THE MONTY HALL PARADOX

This is an oldie. But I still don't get it.

At the end of the game show "Let's make a deal", the big winner chose 1 of 3 doors. He chose door number 1. The big prize is a new car. Behind another door is washer and dryer. And behind the last door is a rusty wagon and a billy goat.

Monty Hall shows everybody what is behind door #3. It's the washer/dryer. The contestant has picked door 1. Hall offers to let the contestant switch his choice of door #1, to door #2. Or keep door #1.

The question is, which door gives him a better chance of winning the car? To me and most others, the car has a 50-50 chance of being behind either remaining door, #1 or #2.

But people far above my pay scale, including highly regarded mathematicians, insist that if the contestant switches from door #1 to door #2, he has a better chance of winning. They see it as his original choice, door #1, has a 1 in 3 chance of winning, but if he switches to door #2, he has a 1 in 2 chance of winning, since he is choosing a new door and there are only 2 possibilities.

Who is right?

Ahh the Monty Hall Problem! I had a statistics/probability professor show us that in college. It's not intuitive (and honestly still isn't to me). The contestant has a 1:3 chance of getting the car when he chooses door1. Doors 2 and 3 then have a 2:3 chance. So when Monty opens door 3 (knowing the goat is in there) it shifts all of that 2:3 probability to door 2 while door 1 is still 1:3.

That's how it was explained in class, but I'm with you River, I don't quite get it. Still seems 50/50 to me but hey I don't have a PhD in Math.

Write on Rachel. I took algebra, geometry and calculus in high school, but I take a back seat to mathematics professors and others well versed in statistics and the like.

Still, I'm not buying it. Logic tells us that before the contestant chooses a door, that each door has a 1 in 3 chance of having the car behind it. I can't see those odds changing just because 1 door was taken out of the equation. Or at least they both still have the same odds, which are now 1 in 2.

I dunno. People smarter than me, say I am wrong. Maybe I need a new stragedy...lol.

Always stay with your gut. Changing from your initial pick usually bites you in the ass. That applies to Let’s Make A Deal, or life in general…

I asked AI. Here is the response:

Search Assist


>>The correct answer to the Monty Hall puzzle is that you should always switch doors after the host reveals a goat behind one of the other doors. This strategy gives you a 2/3 chance of winning the car, compared to a 1/3 chance if you stick with your original choice.<<


I still disagree..My last attempt at logic...

So if the contestant picks door 1 and Hall shows what is behind door 3, suddenly door 2 has the better odds of having the car behind it and he should switch from door 1 to 2. But suppose the contestant had chosen door 2 when choosing a door? Hall then shows the goat behind door 3. That would mean that now door 1 has the better odds and he should switch from door 2 to 1.

But both scenarios can't be true.

Exactly the way I see it too. Any Math Professors here to make sense of this?

And it would be my luck if I were in that situation and switched to door 2, the damn car would be behind door 1. lol

I'm a firm believer in the 50/50/90 rule. Whenever faced with a 50/50 situation there is a 90 percent chance you'll make the wrong choice.

Actually you will be better off switching doors.
Initially each door has 1/3 chance to get the car. Say you choose door 1 so you have 1/3 chance to win the car and there is a 2/3 chance that either doors 2 or door 3 will have the car. Monty shows door 3 had a goat and a wheelbarrow so now door 3 has 0 chance of having the car. Being door 1 had 1/3 chance of having the car initially that does not change but door 2 now has 2/3 of a chance of having the car. This is because initially door 2 and 3 had a 2/3 chance of the car but door 3 from the reveal of the goat now has 0 chance the 2/3 is all on door 2. By switching you go from a 1/3 of a chance to 2/3 chance of winning the car.

I think the shy one has hit the nail on the head.

Interesting conundrum.

All things being equal stick with your first choice .

In reality after you are shown door number 3, and know nothing about doors one or two, you have a new problem. What is the probability that a car exists behind one of the 2 doors. The mathematics of 1/3 for door 1 and 2/3 for door 2 is correct, just not applicable to your new problem which is best represented with the toss of a coin. Heads or tails.
You need to read Antifragile and understand the difference between Fat Tony and the math professor.

It's too early in the morning for me to thing about this. Right now, I'm thinking what's wrong with a billy goat?

The MythBusters guys once ran this through a practical display on an episode. The results were staggering in support of changing your original decision. Always change.

Ok, I guess I'll give it another shot. This just doesn't compute with me.

Let's say there are 2 contestants who tie for being eligible to pick a door at the end of the show. So Hall allows each contestant to pick a door.

Contestant 1 picks door #1 and contestant 2 picks door #2. Hall then shows the billy goat is behind door #3. He then allows both contestants to stay with their original choices or switch doors. According to this new math logic, if contestant #1 switches from door #1 to door #2, he now has better odds of winning the car.

But also if contestant #2 switches from door 2 to door 1, he also has better odds of winning a car.

How can both be true?

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I like the old logical math. Each of the 3 doors has a 1 in 3 chance of having the car behind it. When Hall takes door # 3 out of the equation, that leaves only 2 doors, 1 and 2. The fact that door 3 was not the winning door and has now been removed, has no bearing on whether the car was put behind door 1 or door 2. Each door SHOULD now have a 50% chance of having a car behind them or a 1 in 2 chance.

The car was randomly put behind 1 of 3 doors. I don't see how showing that one of the doors does not have the car behind it could possibly improve the odds of either of the 2 remaining doors of having the car behind it.

Could I be wrong? Sure I could. I just don't see how. Perhaps there is a mathematics expert out there who likes coming to this site and looking at boobs (not these kind of boobs>> ruud, ban-anna or buttpirate), who could show me the error of my ways??

from what ive seen on these boards,a lot of people would think the goat is the best prize to win.....

The "pick a door" is played with only one contestant. Once the switch is made the game is over and the reveal is made. Maybe you win the car, maybe you don't. But your odds were improved when you made the switch. Play the game over and over and you will win a parking lot full of cars over what was behind the door you switched from.

I need a beer…..

Doesn't seem possible bonefang.

Let me see if I've got this straight. If the contestant picked door 1 and Hall shows the goat behind 3, the contestant's odds are still 1 in 3. But if he switches to door 2, his odds are now 2 in 3?

Ok, what if from the start, he chooses door 2? The goat is behind door 3. The contestant switches from door 2 to door 1. Did his odds of winning then go from 1 in 3 to 2 in 3? If not, why not?

Yes. The odds improved with the switch but there is still the chance that the contestant made the wrong choice in doing so. Gambling is still gambling but successful gamblers know how to work the odds a little more in their favor.

Or maybe better stated, if the entire Let's Make a Deal show was played with only one contestant doing nothing but "pick a door" over and over, they will ultimately win more times by switching than not switching.

With my luck I would lose either way!

So if the contestant picks door 1, and door 3 has the goat, and he switches from door 1 to door 2, his odds of winning are now 2 out of 3?

At the same time, if he picks door 2, and door 3 is eliminated, then he switches from 2 to 1, his odds of winning is also 2 out of 3?

That doesn't make sense.

I think what most of you are missing is this isn't purely math. Monty knows where the car is and he's attempting to change the outcome to get you to pick a goat.

ok... you had a one and three chance and that was changed when on of the doors was selected.

given the choice again you know you don't want what is behind the door revealed. regardless of what door you initially selected, you get new choice.. either it is the same door as you initially choose or a different one.

you only have two choices the second time around... stay with what you picked or change. 50/50.

it's like flipping a coin. what's the odds of flipping two heads. initially it would be 1/4 but after the first flip is a head, the second flip is 50/50 or 1/2. if you view it from the perspective of no choices then the 1/4 still remains true. but the next flip would still be 1/2. as would the next one and the next one.

your odds improved which is all I think they were saying. i don't see where changing it to a tails would improve you odds.

Yes, the contestant will improve the odds by switching from their original but not yet revealed choice. But only after one of the dud doors has been taken from the mix. As I said, switching doesn't guarantee a win but it does improve the odds. Play long enough and you will win more by changing your pick.

https://youtu.be/4Lb-6rxZxx0?si=Uaw-xoq-o7ujmVSs

When in doubt stick with your original choice....Trust your original instincts...The house wants you to doubt yourself...

It is easier to understand using an example with more doors. There are 1000 doors. You pick one, they open 998 doors showing no prize. Now there are two doors left. Obviously you switch doors, your original door has 1 in 1000 odds

I am totally convinced that whether you keep the door you originally selected (door 1), or you switch to door 2, your odds are the same.

Assuming Monty Hall RANDOMLY puts the car, the goat and whatever behind the 3 doors, then whoever gets to pick a door at the end of th show, has a 1 in 3 chance of winning the car. Each door has a 1 in 3 chance of having the car behind it.

When he picks a door he has a 1 in 3 chance of winning the car and a 2 of 3 chance of not winning the car. However, when Hall shows the goat behind door 3, the odds change. It is no longer a 1 in 3 chance to win and a 2 in 3 chance of losing. There are NOW only 2 doors left. Logic tells you that he has a 1 in 2 chance of winning and a one in 2 chance of losing. How can that possibly be in dispute?

Another way to look at it is when the game starts, there are 3 doors and there is a car behind one of the doors. That means there is a 1 in 3 chance that the car is behind any of the 3 doors, with each door having an equal chance of having the car. That also can't be disputed.

If you take one of the doors away that doesn't have the car (door 3), how could that possibly improve the odds the car is behind door 2 and worsen the odds that it is behind door 1? It can't. All 3 doors had a 1 in 3 chance the car was behind them. Taking away door 3 couldn't possibly influence which of the remaining doors has the car.

I am totally convinced that whether you keep the door you originally selected (door 1), or you switch to door 2, your odds are the same.

Assuming Monty Hall RANDOMLY puts the car, the goat and whatever behind the 3 doors, then whoever gets to pick a door at the end of th show, has a 1 in 3 chance of winning the car. Each door has a 1 in 3 chance of having the car behind it.

When he picks a door he has a 1 in 3 chance of winning the car and a 2 of 3 chance of not winning the car. However, when Hall shows the goat behind door 3, the odds change. It is no longer a 1 in 3 chance to win and a 2 in 3 chance of losing. There are NOW only 2 doors left. Logic tells you that he has a 1 in 2 chance of winning and a one in 2 chance of losing. How can that possibly be in dispute?

Another way to look at it is when the game starts, there are 3 doors and there is a car behind one of the doors. That means there is a 1 in 3 chance that the car is behind any of the 3 doors, with each door having an equal chance of having the car. That also can't be disputed.

If you take one of the doors away that doesn't have the car (door 3), how could that possibly improve the odds the car is behind door 2 and worsen the odds that it is behind door 1? It can't. All 3 doors had a 1 in 3 chance the car was behind them. Taking away door 3 couldn't possibly influence which of the remaining doors has the car.

For what it's worth DTR, I totally agree and would never switch my original guess anyway, historically when I change my mind on a random response my 2nd guess is wrong, so I would hold on door #1 and I call BS on any other conclusion.

I have experienced the same thing Mystic. You learn patterns and other realities over the years. I have learned that my first choice, is usually the right one. Or at least is right more often than when you switch. It's almost like you have intuition. Your mind subconsciously lets you know what decision to make and it is usually right. And when you go against that decision, you usually lose. Not all the time, but usually. Of course picking a random door isn't something your subconscious would know what to do with. But I do agree. I find it's best to stick with my first choice.

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