Three Curtains

[rstrats]

thisnumbersdisconnected,

re: "However, in fairness I have to point out that an analysis of 'Let's Make a Deal' over the years showed that changing your door was a good move 48% of the time, while keeping the original pick was a good move only 37% of the time."


If changing were a good move 48% of the time, then not changing would be a good move 52% of the time - not 37% of the time.

 

[InTheLight]

So you would switch to both #2 and #3 even though you know that at least one of those curtains doesn't have the car behind it.

According to your new scenario I don't know there is no car behind either curtain. You said:

"if after your initial pick of curtain #1 and before any curtain was opened, the host told you that you could switch to both curtains #2 and #3, what would you do?"

Keeping in mind that the host will only open a curtain that doesn't have the car behind it, what is the difference if you personally open curtain #3 or the host opens it for you? Either way you get to look behind both curtains.

You're not making any sense.

 

[rstrats]

InTheLIght,

re: "According to your new scenario I don't know there is no car behind either curtain."

But you know that there is no car behind at least one of the curtains.
　
　

re: "You're not making any sense"

What don't you understand about my question: "...what is the difference if you personally open curtain #3 or the host opens it for you? Either way you get to look behind both curtains."?

 

[rstrats]

thisnumbersdisconnected,

re: "You have a question directed to you in post #13.

 

[thisnumbersdisconnected]

thisnumbersdisconnected,

re: "You have a question directed to you in post #13.

I saw it months ago. It is irrelevant, a no-brainer. He is essentially giving me the win. Who wouldn't do that?

 

[rstrats]

thisnumbersdisconnected,

re: " He is essentially giving me the win. Who wouldn't do that? "


You implied in your post #11 that you wouldn't. You wrote: "There is now a 1/2 chance one of the two remaining doors are the right door...So the odds are now 1 in 2. Purely from a probability viewpoint, stick with the door you've got."

 

[thisnumbersdisconnected]

thisnumbersdisconnected,

re: " He is essentially giving me the win. Who wouldn't do that? "


You implied in your post #11 that you wouldn't. You wrote: "There is now a 1/2 chance one of the two remaining doors are the right door...So the odds are now 1 in 2. Purely from a probability viewpoint, stick with the door you've got."

I'm not sure you understand the mathematical concepts here. Choosing to go for the assured win -- I can choose both doors -- versus having only a 50-50 chance, which is what I have once Monty has revealed one door that does not hide the prize, is, again, a no-brainer.

 

[rstrats]

thisnumbersdisconnected,

re: "Choosing to go for the assured win -- I can choose both doors..."


How can switching from curtain #1 to both curtains #2 and #3 assure a win? There is still a 1/3rd chance that the car is behind curtain #1.

 

[thisnumbersdisconnected]

thisnumbersdisconnected,

re: "Choosing to go for the assured win -- I can choose both doors..."


How can switching from curtain #1 to both curtains #2 and #3 assure a win? There is still a 1/3rd chance that the car is behind curtain #1.

Sorry, I had to reread the original OP scenario. Monty opened curtain #3, and I was thinking he'd opened door #1. So now you're letting me choose both #2 and #3? Why? I already know what's behind #3 -- nothing. Essentially, you're giving me the exact same scenario as being allowed to change from #1 to #2. That's nothing more that restating the already existing mathematical equivalent.

PS: Please learn to use the quote feature. :thumbsup:

 

[rstrats]

thisnumbersdisconnected,

re: "So now you're letting me choose both #2 and #3? Why? I already know what's behind #3..."


I asked what you would do before any curtain was opened, thus you wouldn't have known anything about curtain #3.

 

[thisnumbersdisconnected]

thisnumbersdisconnected,

re: "So now you're letting me choose both #2 and #3? Why? I already know what's behind #3..."


I asked what you would do before any curtain was opened, thus you wouldn't have known anything about curtain #3.

Again, I misread. Maybe its time to get my eyes checked.

Still, it makes no difference in light of the overall mathematical concept. Assuming he has given me at least one of the curtains/doors where the prize definitely is not, I pick #1 anyway. My thought is that he's not going to just give me the car. Him giving me both #2 and #3 still provides me a 50/50 shot at being right with #1, and because I don't believe he's going to make winning the car that easy, it probably puts the odds in favor of #1. The choice between #1 and both #2/#3 is still just one pick.

If he reveals #3 as in the OP scenario, it is still a 50/50 shot because I've got the same scenario as I have from the OP -- #3 is out of play, and I could keep #1 or switch to #2. In your scenario, I'll definitely hold on to #1, because as I said, I don't think he's going to just give me the car, which is what he would be doing if it is behind #2 or #3. His offering of the choice convinces me the car is behind #1.

 

[rstrats]

　
　
thisnumbersdisconnected,

re: " Him giving me both #2 and #3 still provides me a 50/50 shot at being right with #1..."


That would be incorrect. Think of it as two areas: Area "A" contains curtain #1 and area "B"￾ contains curtains #2 and #3. There is a 1/3rd chance that the car is in area "A" and a 2/3rds chance that it is in area "B". Before opening any curtains, you KNOW that at least one of the curtains in area "B" doesn't have the car behind it. So by opening a curtain in area "B"￾that doesn't have the car behind it doesn't change the 2/3rds odds that area "B" still has the car in it. Remember that the host knows where the car is and so will not open that curtain. If it had been behind curtain #3, he would have opened curtain #2. And if it had been behind curtain #1, he could have opened either #2 or #3.

BTW, the host is not trying to influence you in anyway.

 

[thisnumbersdisconnected]

That would be incorrect.

No, that would be very correct. Your reasoning is somewhat faulty but not mathematically off-target. However, one thing that "Let's Make A Deal" has taught us is that intuition does not follow probability. That is why an analysis of the show over the years, even after Hall retired and a new host took over, showed that the guest is nearly twice as likely to win the car if he/she changes his/her original choice, not that the odds change, but because they were never anything other than 50/50 to start.

Think of it as two areas: Area "A" contains curtain #1 and area "B"￾ contains curtains #2 and #3. There is a 1/3rd chance that the car is in area "A" and a 2/3rds chance that it is in area "B".

Sorry, I've changed my thinking on this somewhat over the last six months. I originally said Monte changes the odds, but in reality, they are 50/50 from the very start.

Regardless of which door you pick, regardless of whether we use your scenario of combining two doors as one choice, Monte is going to reveal one of the doors was never in play. It never was hiding the prize, it was never going to be a factor in the game. From the very start, he was giving you a choice between only two doors. The door Monte opens to reveal there is no prize there is one of two that he knew he would elminate, because the prize is only behind one door. The revealed door is never in play.

That means you've either picked the door with the prize, or it is behind the remaining door. Whatever door he eliminates was never in the equation to begin with. All he was doing was giving you a choice between two doors, and a chance to change your mind about which door you pick.

BTW, the host is not trying to influence you in anyway.

Nonsense. It is apparent you never watched the show. :laugh:

Did you know that stock market analysts have applied the "Monte Hall Paradox," as this is known, to investing? Say you want to diversify into a new industry or technology to spread your investment strategy out, but you have limited resources and you can't negate your investment by buying more than one stock, which usually results in one under-performing, one being stagnant, and one rising. There is only one that can be the "best investment."

So you observe the performance of three stocks for the very short term, all apparently good, but you watch them carefully for a few days, then eliminate the most under-performing one -- that is, the one that either drops the most, remains static, or doesn't gain as much as the other two. Now you watch the other two, but not for very long. The choice as to which you keep is 50/50, because you eliminated one of them without ever making a purchase. Of the other two, which do you buy?

The advise is "Buy the under-performing one." It will eventually out-perform the other stock. Why? Who knows? It is the same scenario as with Monte Hall and "Let's Make A Deal." Intuition trumps probability.

 

[rstrats]

thisnumbersdisconnected,

re: "Nonsense."


There is no nonsense. I'm the host, and I'm telling you that I am not trying to influence you to switch or not switch. Regardless of what curtain the car is behind, I'm going to offer you the chance to switch your choice.

OK, then. Let's take this one step at a time. Do you agree that if curtains #2 and #3 are in area "B" that there is a 2/3rds chance that the car is in area "B"?

 

[thisnumbersdisconnected]

There is no nonsense. I'm the host, and I'm telling you that I am not trying to influence you to switch or not switch. Regardless of what curtain the car is behind, I'm going to offer you the chance to switch your choice.

Which is "influence," whether your realize it or not. You've given me a chance to let my anxiety over a wrong choice convince me to change my mind.

OK, then. Let's take this one step at a time. Do you agree that if curtains #2 and #3 are in area "B" that there is a 2/3rds chance that the car is in area "B"?

No. Reread my post, please. The odds are never greater than 50/50. I've done some rethinking on this since this thread started, and I edited that last post for clarity, too. Sorry.

 

[rstrats]

thisnumbersdisconnected,

re: "The odds are never greater than 50/50."


I just don't see how you can think that there is the same chance of the car being in area "A" where there is only 1 curtain, as there is of the car being in area "B" where there are 2 curtains.

 

[thisnumbersdisconnected]

thisnumbersdisconnected,

re: "The odds are never greater than 50/50."


I just don't see how you can think that there is the same chance of the car being in area "A" where there is only 1 curtain, as there is of the car being in area "B" where there are 2 curtains.

You've made them one choice. If I can choose both #2 and #3, I have whatever is behind both, meaning I can keep what I have, or switch to the other. Just as I explained in the lengthier post, that's really the only choice I have in the first place, with the opportunity to change my mind. In the end of either scenario, there are only two choices standing, I've made one, and I can choose to change to the other. Any way you slice it, that's 50/50.

 

[rstrats]

thisnumbersdisconnected,

re: "Any way you slice it, that's 50/50."


OK, let me pose my last comment as a question: Do you think that there is the same chance of the car being in area "A" where there is only 1 curtain, as there is of the car being in area "B" where there are 2 curtains?

 

[thisnumbersdisconnected]

OK, let me pose my last comment as a question: Do you think that there is the same chance of the car being in area "A" where there is only 1 curtain, as there is of the car being in area "B" where there are 2 curtains?

You're still trying to deal with three doors/curtains whatever. In reality there are only two, because whether you pick the door with the prize, or one of the two that doesn't have the prize, the third door is going to be eliminated by the host. As I said, it is never really in play. No matter what you do, one non-prize door is going to be removed from the scene. Therefore, in reality, you are only dealing with two doors: The one you pick, and the one that is left. The third door has nothing to do with the game, other than to distract you.

 

[rstrats]

　
thisnumbersdisconnected,

re: "You're still trying to deal with three doors/curtains whatever."
　
That is because there are three curtains.

If you would just bear with me for a bit and answer my yes or no question I think you'll see why there is never a 50/50 chance for any single curtain. Question: Before any curtain is opened do you agree that if curtains #2 and #3 are in area "B" that there is a 2/3rds chance that the car is in area "B"? Yes or no?