Monty Hall Problem... For Newbies

[Elaedith]

And as Lionking said, until it's properly explained...

Indeed, as I said, I know I've had it explained in the past, got it (or took it on faith...), but still, I see the problem afresh every time. How does the act of confirming something we already knew alter the likelihood that you just picked the car?

Let me further embarrass myself by showing my workings...it starts at 1 in 3, of that I'm quite sure. Three doors; X, Y & Z. A 33.33r% chance of picking the car-door, and a 100% certainty that one of the unchosen doors hides a goat. You pick door X. To nobody's surprise, there is a safe goat (behind Y, let us say, as if it mattered). Now the question becomes 'which of two doors hides the car? Is it X or Z?'. You certainly have a better chance of finding the car from a 50/50 choice than a one in 3, I see that. But why isn't just as likely to be behind the door you originally picked? 50% one door, 50% the other. An equal chance at the start that the car is behind X, and an equal chance after seeing a goat that the car is behind X.

Monty always opens a door. That happens whether you picked the car or a goat. There is always a door available to Monty that hides the goat, the safe goat. The safe goat is a red herring.

The first pick will be correct 1/3rd of the time and incorrect 2/3rds of the time.
If you picked right the first time you will always lose by switching whether you have two doors left or one.
However, if you picked wrong the first time you will win by switching 50% of the time and lose by switching 50% of the time when there are two options left. However, if one wrong option is eliminated, then you will always win by switching.

So after one wrong choice is eliminated, 2/3rds of the time it will be the correct decision to change, and 1/3rd of the time it will be the wrong decision to change.

This all requires the assumption that you see a goat on every trial and there is no special motivation to show a goat on this particular trial (which is not always made clear). If Monty only shows the goat when he knows you chose wrongly, you will always win by switching, and if he only shows it when he knows you chose correctly, you will always lose by switching.

 

[GlennB]

What I already said. He has to know, that's a given, or as you say it would make for dreadful television. I'm not a complete idiot. But his knowing surely has no bearing on the mathematics. Whatever you pick, he knows where to find a safe goat. You know there will always be a safe goat. You are shown a goat, which we all know to be safe. No surprises there. There was a 100% certainty that he could, and would, show you a goat. (I wouldn't, by the way, claim any of this was great television, but we're in the wrong sub-forum for that discussion).

As you said earlier, there was a 1/3 chance of your first choice being the car. That's still true. It means also that there was 2/3 chance of the car being behind one of the other 2 doors. When Monty eliminates a goaty-door that 2/3 chance resides in the one remaining door.

 

[p0lka]

As I understand it (explained by Derren Brown) You start with a car and two goats (1:3 car - 2:3 goat) you pick one, let's say door 1 and are left holding your choice (1:3) while MH opens door 3 and shows a goat. If you maintain your original choice, door 1 (1:3) your odds are still as they were from the outset. However, if you swap, you are now picking from a 1:2 or fifty - fifty option, thereby improving your chances of winning the car.

I think!

I must be thinking about it wrong, I treat it as the odds are always as they were from the outset.

If you maintain your original choice, door 1 (1:3 that its the car) then its 2:3 that one of the other two doors is the car,
once you have seen the goat behind one of the other two doors, then its still 2:3 that one of the other two doors is the car, but there's only one door left to swap to.
so by swapping, you are swapping from a position of 1:3 that its the car to a position of 2:3 that its a car, making it more likely to get the car by swapping than not swapping, and not fifty-fifty as you got.

EDIT:
Oh, more people posted as I was typing which I have just read, so never mind this post.

 

[Zax63]

A friend of my mine came up with this way to help explain the answer:
Imagine there are 100 doors instead of 3 and still 1 prize. You make a choice which is obviously 1% chance to be correct. If you could switch at this point and take the all of the other 99 doors or keep your original choice you wouldn't hesitate to switch. The host, with knowledge of where the prize is, can reveal 98 goats and you are back to 2 doors and the choice to stay or switch. Your original choice is still 1% to win and switching is 99%. While you seem to be being offered a choice between 2 doors at the end, you are in effect being offered the choice between 1 door(your original choice) and all of the other doors originally available.

In the 3 door case of the problem it is much harder to see through the illusion of a 50/50 chance in picking between the 2 final doors.

 

[One Tenth]

Yeah that's the explanation I go with when people don't get it. Or imagine if Monty didn't reveal the goat, and after you picked door A, he said "Okay, you can stick with door A, or you can open BOTH door B and door C."

 

[GlennB]

Yeah that's the explanation I go with when people don't get it. Or imagine if Monty didn't reveal the goat, and after you picked door A, he said "Okay, you can stick with door A, or you can open BOTH door B and door C."

That's a good way to look at it. And on the actual show all Monty is doing is tossing out a guaranteed loser.

 

[FenerFan]

For me, the solution becomes much more easier to understand and explain if you start with a number of doors larger than 3. Lets's say 100 doors. You pick your door and obviously have a 1/100 chance of picking the car. Monty then opens 98 doors each of which has a goat. Now you are left with your door and the door with the car. Now it's 50-50 right? Clearly you would change your door. Monty's knowledge does change the odds because he is always going to leave you with a goat door and a car door. The odds change from1/3 to 1/2. What about a million doors? 1/1000000 to 1/2. As long as Monty knows where the car is, you should change your door.

(Didn't see the Zax63 post.)

 

[welshdean]

I must be thinking about it wrong, I treat it as the odds are always as they were from the outset.

If you maintain your original choice, door 1 (1:3 that its the car) then its 2:3 that one of the other two doors is the car,
once you have seen the goat behind one of the other two doors, then its still 2:3 that one of the other two doors is the car, but there's only one door left to swap to.
so by swapping, you are swapping from a position of 1:3 that its the car to a position of 2:3 that its a car, making it more likely to get the car by swapping than not swapping, and not fifty-fifty as you got.

EDIT:
Oh, more people posted as I was typing which I have just read, so never mind this post.

Thanks, I know I understood the concept, it's the fact that I'm. ahem, 'hard of counting' that let me down.

 

[jiggeryqua]

But after your original pick (indeed, even before it), we know that one of the doors you haven't picked hides a goat. Seeing the goat changes nothing. Your first pick is 1 in 3, and yes, I get that that means it is more likely that the car is behind a door you didn't pick. What I don't see is why the presence of a guaranteed goat shifts the odds on one of the remaining doors and not the other. What distinguishes the two remaining doors, that one of them should now be more likely to hide a car than before an event that must be possible and in no way comes as a surprise. One of the doors you didn't choose hides a goat - we always knew that. Why does it then make only one of the other doors more likely to hide a car? Why doesn't the 2/3 shift to the door you picked? Your choice cannot have affected the odds...

And for the record, the hypothetical 'you could open B and C!' hasn't helped (it isn't part of the problem - we know there's a goat, we see a goat, we merely confirm what we already knew. I do get that a choice to open two doors improves your chances, but seeing a safe goat isn't the same as opening both of them - the odds you'll see two goats remain the same) Imagining 100 doors doesn't help either. It doesn't matter (to me, at present) how many doors you start with, the question is why the odds shift to one of the doors and not the other?

 

[jiggeryqua]

Now it's 50-50 right? Clearly you would change your door.

Hold on, haven't I just been assured it's not 50/50? Even if it were, why would you 'clearly' change your 50/50 choice of two things?

I await Lionking's correct explanation...

 

[jsfisher]

How can it matter whether it's SOP? In this instance, it happens, regardless of how Monty may have played it in the past. It surely can't affect the odds whether he does it on a whim or to a script?

Without the parenthetical "this is standard procedure for the show", you'd have the situation where the host didn't have to open a door, but he did. His rule could have been that if you picked the door with the car then he would open one of the other two doors, otherwise he'd leave things as they were.

If you knew that that was his rule, you'd be a fool to switch (unless you had a thing for goats).

 

[Cuddles]

Once we vote, will you reveal one of the incorrect answers and give us the chance to change our vote?

But after your original pick (indeed, even before it), we know that one of the doors you haven't picked hides a goat. Seeing the goat changes nothing. Your first pick is 1 in 3, and yes, I get that that means it is more likely that the car is behind a door you didn't pick. What I don't see is why the presence of a guaranteed goat shifts the odds on one of the remaining doors and not the other. What distinguishes the two remaining doors, that one of them should now be more likely to hide a car than before an event that must be possible and in no way comes as a surprise. One of the doors you didn't choose hides a goat - we always knew that. Why does it then make only one of the other doors more likely to hide a car? Why doesn't the 2/3 shift to the door you picked? Your choice cannot have affected the odds...

Try a variation of the game. Instead of three separate compartments with three separate doors, you have one large area partitioned into three sections with removable walls. After you pick one section, the partition between the two other sections is removed. You now have the choice of sticking with your original section, or changing your pick to the other section which now contains two prizes and therefore has twice as much of winning.

 

[wollery]

Let's work through the possibilities. There are 3 doors, A B and C. The car is behind door A. If you pick door A then Monty can open either door B or door C.

A B C
Car Goat Goat

Pick A. B opened. Change to C. Lose.
Pick A. C opened. Change to B. Lose.
Pick B. C opened. Change to A. Win.
Pick C. B opened. Change to A. Win.

Pick A. B opened. Stick with A. Win.
Pick A. C opened. Stick with A. Win.
Pick B. C opened. Stick with B. Lose.
Pick C. B opened. Stick with C. Lose.

Now from that it would appear that it makes no difference if you stick or change, but it's slightly misleading. Having picked door A you don't care which of door B or C is opened, so actually the first 2 possibilities reduce to one possibility.

We therefore have

Pick A. Change. Lose.
Pick B. Change (to A). Win.
Pick C. Change (to A.) Win.

Pick A. Stick with A. Win.
Pick B. Stick with B. Lose.
Pick C. Stick with C. Lose.

This means that changing gives you a 2/3 chance of winning, while sticking gives you a 1/3 chance of winning.

 

[phunk]

Either way, he knows where one goat is and shows you one goat. It's a math's question, not a body language question - he might be avoiding the car, he might merely be picking one of the two available goats, there's no way of telling.

It doesn't matter which of those options it is. As long as we know he won't show the car accidentally, the problem remains the same.

 

[bonzombiekitty]

The easiest way of explaining the monty hall problem to people who have trouble understanding is to use a million doors instead of 3. I find that people grasp it better when you can say "If you picked the right door at the beginning, you win if you stay with the door you chose. When you chose a door, you had a 1 in a million chance of being right..."

Something about it seems to make it click in people's heads

edit: Fener has already mentioned this.

 

[bonzombiekitty]

edit: nevermind

 

[epeeist]

I get how if you just stay your chance is 1/3, and if you switch it's 2/3.

But now, let's say when you have to decide whether to stay or switch, you flip a coin. Heads you switch, tails you stay. That is, you are now choosing between 2 choices, 50/50, even if what you end up picking happens to be the same pick you had made originally? Or is it still 1/3 and 2/3?

 

[jsfisher]

A friend of my mine came up with this way to help explain the answer:
Imagine there are 100 doors instead of 3 and still 1 prize. You make a choice which is obviously 1% chance to be correct. If you could switch at this point and take the all of the other 99 doors or keep your original choice you wouldn't hesitate to switch. The host, with knowledge of where the prize is, can reveal 98 goats and you are back to 2 doors and the choice to stay or switch. Your original choice is still 1% to win and switching is 99%. While you seem to be being offered a choice between 2 doors at the end, you are in effect being offered the choice between 1 door(your original choice) and all of the other doors originally available.

In the 3 door case of the problem it is much harder to see through the illusion of a 50/50 chance in picking between the 2 final doors.

Another way to look at it is to not have Monty open any doors at all. Instead, after picking the first door, Monty gives you the option of taking both of the other two doors instead of the one you selected.

You already know at least one of the two doors has a goat behind it, and so Monty showing you a goat changes nothing. 33% for the car if you stay with the one; 67% if you switch to the two.

 

[Soapy Sam]

Hold on, haven't I just been assured it's not 50/50? Even if it were, why would you 'clearly' change your 50/50 choice of two things?

I await Lionking's correct explanation...

Call the doors ABC.
I'm Monty. I know the car is behind C.
Scenario 1.
You choose A - a goat.
I have only one choice, which is to open B- goat 2.
You are left with a choice- Stick or change. If you stick, you lose.

Scenario 2.
You choose C, the car.
I now have a choice. I can show you A or B, both goats.
I choose A.
You must now choose to stick, in which case you win, or change and lose.

Scenario 3.
You choose B - a goat.
For me, this is functionally the same as scenario 1.
I can only open one door- in this case A.
You now have to choose Stick or change. If you stick, you lose.

These are the only three possible scenarios and in two out of three you lose if you stick and win if you swap.

That's really it.

ETA Woops! Wollery just beat me to it.

 

[jsfisher]

I get how if you just stay your chance is 1/3, and if you switch it's 2/3.

But now, let's say when you have to decide whether to stay or switch, you flip a coin. Heads you switch, tails you stay. That is, you are now choosing between 2 choices, 50/50, even if what you end up picking happens to be the same pick you had made originally? Or is it still 1/3 and 2/3?

Ok, so there's 1/3 chance the car is behind Door A and 2/3 behind Door B, and you are going to select A or B based on a 50/50 event.

So, 1/2 chance you stay with Door A; 1/3 chance there's a car. 1/2 * 1/3 = 1/6.

1/2 chance you switch to Door B; 2/3 chance there's a car. 1/2 * 2/3 = 2/6.

The combined result from your coin flip, then, is 1/6 + 2/6 = 1/2.

The conclusion: relying on a coin flip improves your chances form 1/3 to 1/2 over just staying with your first choice, but your best option is to always switch.