Monty Hall Problem... For Newbies

[Soapy Sam]

Yes. Took me an embarrassingly long time to see it though.

 

[Brian-M]

LOL, it takes a lot of silliness to believe this. As soon as you know what is behind one door, it becomes a 50/50 chance, regardless of whether you switch your choice or not. There are only two doors remaining, a goat behind one and a car behind the other. The door that was opened ceases to be a factor as soon as it is eliminated (opened), and has nothing to do with your choices.

No, it isn't 50%.

These are the odds for each scenario...
16.67% chance you picked the car and Hall revealed Goat A
16.67% chance you picked the car and Hall revealed Goat B
33.33% chance you picked Goat A and Hall revealed Goat B
33.33% chance you picked Goat B and Hall revealed Goat A

There is no 50% chance in there.

Notice that 66.7% of the time Hall is forced to open a specific door because of your choice. Which door he's forced to open tells you exactly where the car is. Whenever Hall is forced by your choice to open a specific door, there's a 100% chance the car is behind the other door.

Of course you have no way to tell when he's being forced to open a specific door or when he can choose freely. But since his choice if forced 66.7% of the time, this means that if you always assume that his choice is forced you'll be correct 66.7% of the time.

 

[UKBoy1977]

Yes, it is a strawman. My post #43 lays out the obvious switching strategy where Monty randomly opens a door, which includes cases where he happens to open the door with the car. And my claim there is that your probability of success by switching is still 2/3.

That is what you disagreed with.

And then sneakily narrowed the focus.

Reread those posts. You're treading close to wilfully lying simply about what was written.

I've read through your counting scenario again and here is the issue:

You are right that if we had to pre-commit to a strategy before the game started and we chose to switch whatever happens (to the car if he shows the car, to the 3rd door if he shows a goat) then you would be succesful 2/3 of the time. 2/3 is the correct probability for that strategy when viewed from before the moment Monty has made his choice.

But once Monty has made his choice the probabilities now shift. Clearly they shift if he shows the car: our probability of success is now 1 (as we will of course switch to the car). If he shows a goat, then our probability of getting the car if we switch is now 1/2 (as I showed before). So basically your pre-game probability of 2/3 is a weighted average of these two mid-game probabilities (1 and 1/2).

Another way of saying it is that if you chose to change your strategy so that if he shows the car you switch (obviously), but if he shows a goat then you stay with your original choice, you would still win 2/3 of the time. On being shown a goat, it would make no difference to your chances if you switch or not. Here is your counting table to test this strategy of 'Switch if shown car, stick if shown goat' :


As before, let's call the doors A, B, and C, and fix the car behind door "C". That gives 6 scenarios:

First choice Monty opens Second choice Success
A B Stick with A no
A C C yes
B A Stick with B no
B C C yes
C A Stick with C yes
C B Stick with C yes


Still 4/6 successes or 2/3. So choosing to stick after being randomly shown a goat yields the same long term success rate as switching at that stage. This is because after being shown a goat, the probabilities change to 50/50, there is no difference between the strategies once you've been shown a goat.

Edit: On posting, the site is collapsing the table, hopefully it still makes sense.

 

[jt512]

A solution to what?

I am trying to make it clear that there is no problem here.

You can choose a car or one of two goats. You can then choose to stick or swap. If you chose a goat and swap you win, if you chose a car and swap you lose.
2 out of every 3 swap choices gets you a car. (Dammit, I just typed "goat").
What further analysis do you feel is required?

How is "2 out of every 3" not a probability? You might want to look up the definition (specifically, the classical (ie, frequentist) definition.

 

[AdMan]

How is "2 out of every 3" not a probability? You might want to look up the definition (specifically, the classical (ie, frequentist) definition.

That's what I was thinking.

 

[Modified]

How is "2 out of every 3" not a probability? You might want to look up the definition (specifically, the classical (ie, frequentist) definition.

Well, simple probability that most people understand.

 

[GlennB]

How is "2 out of every 3" not a probability? You might want to look up the definition (specifically, the classical (ie, frequentist) definition.

I think what Soapy Sam is saying is that it doesn't require any deep mathematical analysis. Sure, "2 out of 3" relates to probablility theory but it's an idea that's also available at an everyday level. No maths required, really.

It strikes me that the simplest answers to this puzzle are the best. "In effect, Monty is offering to swap your original one for his original two. But he throws away some irrelevant waste just before that, and that confuses people"

 

[boooeee]

'What is the probability we have the car given that Monty has picked the goat?' which we can see is 2 out of the 4 remaining outcomes, or 1/2.

ddt - Do you disagree with this statement? It seems pretty clear to me, assuming that Monty picks a door at random.

UKBoy1977 is not trying to calculate an a priori probability, he is calculating a conditional probability.

 

[GlennB]

ddt - Do you disagree with this statement? It seems pretty clear to me, assuming that Monty picks a door at random.

UKBoy1977 is not trying to calculate an a priori probability, he is calculating a conditional probability.

This is a separate problem. Monty only 'picks a door at random' when he started with 2 goats. Otherwise he must show the goat from a goat/car combination. It's a huge digression that has nothing to do with "The Monty Hall Puzzle"

 

[UKBoy1977]

This is a separate problem. Monty only 'picks a door at random' when he started with 2 goats. Otherwise he must show the goat from a goat/car combination. It's a huge digression that has nothing to do with "The Monty Hall Puzzle"

I think it's a very informative digression as it demonstrates how important it is, when stating the problem, to emphasise Monty KNOWS where the prize is.

 

[GlennB]

I think it's a very informative digression as it demonstrates how important it is, when stating the problem, to emphasise Monty KNOWS where the prize is.

Fair enough, but in a fairly long and sometimes contentious thread it isn't totally obvious that this is a digression. Maybe people could put Digression in their post header?

 

[Soapy Sam]

How is "2 out of every 3" not a probability? You might want to look up the definition (specifically, the classical (ie, frequentist) definition.

If looking at it that way makes it simple for you, fine.
The point I was apparently failing to make is that it's not a requirement.
All you need to know is that of the three possible picks, swapping will give twice the success rate of not swapping.
Personally, I find that expressing this extremely simple fact in terms of probabilities just clouds the issue. I think I'm not alone in this, but each to his preferred method.

 

[Soapy Sam]

I think what Soapy Sam is saying is that it doesn't require any deep mathematical analysis. Sure, "2 out of 3" relates to probablility theory but it's an idea that's also available at an everyday level. No maths required, really.

It strikes me that the simplest answers to this puzzle are the best. "In effect, Monty is offering to swap your original one for his original two. But he throws away some irrelevant waste just before that, and that confuses people"

Exactly.
Thing is, your simplest way of viewing it may be very different from mine,
so some apparently simple explanations actually confuse other people, leaving the explainer even more confused by their confusion.

 

[IXP]

LOL, it takes a lot of silliness to believe this. As soon as you know what is behind one door, it becomes a 50/50 chance, regardless of whether you switch your choice or not. There are only two doors remaining, a goat behind one and a car behind the other. The door that was opened ceases to be a factor as soon as it is eliminated (opened), and has nothing to do with your choices.

Yes, your chances are improved, but through no action of your own. It doesn't matter what you do, the statistics are independent of your choice.

If you believe that, I'll tell you what. Let's play a game. You put up $100 to play, then I will, unobserved put my $100 in one of three numbered envelopes. You will pick one of the envelopes. I will then open one of the other 2 envelopes showing you that it is empty. And you will not be allowed to change your choice. If you get the envelope with the $100 you keep that and your original $100. If you get an empty envelope I keep both $100. Then we will reverse the roles, but I will be allowed to switch after you show me the empty envelope (and I will every time.)

According to you, the bet will be 50/50. But I guarantee that I will take all your money pretty quickly if we continue to play this game.

IXP

 

[bruto]

Exactly.
Thing is, your simplest way of viewing it may be very different from mine,
so some apparently simple explanations actually confuse other people, leaving the explainer even more confused by their confusion.

The problem is made to appear more difficult by having only three elements. If it had more, I think it would be more obvious. You choose one out of a hundred doors. You're then offered the option of abandoning your one door and choosing instead, all 99 others at once. It's the same problem.

 

[Soapy Sam]

The problem is made to appear more difficult by having only three elements. If it had more, I think it would be more obvious. You choose one out of a hundred doors. You're then offered the option of abandoning your one door and choosing instead, all 99 others at once. It's the same problem.

See, for some reason, that does not help me to see what's going on.
This is why I feel the more ways something can be explained , the more people will find one that works for them.

Oddly, one aspect nobody ever seems to mention re the MHP is the psychology of linking the best strategy to the need to change the original choice. How many folk choose to stick, simply because they feel swapping involves coercion?

 

[bigred]

When the goat door is opened, the odds of the car not being your choice are still 2/3

Nope. There are now 2 doors. One is the right choice, one isn't. Your odds just changed.

 

[dlorde]

Nope. There are now 2 doors. One is the right choice, one isn't. Your odds just changed.

There may now be two doors to choose from, but the chances of the door you have already chosen being a car are still 1 in 3; once you've made your initial choice, subsequent events can't change the odds for that choice. You can only change your chances of success by choosing another door.

An alternative way to look it is that if you switch each time, you'll always get the car if you happened to initially pick a goat, and you'll always lose the car if you initially picked the car. You have a 2 in 3 chance of picking a goat first, and a one in three chance of picking the car. So your chances of getting the car after switching are 2 in 3 and the chances of not getting it (losing it) are 1 in 3.

 

[lionking]

ddt - Do you disagree with this statement? It seems pretty clear to me, assuming that Monty picks a door at random.
UKBoy1977 is not trying to calculate an a priori probability, he is calculating a conditional probability.

Aarrggh!

He doesn't pick at random. He knows where the car is.

Too many people over-complicating a simple problem, best explained by Soapy Sam.

 

[boooeee]

Aarrggh!

He doesn't pick at random. He knows where the car is.

Too many people over-complicating a simple problem, best explained by Soapy Sam.

In the standard formulation of the problem, you are correct. And it's crucial to why the odds for switching are 2/3. Which is why it's important to acknowledge that when Monty doesn't know where the car is, AND just picks a door at random, AND happens to pick a goat, you've got a 50/50 shot of having the car, whether you switch or not.

A common source of disagreement on the Monty Hall problem is because the rules don't get defined up front. This isn't a digression, it's crucial to understanding why 2/3 is the correct answer under the standard formulation of the problem.