Monty Hall Problem

[CFLarsen]

You did, Claus-- in post 535--in your everlasting quest to prove me wrong by knocking down a straw man version of what I actually said.

Nonsense.

I think anyone can see that if you always stick with your current choice--you will have 1 in 3 chance of winning-- the only way to increase those odds is to change your choice after a goat is shown. The host can always show a goat. The host does not always need to offer you the switch.

Yes, he does. That's the whole idea of the Monty Hall Problem:

The host knows what is behind the doors.

The host will offer to switch.

If the host doesn't know, and doesn't offer to switch, it isn't the Monty Hall Problem.

You are miseducating your students if you claim otherwise.

However, in the scenario--with no added information-- your odds increase by switching. Moreover, even if you add information-- you always have 1 in 3 chance of winning by not switching--that doesn't change with the added information. In your made up scenarios you are only switching whether the host is going to be offering you a switch or not.

That is not a "made up scenario". That is the Monty Hall Problem.

Because the only information we have as the scenario is laid out is that he's shown a goat and offered a switch. That is the only information in the scenario. But no matter how much you add information, you are never increasing the odds over 1 in 3 by staying. If the host only offers a switch if you have the car... then the information you get when if he asks you to switch, is that you have the car. But that means that in 2/3 of the cases, he'll never ask you to switch. And because you don't know this is the case, it's more likely that he's offering you to switch because he always does (the original scenario) or because he blindly chose a goat... (meaning that if he blindly chose the car... he'd not be offering you the chance to switch.)

The salient part of this exercise is that your odds if you stay are not more than 1 in 3-- ever. The only way to increase those odds is to switch if given the opportunity. (Of course 1 in 3 isn't bad odds)... and even with the 2/3 increased odds with a switch-- you still lose 1/3 of the time.

Please provide evidence that the Monty Hall Problem does not include these two:

The host knows what is behind the doors.

The host will offer to switch.

Evidence. Not your own explanations.

 

[articulett]

Now I see what was worrying me in what you said earlier. You said

It would be clearer to say

"The only way to change those odds is to switch if given the opportunity".​

Obviously if your strategy is to choose a door and never to switch, you will, in the long run, win 1/3 of the games. But if your strategy is always to switch when given the chance, you may increase or decrease your odds, depending on what strategy Monty is using.

For instance, if Monty (unknown to you) only gives the opportunity of switching when there is a car behind your chosen door, the strategy of switching if given the choice will ensure that you never win: you have decreased your odds from 1/3 to 0.

A real-life game show host will probably use a mixture of strategies, to keep up the suspense and make sure that nobody can work out how to cheat him. You can only work out if it's better for you to switch or to keep your door if you know precisely what mixture of strategies he uses.

I was correct in saying that the only way you can increase your 1 in 3 chance of winning a car is by switching. This does not mean that you always increase your odds by switching-- it means that it's the ONLY way you can increase your odds from the 1 in 3 chance you have by staying. If Monty is only offering you the choice because you chose the car-- then your odds do not increase by switching.-- but he'd only be giving you that option in 1 out of 3 cases! In the other cases you'd just be revealed to be a loser without being offered a choice. Your odds of winning by keeping your first choice are never more than 1 in 3-- you are just changing whether you are being offered a chance to switch. Yes, Monty could be using such a strategy... and if you believe that, you will keep your first choice and win the car 1/3 of the time. If it's true, he won't offer you a chance to switch 2/3 of the time-- if it's not true, he will offer you a chance to switch and you won't do it, making you a winner 1/3 of the time. Your odds don't increase by keeping the door. You CAN increase your odds... but only by switching if given the opportunity.

 

[articulett]

Claus... the evidence IS saying what I"m saying. Go ask any expert. Your odds are never more than 1 in 3 by staying with your first choice.

If the host must give you a choice and must show a goat then you double your chance of winning the car by switching. Period. Your odds remain 1 in 3 if you keep your first choice. It never becomes 50-50 even though there are 2 choices.

You are wrong. Embarrassingly so. Again. As usual. No matter what the motives of the host-- you never have more than 1 in 3 chance of winning the car if you stick with your first choice given that the host must offer a choice and must show a goat. Yes, you can still win-- and you will-- 1/3 of the time. 2/3 of the time you'll be the sorry loser saying "I thought my odds were 50-50"

 

[CFLarsen]

Claus... the evidence IS saying what I"m saying.

OK - in the absence of evidence...

You mentioned that you regularly teach this to 15-year olds. How do you present this problem to your students?

I'm not talking about the explanation that follows.

I'm talking about how you describe the scenario and the rules, before you let the solve the problem.

Please write that down and post it here.

 

[articulett]

Exactly as you did Claus... that I will always show the goat and always offer a choice-- I use cups and little plastic cars and goats. There are also computer applets that they can use.

Like you, they think their odds are 50-50 when they have two choices left. But when they run the statistics they always learn that they win 1/3 of the time by keeping their original choice and 2/3 of the time by switching-- that is, they learn that despite their convicetion that their odds are 50-50 because there are two choices-- their odds of winning actually double from 1/3 to 2/3 by switching... the whole point of the lesson. The more trials they run the more clear these statistics become.

Even you could learn this.

 

[AntiTelharsic]

Articulett, are you addressing the case where Monty chooses randomly? I think that's what CFLarsen is addressing, not the standard problem. Apologies if I'm mistaken here, but I think that's where the disconnect is (I fear there are several different versions being simultaneously discussed).

 

[CFLarsen]

Exactly as you did Claus... that I will always show the goat and always offer a choice-- I use cups and little plastic cars and goats. There are also computer applets that they can use.

Like you, they think their odds are 50-50 when they have two choices left. But when they run the statistics they always learn that they win 1/3 of the time by keeping their original choice and 2/3 of the time by switching-- that is, they learn that despite their convicetion that their odds are 50-50 because there are two choices-- their odds of winning actually double from 1/3 to 2/3 by switching... the whole point of the lesson. The more trials they run the more clear these statistics become.

Even you could learn this.

Then, why are you going on and on and on about something that is not the Monty Hall Problem?

 

[articulett]

Then, why are you going on and on and on about something that is not the Monty Hall Problem?

because other people proposed other scenarios.

My first response to you was because you claimed that your odds change when you are shown the goat-- they don't-- the only thing that changes is that you increase your odds by switching. That's it. Then you denied saying that and kept with your quest that I was saying something false or wrong-- when I was not.

Then other people proposed various scenarios but no matter what the scenario, you never increase your odds above the 1 in 3 chance of having picked the car the first time no matter what is or isn't revealed or what choices you are and aren't offered. You are only gaining information as to whether it's better to switch if offered.

You can't make your offer more likely to be a winner by choosing it again. But it sure feels like you can. Choosing an option gives a person a conviction that wasn't there before the choice causing them to choose irrationally. It's a known psychological quirk of humans. Moreover, we see 2 choices and think 50-50. God or "no God "seems like a 50-50.

But is Xenu vs "no Xenu "a 50/50? Unicorns vs "no unicorns"? I have to teach my students how they get things wrong before I can teach them the methods for correcting it.

 

[CFLarsen]

because other people proposed other scenarios.

Then, stop telling people they are wrong about the Monty Hall Problem when they're not.

My first response to you was because you claimed that your odds change when you are shown the goat-- they don't-- the only thing that changes is that you increase your odds by switching. That's it.

Whatever gave you the idea that I was talking about other scenarios than the Monty Hall Problem?

Then other people proposed various scenarios but no matter what the scenario, you never increase your odds above the 1 in 3 chance of having picked the car the first time no matter what is or isn't revealed or what choices you are and aren't offered. You are only gaining information as to whether it's better to switch if offered.

Nobody has argued otherwise.

 

[articulett]

Don't give me advice Claus... I would never take advice from someone I would never want to be like. You are the one who told me I was wrong when I was not. I never told someone they were wrong, when they were not.

My words are there for everyone to see. If I said something wrong, it's easy to highlight it... it should also be easy to show that I am wrong with an opposing version on the page you linked.

I didn't think YOU were talking about anything different-- I was only answering your allegations that I was wrong in saying that your odds don't change from 1 in 3 just because a goat was revealed. You contended that they do. That makes you wrong.

I quoted your words in post #535. You accused me of being wrong-- you were wrong. And now you are trying to weasel your way out of your wrongness by tsk-tsking me. I was never wrong in any of my statements. You were wrong... as you often are-- and yet, I've never seen you admit it. Instead you attack the people who reveal you gaffes and change the subject and pretend you have nothing to do with the derail.

 

[CFLarsen]

That's precisely what the "Monty Hall problem" shows: That your odds do change when you pick a certain door.

Wrong. The odds of the car being behind the door you originally chose is never more than 1/3. It doesn't change. The car always has a 2/3 chance of being behind one of the other doors.

I was talking about the Monty Hall Problem.

If you focus on the first part and leave out the parts where the host opens the door to the goat and asks the person if he wants to switch, then you are not talking about the Monty Hall Problem.

The Monty Hall Problem isn't the Monty Hall Problem without these two things. Without them, it's just a trite probability question.

 

[articulett]

Yes... and you are still wrong. Your odds don't change when you pick the door. Your odds of being correct are 1 in 3 if you keep the door--the same as if you never made the switch... and 2 in 3 if you switch.

Thanks for finally admitting your error and for highlighting it so everyone else can see.

 

[CFLarsen]

Yes... and you are still wrong. Your odds don't change when you pick the door. Your odds of being correct are 1 in 3 if you keep the door--

That isn't the Monty Hall Problem. That's just a trite probability problem.

the same as if you never made the switch... and 2 in 3 if you switch.

That is the Monty Hall Problem.

You can't talk about the first part, leave out the other, and claim you are talking about the Monty Hall Problem. You are not.

 

[articulett]

I'm talking about the Monty Hall problem the whole time. You never increase your odds from 1 in 3 by keeping the same door. You can only increase your odds by changing. If you don't agree with this-- you are wrong.... if you don't understand this... your English is the problem. If you doubt it-- ask someone you trust that is smarter than you on one or both topics.

 

[AntiTelharsic]

Thing is, there's more than just the classic Monty Hall Problem being discussed here. Since the original statement has been interpreted by some to be ambiguous -- with Monty not necessarily revealing a goat each time, but having simply revealed a goat that time -- they have put forth for consideration the situation where Monty chooses randomly. But to keep it consistent with the original statement they have been considering only those situations where Monty chose randomly but revealed a goat.

Certainly, in the classic Monty Hall problem, switching gives you a 2/3 chance of winning.

But in the variant which is also being discussed, switching gives you a 1/2 chance of winning (for rationale, see my and others people's earlier posts; for experimental proof see Vorpal's program or I could post mine or someone else could write another).

Since the argument seems to be between 2/3 and 1/2, I wonder if people aren't arguing about those two different scenarios.

To help clarify:

Is there anyone here who believes that in the classic Monty Hall Problem (You pick a door, Monty always opens a goat door, you're always allowed to choose the remaining door) you DO NOT get a 2/3 chance of winning if you switch?

I'm not convinced anyone here believes that. I think many people here are failing to clearly specify which version they are talking about.

Personally, I don't find the Monty-choosing-randomly variant very interesting, but it does appear to have been contentious here, so it's worth being careful to specify which variant one is addressing.

 

[Herzblut]

To help clarify:

Is there anyone here who believes that in the classic Monty Hall Problem (You pick a door, Monty opens a goat door always, you're allowed to choose the remaining door) you DO NOT get a 2/3 chance of winning if you switch?

No. There is no need to clarify the obvious.

It was in Egypt, I think, where they recently found a 5.000 years old papyrus bobbin. Guess what, the hieroglyphics revealed exactly this explanation!

 

[articulett]

Yes, A-T, I agree... my explanation at first was geared towards Claus because he said you odds change in the middle of the game... they don't.

And then other people brought up other scenarios... but they either involved the possibility that a car could be shown (rather than a goat) and/or the possibililty that you wouldn't be allowed to make a choice... so that changes the game... it doesn't change the fact that you have a 1/3 chance of winning the car by staying with your first choice.

More here:

http://www.shodor.org/interactivate/activities/
http://www.shodor.org/interactivate/activities/SimpleMontyHall/
http://www.cut-the-knot.org/probability.shtml

(interactive on probabilities.... if you are really curious I'll be tossing punnett squares up there.)

 

[CFLarsen]

Yes, A-T, I agree... my explanation at first was geared towards Claus because he said you odds change in the middle of the game... they don't.

That's precisely what they do: When you are given the choice of changing doors.

 

[articulett]

What odds change, Claus? Your odds are the same - 1 in 3 if you don't change and your odds are alway 2 in 3 if you do. Period. That doesn't change in the basic Monty Hall problem. Your odds of winning the car do not suddenly become 50-50 just because there are two choices left.

Saying it, doesn't make it so.

 

[CFLarsen]

What odds change, Claus? Your odds are the same - 1 in 3 if you don't change and your odds are alway 2 in 3 if you do. Period. That doesn't change in the basic Monty Hall problem.

The odds change when you are given the option of switching doors.

You simply don't understand the Monty Hall Problem, if you don't understand this. This is the key element in the problem.

Your odds of winning the car do not suddenly become 50-50 just because there are two choices left.

Saying it, doesn't make it so.

Huh? Where did I say that the odds suddenly became 50-50?