Understanding the Monty Hall Problem

Blog Articles
821

@PalmerEldritch

+5 for the top portion

In regards to suggesting the problem was offering 2 doors to switch to, I was never saying that…I was only trying to make these other people think of why 1 of those 2 doors would be preferred over each other as a “switch” door, given that its in your best interest to switch, and the fact that his door open is a necessary step to taking 3 unknown doors and making them only 2 unknown doors.

I think because multiple things happen in this is where the concusion lies.

I was trying to say the same thing you just said. Maybe im just seeing this visually rather than mathematically.

822
  1. ‘The point I have been trying to make is that the contestant knows going into the game that one of the doors they don’t choose hides a goat, and that the host will show that door to them. Therefore no relevant information is gained when that happens, and after it happens the odds that the contestant chose the wrong door initially are still 2/3.’

This is incorrect. It is plain wrong. The MHP does NOT state that the contestant ‘knows going into the game’. The MHP states that the host reveals a goat; it does not state that the contestant is aware beforehand that this will happen. In the MHP question originally posed in American Statistician and Parade, the contestant is NOT aware beforehand that a goat will be revealed. In the actual TV show Monty used various strategies, too variably to be reliably predicted by contestants.

The reason the version at the head of this thread uses the phrase “… and always opens one of them with a goat” is to establish the rules for those putting themselves in the shoes of a hypothetical contestant in order to address the specific question at issue; this is a rhetorical device, as is the question having been worded in the first person.

Monty’s reveal of the goat is the point at which the contestant becomes aware of this new information. The same effect could be achieved by informing the contestant beforehand that this will happen, but to debate the method by which the information is given, rather than accepting the method given in the MHP, is spurious and irrelevant.

After the goat reveal occurs, the chance of the remaining door increases from 1/3 to 2/3. The reveal is essential and is material to the calculation of chances (and thus the solution). It is incorrect to suggest that to switch regardless is advantageous - Monty’s goat reveal informs the decision to switch.

The contestant does not get to swap to two doors. The only usefulness of the ‘two-door’ argument is that having been shown a goat the contestant is reminded that a 2/3 chance of car resides elsewhere than in the door they first chose. The solution to the MHP is that the 1/3 chance of the remaining door changes to 2/3 after the goat is revealed. One door, 2/3 chance. That door, that one there. Not the goat door, or the first choice door, the other one. 2/3 chance in that individual door. One door, not two.

823

Not only does Jorgensen misrepresent how the MHP is expressed, he makes inconsistent statements within the the one post when he writes (@#803), on the one hand:

“I don’t see how [people posting under multiple names] … matters.”

but on the other (in the very next paragraph):

“I’m not hiding behind a first name only, or a character’s name from a Philip Dick novel.”

So posting under multiple pseudonyms “does not matter” but using only your first name or a single pseudonym is “hiding”. Right. As if using your full name would tall you which of the many people who use that name you actually are.

This reasoning (that faking multiple IDs does not matter but using only a first name or pseudonym is hiding) is consistent with Jorgensen’s other logical inconsistency that:

Being offered to switch to one door is actually being offered to switch to both doors and the opening of the goat door reveals “nothing” or “no new information” according to Buxton and Jorgensen. In this case one has to wonder why Jorgensen et al need the goat door opened at all. Why not ask the host to leave all doors closed and instead insist on being allowed to swap to the two other doors? Probably because anyone who dug their heels in on this point would end up leaving the “studio” in a straight jacket or handcuffs.

824

In 813, I should have said the MHP is worded in the second person (not the first person). However, whether they are actually different persons or merely the same under a pseudonym …

825

“our decision: Do you want a random door out of 100 (initial guess) or the best door out of 99? Said another way, do you want 1 random chance or the best of 99 random chances?”

Now, let’s say Pitcher A is a rookie, never been tested, and Pitcher B won the “Most Valuable Player” award the last 10 years in a row. Would this change your guess? Sure thing: you’ll pick Pitcher B (with near-certainty). Your uninformed friend would still call it a 50-50 situation.

I still have a problem with the explanation. In the first example Monty is randomly eliminating choices.

In the second the elimination is not random. there is a connection between all of them, their past histories as pitchers.
So in the second example the choice is not the best out of random choices.
It is the best of choices that share a common atribute, their record as pitchers

But in the many goats one car choice the elimination of goats is purely random. there is no connection among the doors.

826
  1. ‘But in the many goats one car choice the elimination of goats is purely random. there is no connection among the doors.’

Doesn’t the ‘pitcher’ analogy relate to the contestant’s knowledge of the doors after the goat reveal? One, the ‘rookie’, has the lesser chance of the car while the other, the ‘Most Valuable Player’, has the greater chance. The other, now showing a goat, isn’t even on the team. Monty’s actions are based on insider knowledge and are not random. The original distribution of goats and cars was what was random.

827

To get it, you have to understand that MH isn’t randomly picking one of “his” two doors to show what’s there.

IF that were the case then each remaining door (your original and the one he didn’t show you) would indeed each have a probability of 1/3 of having the car so switching would only work 50% of the time if the one he showed had a goat.

But he’s not picking one of his two doors randomly. He doesn’t want to spoil the game. THAT’S THE KEY. He is always going to show you a goat and that SET of two doors that he controls retains a 2/3 probability of having the car. He’s done you a favor by eliminating a door for you and his SET of two doors (now one door you have an option to pick) is right there for you to choose.

SWITCH and you’ll win 2/3 of the time.

828
  1. ‘that SET of two doors that he controls retains a 2/3 probability’

Quite so. Please may I emphasise the following:
Step 1 - the set of doors not chosen has 2/3 chance of the car.
Step 2 - within that set, the door revealing a goat has 0/3 chance and the remaining door has 2/3 chance. ONE door = 2/3.

829

I am in math right now with Justin Bieber and we are trying to work out the answer to this problem. I’ll make sure that I DON’T FORGET this problem, but next time put them behind SKYSCRAPERS. You have a WORLD OF CHANCES if you pick to change the door.

PS - Follow @DDLovato and @DemisBuddy on Insta and Twitter

Stay Strong <3

830

Just when you thought you were out, they pull you back in, Palmer. You’ll have to turn off updates.

As for comment #832, it made perfect sense except the bit after “tommyt says”.

831

If the odds of door #3 change after Monty picks door #2, then why don’t the odds for door #1 change at the same time? In reality, you are never picking one of three because Monty is ALWAYS going to show you one of the three that is not correct. So there is a way in which you are only ever picking one of two. One of the three will ALWAYS be eliminated before the end of the game.

This makes perfect sense to me, but I cannot reconcile it with a literal trial that demonstrates it to be wrong.

832

"If the odds of door #3 change after Monty picks door #2, then why don’t the odds for door #1 change at the same time?"
Because the odds for #3 are affected by Monty’s actions, whereas he is prevented from any chance of opening #1 by the rule that the contestant’s choice is left alone.

833

How I visualised this :

We have 3 doors.

                              *                      *                   *

I pick the first one, and put the other two in brackets

                            *                      ( *                     * )

                           1/3                                2/3

Chances of being right are 1/3 for the door we picked and 2/3 for everything else which we put in brackets.

Collapsing what is inside brackets to just one door gives

                           *                                   ( * )

                          1/3                                 2/3

In essence , whatever is inside the brackets will always have a 2/3 chance of being right, be it one or multiple doors. Please correct if wrong. Thanks.

834
  1. "Therefore the probability that the car lies behind the door you haven’t picked is 2/3."
    For a moment in 822 Seph, I thought you might be applying Buxtonian logic by claiming two doors as the final solution. This clears it up: one door has 2/3 chance.
835

This is not a probability problem it is a filtering problem. Those who use the 1/3 vs. 2/3 probability explanation are flat out wrong. Here’s why.

Before opening any doors there are three possible cases:

           Door A     Door B    Door C

case#1 car goat goat
case#2 goat car goat
case#3 goat goat car

Using these cases to illustrate the point of (staying = 1/3) and (switching = 2/3) typically looks something like this:

           Door A     Door B    Door C     Stay    Switch

case#1 car goat goat car goat
case#2 goat car goat goat car
case#3 goat goat car goat car

What gets overlooked is that opening a door reduces the number of possible cases. If, for example, Door C is opened revealing a goat, case #3 is no longer valid because we now know that there is no car behind Door C. So case #3 must be eliminated from the set of possible cases, leaving only two possible cases:

           Door A     Door B    Door C     Stay    Switch

case#1 car goat goat car goat
case#2 goat car goat goat car

This clearly shows a 1/2 probability.

But then why don’t we see a 50/50 split in real life? Because it’s HOW the decision was made as to which door to open that explains the outcome, not probability.

Don’t believe me? Then perhaps if we approach this from another perspective you’ll understand.

If you are presented with 3 doors and told only that behind one door is a car and behind the other two are goats and one door is already open revealing a goat, the car probability of each of the closed doors is 1/2. The only difference between this case and our Monty Hall problem is that in Monty’s case we know something about HOW the decision was made to open the door. Probability is not concerned with HOW something has come to be, just that it exists in a certain state. The state of each of these cases is the same - two closed doors and one open door with a goat. So, as expected, the probability calculations alone yield the same 50/50 result. It’s only after one applies the HOW that we are able to explain the actual outcome. Hence this is not a probability problem and it is wrong to explain it as such.

836

As already established, when you select a door there is a 1/3 chance the car is behind the selected door and a 2/3 chance it is behind the other doors. If after my selection Monty offered a swap to the other two doors and I get to keep both prizes then I would have the following possible results if I swapped- Car/goat or goat/Car or goat/goat. I.e in two of the three options I get a car. So it will always be sensible for me to swap to the two door option as I get a car 2 out of 3 times. I’ll always get at least one goat and revealing one before Monty offers the swap does not change the probability of the car being in the two door option as I have already made my initial choice under the initial probabilities.

837
  1. “If after my selection Monty offered a swap to the other two doors …”
    … you would be playing a different game to the MHP.

Yes Seph, the much-loved hypothetical ‘swap to two doors’ scenario is indeed a useful first step in understanding the solution to the MHP. However a fuller explanation is available which quantifies the car/goat chances of the single remaining door, not the set of two doors containing it but that one specific, individual door. Would you care to have a go at it?

838
  1. "Please correct if wrong."
    No need - spot on and nicely put.
839

awesome!

840

824 - Good representation. I gave an answer in 623. To expand on my explanation in 822. The two doors not selected have a 2/3 chance of containing the car, 1/3 from each door (1/3+1/3=2/3). The combined probability will always be 2/3 until all doors are revealed. When one door is opened the probability for that door drops to 0. However the combined probability must still be 2/3. So X+0=2/3.
X=2/3. Therefore the probability that the car lies behind the door you haven’t picked is 2/3.
Going back to my original post you are ALWAYS offer a swap of your initial pick no matter how many doors there are (>1). If I had 100 doors with one car. My chance is 1/100 for success. The others doors have 99/100 chance for success. If I reveal 98 doors one at a time the remaining unopened doors increase their chances of containing the car for each opening until I get to the last door. However the total of all non-picked doors is always 99/100 but the distribution changes as the open doors drop to 0. In this instance I will have a 99/100 chance of winning the car if a always swap.