Monty Hall Problem... For Newbies

[UKBoy1977]

Not my fight, so I'm not going to get in to it beyond this, but he's not wrong - you are. His initial post made one statement that you responded to by saying "I disagree", and for which you provided incorrect reasoning. It was pointed out to you that your reasoning didn't support your "I disagree" and you changed what it was you were arguing against to something that ddt hadn't claimed to be the case.

If you do disagree with what ddt actually said, then you have yet to explain why what he said was wrong. If you don't disagree with what ddt originally said, then you need to retract your original disagreement and admit that what ddt said was correct.

As far as civility goes, if you have an issue, then report it to the mods. If you think this latest post of yours puts you on any kind of higher moral ground, then you are mistaken. That's my last word on the subject.

Yes, reading back once more my last post was a bit unnecessary as I can see that we were ultimately talking about different things (pre-game vs mid-game probabilities) as I summarised in post #123. He also threw me by arguing that 'randomly picking a goat' means something different to 'randomly picking a door behind which there happens to be a goat'.I still maintain his tone was unnecessarily aggressive. However I wouldn't report it on account of the fact I'm not 5.

 

[jiggeryqua]

At the start of the game, there are only 9 possible setups.

The car is behind Door A. I pick Door A.
The car is behind Door A. I pick Door B.
The car is behind Door A. I pick Door C.
The car is behind Door B. I pick Door A.
The car is behind Door B. I pick Door B.
The car is behind Door B. I pick Door C.
The car is behind Door C. I pick Door A.
The car is behind Door C. I pick Door B.
The car is behind Door C. I pick Door C.


If I stick with my original choice, I win in three of these cases. If I switch, I win in six. Easy peasy.

I've had a busy weekend, and I'm still not through properly reading the posts addressed to me on page 1, let alone got to page 3, but this is the first explanation that's made conceptual sense (to me)...by ignoring the inevitable (and therefore irrelevant) goat. Which I think is what some posters were trying to convey with the whole 'it's one big door' idea. Bravo Mr Brady.

 

[toto]

I voted how I would have done when I first heard about this. (Voted didn't matter -odds even). I only really accepted the truth when I had a go on automated device on the web.

 

[Red Baron Farms]

The problem is easier to understand if you change it slightly so it becomes intuitive.

Lets have 100 doors, 99 goats and 1 car. If you choose a door you only have 1% chance at the car. Monty removes 98 doors as choices. Now the remaining door has a 50% chance of containing the car.

Which odds would you prefer? The original 1% chance from your original pick? Or changing doors to the 50% chance? Change doors every time. It improves your odds.

Same thing with just three doors. But less intuitive.

 

[jsfisher]

The problem is easier to understand if you change it slightly so it becomes intuitive.

Lets have 100 doors, 99 goats and 1 car. If you choose a door you only have 1% chance at the car. Monty removes 98 doors as choices. Now the remaining door has a 50% chance of containing the car.

Which odds would you prefer? The original 1% chance from your original pick? Or changing doors to the 50% chance? Change doors every time. It improves your odds.

Same thing with just three doors. But less intuitive.

You were good up until the highlighted part. The other door has a 99% chance of having the car behind it.

 

[Red Baron Farms]

You were good up until the highlighted part. The other door has a 99% chance of having the car behind it.

hahaha yes of course, but not intuitively. You are right, taken as a whole there was still 99% chances the car was behind ONE of the other doors. Yet intuitively we know when you are down to two it is a 50/50 chance for one or the other. One might be tempted to hold onto their door unless they realize the 50% beats their 1% nearly every time, and the only way to "reset" their 1% is to change doors.

 

[Brian-M]

hahaha yes of course, but not intuitively. You are right, taken as a whole there was still 99% chances the car was behind ONE of the other doors. Yet intuitively we know when you are down to two it is a 50/50 chance for one or the other. One might be tempted to hold onto their door unless they realize the 50% beats their 1% nearly every time, and the only way to "reset" their 1% is to change doors.

If there's a 1% chance the car is behind the originally selected door, and a 50% chance that it's behind the only other remaining unopened door, where is it the other 49% of the time? In the parking lot?

 

[steve s]

Yet intuitively we know when you are down to two it is a 50/50 chance for one or the other.

No. It's still 1% vs 99%. If it were 50/50, it wouldn't matter whether you switched or not. You'd win the same number of times regardless of what you choose. As the Mythbusters showed, it's clearly in your advantage to switch.

Steve S

 

[Red Baron Farms]

hahahaha You guys crack me up.

 

[jbm]

The problem is easier to understand if you change it slightly so it becomes intuitive.

Lets have 100 doors, 99 goats and 1 car. If you choose a door you only have 1% chance at the car. Monty removes 98 doors as choices. Now the remaining door has a 50% chance of containing the car.

Like so many already posted, it is really pretty simple. Your suggestion of using 100 doors is good, so let's use that:


You have 100 doors, behind one there is a car, the rest hides a goat each.

You choose 1 door, so you have a 1% chance of having chosen the car. I think we all agree on that percentage.

The other 99 doors together have a chance of 99% that the car is behind one of them. I think we also agree on that percentage.

You also know that either 98 or 99 of those doors hide a goat (99 if you actually chose the door with the car).

Monty, knowing which door hides the car, opens 98 of those 99 doors and shows that each has a goat behind it. We already knew that. So nothing changes the fact that the 99 doors you did NOT choose TOGETHER have a 99% chance of having the car behind one of them.

Since of those 99 other doors only one is left unopened and the 98 opened ones all had a goat, the one door left unopened obviously has a 99% chance of hiding the car, because "all the other doors together have a 99% chance" still is valid. The car has to be SOMEWHERE, and since your first pick was a 1% chance and of the 99 doors with together 99% chance only one door is left, well, that gives that one other unopened door the 99% chance.


An explanation with less text would be:

100 doors, you open one. Monty asks you "would you like that door you chose, or would you like to open ALL of the others and get the car if it is behind one of those doors? Here, I open 98 of those "all others" which hold the goats already, if you want "all others", you just choose that one unopened door left here."

That makes it a bit clearer that the choice is between 1% and 99%.

 

[jt512]

An explanation with less text would be:

100 doors, you open one. Monty asks you "would you like that door you chose, or would you like to open ALL of the others and get the car if it is behind one of those doors? Here, I open 98 of those "all others" which hold the goats already, if you want "all others", you just choose that one unopened door left here."

That makes it a bit clearer that the choice is between 1% and 99%.

That was very nicely explained.

 

[Red Baron Farms]

Monty, knowing which door hides the car, opens 98 of those 99 doors and shows that each has a goat behind it.

Exactly. I agree. Since 50/50 is 1 you get no change. You go to 99% if you switch doors and stay at 1% if you don't switch your door. So even though a 50/50 random chance between two doors, because your initial chance is stuck at 1% until you change, and 50/50 =1, the 99% chance stays at 99% when you switch. I was just trying to break the intuitive block some people have with my first post. All is cool for you math guys.

 

[boooeee]

So even though a 50/50 random chance between two doors, because your initial chance is stuck at 1% until you change, and 50/50 =1, the 99% chance stays at 99% when you switch.

What.

 

[GlennB]

Exactly. I agree. Since 50/50 is 1 you get no change. You go to 99% if you switch doors and stay at 1% if you don't switch your door. So even though a 50/50 random chance between two doors, because your initial chance is stuck at 1% until you change, and 50/50 =1, the 99% chance stays at 99% when you switch. I was just trying to break the intuitive block some people have with my first post. All is cool for you math guys.

Yeah, what???

The number 50 is not involved at the end, either mathematically or intuitively. Why do you keep mentioning it?

 

[This is The End]

From one of the older threads:

There's two goats and one car. I have 2/3 chance of picking a goat. Monty shows me the other goat. So now I'm pretty sure I've found both goats. The one remaining door probably has the car. That's about as far as my math goes, or needs to go, to solve this problem.

The rest is up to Monty.

That is a good explanation.

A good thing to tell people is that they do NOT want to pick the car on the first guess (1/3). Switching will then get them a goat. Because picking a goat is more likely (2/3), if they do pick a goat on the first guess, switching will get them a car!

After you tell them they DON'T want to pick the car on the first guess, it usually clicks. It's hard for people to not think: must pick car, must pick car.

 

[This is The End]

Another good visual:

You are wealthy, and have 3 garages, a car and a chauffeur, Jeeves, who normally drives the car, and randomly parks it in one of the 3 garages.
You sometimes drive it, so open one of the garages.

<snip>

your first choice is the first garage, but the light is off, so you don't see it, Jeeves shows you an empty garage. If you then turn the light on, 1/3 of the time you will find the car, if you don't, and try the remaining one, you will find it 2/3 of the time.

 

[This is The End]

One more good post from the old thread... (sorry )

Ordinary:

Column|A|B|C
You pick|1|1|1
Car behind|1|2|3
Monty shows|2 or 3|3|2
Decline|C|G|G
Accept|G|C|C

Chance of car if accepting gambit is 2/3. Chance if declining 1/3.

 

[jiggeryqua]

I'm more fascinated by the interest in explaining this problem than I am by the problem itself. I definitely don't intend to sound snarky, but I am amused by all the 'this is how you do it' posts, especially as so many haven't worked for me. If only all learners were identical sponges, education would be a breeze...

Adding doors isn't helping with the conceptual blocks I stumble over. 100 doors, a milliard doors, it really doesn't matter to me. We know there's an inevitable (irrelevant) goat. 98 irrelevant goats are still irrelevant.

The 'one big door' analogy doesn't help. We know our thoughts do not affect the material world. Why does deciding on one door affect the mathematics? If there's probability to be shifted, why to the door you haven't picked? Your picking it hasn't affected the material world nor, surely, has it affected the mathematics underpinning the material world. I understand that you'd all like me to shift the whole of the 'spare' probability to the other door, but what distinguishes the two available doors??

I've seen lots of table, most of which have different numbers of rows and columns. If yous can't all decide what the problem is, why should anyone be expected to understand it. Some tables are particularly unhelpful (I'm looking at you, #177...)

I will take a stab at recreating on here the paper and pen table I mentioned a few pages ago. It doesn't do what #177 and several other tables and examples have done - to compress two choices into one. Given that the 'convincer' is common and is counting, then counting two choices as one so you end up with the result you 'want' rings alarm bells...

Picked Door|Car Is Behind|Monty Opens|Change|Stick
X|X|Y|LOSE|WIN
X|X|Z|LOSE|WIN
X|Y|Z|WIN|LOSE
X|Z|Y|WIN|LOSE
Y|Y|Z|LOSE|WIN
Y|Y|X|LOSE|WIN
Y|X|Z|WIN|LOSE
Y|Z|X|WIN|LOSE
Z|Z|X|LOSE|WIN
Z|Z|Y|LOSE|WIN
Z|X|Y|WIN|LOSE
Z|Y|X|WIN|LOSE

Evidently, I'm missing something somewhere, and if I'm missing something obvious in that table, sing out...but don't I see all the possible situations with an equal number of change-wins and change-loses?

 

[GlennB]

Picked Door|Car Is Behind|Monty Opens|Change|Stick
X|X|Y|LOSE|WIN
X|X|Z|LOSE|WIN

Evidently, I'm missing something somewhere, and if I'm missing something obvious in that table, sing out...but don't I see all the possible situations with an equal number of change-wins and change-loses?

I reduced your table to the important part. You're counting that as two separate events, but really they're one in terms of probability.

Picked Door|Car Is Behind|Monty Opens|Change|Stick
X|X|Y or Z|LOSE|WIN

is how it should read. Half the times the setup is X-X Monty will open Y, and half Z.

 

[Red Baron Farms]

What.

Yeah, what???

The number 50 is not involved at the end, either mathematically or intuitively. Why do you keep mentioning it?

Come on guys, don't get your panties in a bunch. I said from the beginning, better to change and why, with the correct %.

But the intuitive answer is 50/50, when you are down to 2 doors. I was just showing that the intuitive (incorrect) answer should also be to change doors, for those stuck on it. And don't pretend that isn't part of the intuitive answer, because that's why it is on the poll and actually many people when first confronted pick 50/50. I would guess that if you checked the general population the poll would be more strongly in favor of 50/50.

I personally never saw this exact problem before now, but I do play, (and teach) poker. And a similar thing happens in poker. You have to be able to break your intuition (when incorrect) without pencil and paper or exact detailed calcs in poker. Yet still make the correct choice. Because there is a bunch of other important things to think about besides the rough calcs. So I probably approach this problem differently than an actual math guy.

But I will point out, I still made the right choice when confronted with the problem the first time. No matter how screwballed the thinking might seem to a precise math type brain.

If I communicated this poorly before, (and I looked up and saw that I did) I sincerely apologize.