Three Curtains

No, there really are only two.

No -- and I do fully understand what you've been saying. You're just not grasping the reality of the game, but allowing a door that is never in the equation to become a distraction for you.

First, let me deal once again with your altered version. In your scenario, Area "A" is one choice with one door in it. Area "B" is one choice, even though it has two doors in it. That is two choices -- a 50/50 chance of winning the car.

Now, to the real game scenario. Just as with your altered scenario, there are never more than two doors in the running. You need to understand how the game works. Standing on the floor in front of you when you're next to Monte Hall, there appear to be three choices -- but only apparently.

Now, let me explain why your scenario is the same choice Monty gives me, and why both represent a 50/50 chance of winning the car. From the very start, Monty is going to eliminate one of the non-prize doors. Before you ever make a choice, one of those doors is going to be tossed. He always does that. You know he always does that. No matter which door you pick, before you see where the car is, one of the non-prize doors will be removed from the equation.

Therefore, it is never in the equation to begin with. There is not a one-in-three chance of anything. There is a one-in-two chance. The third door is an illusion, a "false flag," a distraction -- whatever you want to call it. Let me illustrate.

Here are the three doors:

X X X

You pick Door #3:

X X X

Monty removes Door #1:

X X

Now, the same scenario, only Monty removes Door #2 this time:

X X X

X X X

X X

I can do the same thing with you picking another door to show you that Monty is going to remove one of the other doors and while you might get this ...

X X

... that's still the same thing as either of the other two scenarios.

In all the scenarios, how many doors are left? Answer: 2

Was Monty ever going to remove the door that has the prize behind it? Answer: No

Which door, therefore, has the prize been behind all along? Answer: Either your door, or the other one.

Was the door Monty removed ever a factor? Answer: No.

Why was the door Monty removed never a factor? Answer: Because it did not have the car behind it, and you did not pick it.

No matter where the car is, Monty is always going to take one door out of the equation before revealing the prize. Also, before revealing the prize, he is going to barrage you with enticements to trade your door for cash, another prize, or the other door. He ignores the third door, never offers it to you, never speaks of it except to eliminate it.

It doesn't matter if you already have the door with the car, or the remaining door has the car, he's going to do the same thing: Take away one door, entice you to change, and regardless of your decision, reveal the door hiding the car. The fact he removes one door before doing anything else means that door was never in the equation. Mathematically, you had a 50/50 chance from the very beginning. However, because intuition trumps probability, it is nonetheless prudent to trade your door for the remaining door.

 

It doesn't make any difference!! The outcome is the same no matter what scenario you weave around the facts.

How many times do I have to answer that question? This will be number 4 (I think -- I didn't really count):

1/2​

... for the same reasons as I gave in the last post. How many "areas" do the "A" and "B" represent? Two, or three?

You seem completely unwilling to accept that in your scenario or the "Let's Make A Deal" scenario, the contestant always has a 50/50 chance, a 1/2 chance, however you want to express it. The odds are never lower than one-in-two. There is no "1/3" and "2/3" or one- vs. two-in-three. That never exists in any of the scenarios. Get over it, please.

And by the way, the scenario, as I pointed out, is called the "Monty Hall Paradox" so it's kind of hard to leave Monty Hall out of it. :laugh:

 

No, I'm not. By making #1 one choice and combining #2 and #3 into one choice, you've eliminated three options and reduced it to just two -- just like Monty Hall does in the Paradox. You can't understand how I think that? I can't understand how you can't!!

 

Very good. You are correct. Now tell me why you've been trying to turn two choices into three?

 

I know. So why do you want to make it a 1/3 vs. 2/3 chance for winning?

 

Sorry, but that's ridiculous. You admitted in the previous post that ...

You have previously admitted that the "A" circle and the "B" circle are only two choices.

Please note the last two questions you asked in that post. You have established the criteria, your own words, for nothing other than a one-in-two chance.

Yet you insist there is a 2/3 chance the prize is in area "B." That's really really bad math. If you include two curtains in one choice, the number of curtains is irrelevant. The choice is what is relevant, and that choice becomes a 50/50 chance. We're done here. You've stated it unequivocally that there are only two choices, yet insist on posting a wrong conclusion. We're done here. This has become a massive waste of time.

 

OK, I "looked in" and I can't believe you haven't gleaned the answer to that question from the multiple times I've explained it. You really are stubborn, hm?

Yet again, one final time, for perpetuity: Two choices cannot be turned into a 1/3 vs. 2/3 chance of success. Please wrap your head around that. Thanks for playing.

 

Interesting solution from Khan Academy https://www.khanacademy.org/math/pr...ty/dependent_probability/v/monty-hall-problem

And HERE is a simulator. I would have been wrong it appears.

 

Now that I have seen it illustrated it makes perfect sense

I ran the simulation 10,000 times switching and 10,000 times staying. Switching resulted in a win 2/3 of the time and staying won 1/3 of the time.

Your 1/3 vs 2/3 explanation makes total sense now.

I actually saw this on Khan's FB feed and STILL thought I was right until I watched their video and ran the simulation.

Thanks for the lesson

 

One, I would say the simulator is weighted to favor the alleged 1/3-2/3 choice, because continually it breaks the wins and losses down along those lines, whether you keep the door chosen or change it.

Analysis of the show "Let's Make A Deal" over the many, many years it was on with host Monty Hall revealed that switching your original choice won the prize 48% of the time, standing pat won the prize 37% of the time. The simulator and the various mathematical equations that have been devised to prove the odds remain the same don't seem to be born out by the statistics. In fact, the 48% of wins in changing doors after the dud door is revealed would tend to prove they do change to 1/2.

The element that is nearly impossible to factor in, and I'm certain can't be quantified in an equation or a simulation program, is the free will of the human mind to choose the door that is initially chosen. There has to be a significant "don't-know-because-we-don't-understand" dynamic that influences the outcome of any given Monty Hall scenario.

In short, you haven't convinced me.

 

Like any things for which there are very plausible theories, it doesn't prove out in real life. So "understanding" has nothing to do with it, although I do assure you I understand what you're trying to say. It just doesn't work in real life.

 

I am curious as to why it played out the way it did on the show. Do have a link to the stats on the actual outcomes on the show?

I have a hard time believing that the online simulators mentioned were all designed to cheat.