the salient question becomes, how many times out of how many times is the prize gonna be the door you selected door 1?, for repeated renditions of this game? and how many times is the prize gonna be the door you refused to switch to from door 1 over those repeated renditions of this game ?

for repeated renditions you expect your door #1 to have the prize one out three of those repetitions. same for each of the other two doors.

now we introduce the show and switch offer

you have door 1 and keep it, so door 2 is opened and no prize. so it’s just your door 1 and door 3 rendition 1

you have door 1 and keep it, so door 3 is opened and no prize. so it’s just your door 1 and door 2 rendition 2

for whatever door you keep, it’s always your door against two other possible doors as the renditions progress, no?

we can do this over and over and over, but guess what the door you choose is never switched so it’s always the door you chose standing against one of either two different doors.

as the renditions continue it’s a particular door, that door you chose against two other doors.

over and over and over again…..

your one against two

your one against two

your one against two

…….

does that translate to one chance against two chances?

one chance of containing the prize against two chances of containing the prize.

is that a total of three chances and is that 1/3 chance against 2/3 of being the prize?

and what is the chance we theoretically say the door switched to has at the end stage of the game? 2/3 no?

but you didn’t switch, so you just have a 1/3 chance

this refusal to switch, the keeping of the original choice forces the non switcher to always be one against two, as renditions add up, over and over and over ad infinitium.

but what if you switch?

you have door 1 and door 2 is opened and no prize. so you switch to door 3, same as saying door 1 & door 3 versus door 2 rendition 1

you have door 1 and door 3 is opened and no prize. so you switch to door 2, same as saying door 1 & door 2 versus door 3 rendition2

so you could originally choose other doors but these renditions essentially can be repeated over and over and over

and as renditions progress its your two doors against one door

over and over and over again

your two against one

your two against one

your two against one

…..

does that translate into two chances against one?

two chances for the prize against one chance for the prize

two out of three chances for the prize against one out three chances for the prize

your 2/3 chances for the prize against 1/3 chance for the prize

and what is the chance we say the door switched to has at the end stage of the game? 2/3 no?

Fact 13 - The exposed door (Door 2) does not reveal the prize; no it doesn’t, but it reveals the fact that door 3 now has a 2/3 value since door 2 has a zero value far as the prize goes

That is circular reasoning. Just because the prize is not behind Door 2 (now exposed), does not mean that its value automatically should be transferred to Door 3. That has been my point from the gitgo, and has not yet been disproven through objective method. Why can't its value be transferred to my initial selection of Door 1?

Fact 15 - The odds/probability of Door 2 (1:2 or 33 1/3%) for success no longer applies to Door 2 (proven not to contain the prize); no, we now know how much door 2 contributes to the 2/3 sum of door 2 & door 3, it contributes 0

Yes, it now contributes 0 to door 2, but likewise, before being exposed, it equally contributed to the 2/3 sum of Door 1 & Door 2.

Fact 18 - If we nullify Door 2's chance/probability for success, Door 1 and Door 3 each have 1:1 odds or a 50% chance/probability for successfully selecting the prize; NO, NO, NO! door 1 has a 1/3 chance when we chose it, door 2 and door 3 have a 2/3 chance together. just because door 2 ends up having a goat doesn’t mean all of a sudden that door2 and door 3 chances don’t equal 2/3.

You continue to refuse to acknowledge that although Door 2 & Door 3 had a 2/3 chance together, so did Door 1 & Door 2 OR Door 1 & Door 3. So, it means that there may be a goat (cow, sheep, lame horse, etc. or Monty Devil might kill you) behind either Door 1 or Door 3.

Fact 19 - If we split Door 2's chance/probability for success, Door 1 and Door 3 each receive 50% of Door 2's 1:2 odds or 33 1/3% chance/probability for successfully selecting the prize, resulting in each now having 1:1 odds or a 50% chance/probability for successfully selecting the prize; and there is no justifiable reason for us to do that math in fact we have information that makes our brains unable to justify adding any value to door 1

And you questioned my being a CC??? We always do the math! As for your conclusion, it is just that, a conclusion probably without any underlying solid foundation. The value from Door 2 can be added to either Door 1 or Door 3. You continue to refuse to recognize that possibility.

As for the fourth paragraph, how do we know our initial selection was "unwise"? How was that selection being taken from a "mathematically disadvantageously large set" different from whether we initially selected either of the other two doors? Or all you saying selecting any of the 3 doors initially was unwise? YES! or no, or maybe lol, but yes If so, then why would you reward any selection with Door 2's odds/probability for success? because one is either an advantage player to begin with and knows the score about monty, or one is trapped like a rat in a stupid move, but still is a smart enough rat to figure out how to maybe escape the trap. the trapped rat realizes his mistake and fixes it as best he can.

But the rat might have unknowingly fixed "it as best he can" when he made his initial selection of Door 1, and not switching to the possibly deadly decision awaiting behind Door 3. One thing you repeatedly disregarded in replying to my questions is that I assumed from the start that the game show host did not know the identity of the door behind which the prize (or life and $1 million) rested. Alternatively, I posited that it is irrelevant whether Monty knew where the prize rested, provided that he exposed 1 of the 2 doors behind which the prize was not located, and left my initial selection (Door 1) and the other door. Based upon that hypothetical that I posited, the rat cannot know behind which door the prize is located, and cannot know that switching would improve his odds (as his odds would grow to 1:1 or 50%) when the value of Door 2 is split between Door 1 and Door 3.

Even if you expand the set of options from 3 to 1000 doors, and expose 998 doors without showing the prize, each of those options initially were from (your words) a "mathematically disadvantageously large set" of options. But once the door(s) is(are) exposed, we have a relatively small set of options (Door 1 and Door 3, or if a 1000 Doors, Door 1 and Door 537), just 2. true, the example just dramaticizes the error inherent in picking from a mathematically disadvantageously large set.

I will leave this one alone, other than to state my objection to your characterization of the "mathematically disadvantageously large set" on the record based upon relevancy grounds.

I understand your points, but disagree that you are arguing from a point of relevant facts. So, why does our "unwise" initial choice from a ""mathematically disadvantageously large set" deserve to be recognized and be taken into account (punished, definitely not an objective process)? this is why i suggested the devil and his goat angels scenario. such a scenario proves the error of initially playing the game. that error may not be 100% fixable but one can improve one’s odds by using the information monty provided, and switching doors. but what? come on, have you never made a mistake, realized it was a mistake and went back and corrected it? that’s the idea of not putting any more value onto door 1, we realize it was a mistake in the first place. and on top of that we realize that there is no compelling reason to remove any value from the combined values of door 2 and door 3, just because door 2 ends up having zero value doesn’t change the combined value of door 2 and door 3.

Once again, you keep assuming that switching will fix the "mistake", but have failed to prove that the initial selection was indeed a mistake, or that it needed to be corrected. The assignment of Door 2's value solely to Door 3 is an assumption, without proof that it cannot be assigned to Door 1. Also, "just because door 2 ends up having zero value doesn't change the combined value of door 2 and door [1]". Come on, we're playing for my life!

I truly would like to understand your reasoning to attempt to internalize your position and see if I change my approach (an analytical process akin to selecting Door 1 versus Door 3, albeit there does not appear to have been a Door 2 in our discussion and analysis).

ok, tell yah what, i’ll switch, you are correct.

Au contraire. Maybe I'll switch, because your arguments appear on the internet! Any chance you are a french model?

Thanks for trying, but I am still looking for proof or disproof (is that a legitimate word?) of my understanding of the Let's Make a Deal hypothetical.

PS - Obviously, yes I am a lawyer, and I represent working people, their unions, pension plans, health and welfare plans and apprenticeship training programs. It's just the David in me that wants to defeat Goliath, just like when I take money away from the thieves, oops, meant casinos.

edit: don't like my attempt above? try reading this:

Solution:

Assume that you always start by picking Door #1, and the host then always shows you some other door which does not contain the car, and you then always switch to the remaining door.

If the car is behind Door #1, then after you pick Door #1, the host will open another door (either #2 or #3), and you will then switch to the remaining door (either #3 or #2), thus LOSING.

If the car is behind Door #2, then after you pick Door #1, the host will be forced to open Door #3, and you will then switch to Door #2, thus WINNING.

If the car is behind Door #3, then after you pick Door #1, the host will be forced to open Door #2, and you will then switch to Door #3, thus WINNING.

Hence, in 2 of the 3 (equally-likely) possibilities, you will win. Ergo, the probability of winning by switching is 2/3

source: http://www.probability.ca/jeff/writing/montysimple.html

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