MORE Riddles a la Batman

[katanabob]

If a brick weighs a pound and a half plus half a brick, how much does a brick and a half weigh?

 

[TonyC]

If a brick weighs a pound and a half plus half a brick, how much does a brick and a half weigh?

B = 1.5# + ½B
B - ½B = 1.5# + ½B - ½B
½B = 1.5#
2(½B) = 2(1.5#)
B = 3#

a brick and a half???

1½B = 1.5(3#) = 4 and a half pounds

 

[DCitrus9]

Tony, you get a gold star for showing your work. Your teachers must be so proud!

 

[Snoopy58]

An oldie but still a goody...

The show is "Let's Make A Deal." The host tells you that behind one of the three curtains is a new car, while behind the other two are billy goats. Not being a farmer, you're hoping for the car, of course. So, the host asks you to choose which door you want, and you name one of the three. He then has his off stage assistant open one of the other doors, revealing a goat. With great dramatic fanfare, he asks if you'd like to switch your choice.

Would switching your choice improve your odds of getting the car? Why or why not?

And, along a similar line...

You're taking a multiple choice test. You get to question 17, and you have utterly no clue which of the three answers is correct. No idea at all. So you make a guess that it's "A." Just after you've marked your answer, the teacher announces that there is a misprint on some of the tests, and, just to make things equal for everyone, on question 17, "C" is NOT the correct answer.

Now, would erasing your answer and choosing "B" improve your odds of getting the right answer to question 17? Why or why not?

And finally, did you answer the same thing to both of the above questions? If not, how can you justify the difference?

Cheers!

 

[VNugget]

And the gameshow situation is different from the test, because you are changing your choice from one that you just learned is wrong. Your chances went from 1/3 to 1/2. On the test, the chance went from 1/3 to 1/2, but you changing your answer from one possibility to another has nothing to do with the eliminated answer.

 

[Snoopy58]

Just to clairify, in case I wasn't clear:

In both cases, you make your choice, one of the options you DIDN'T chose is revealed to be an incorrect choice, and you have the option to stay with your first answer or switch to the remaining option. In neither case did you learn anything about the option you chose.

Nugget, I don't follow your answer here; could you explain further?

 

[TonyC]

I'll take a stab...

First, the second:

In this case, the teacher proctoring the test announces that "C" is not the answer. This advice is offered unsolicited and without any apparent bias. It would appear to leave, then, just two answers, and neither has any prejudice attached. Consequently, the odds of choosing the correct answer when choosing between "A" and "B" are even, 50/50, or 1 in 2. Changing the answer from "A" to "B" would NOT change the odds of either random choice being correct.

Such randomness and freedom from prejudice does NOT appear to exist in the first scenario. The setting is a game show focused on profit, concerned not only with ratings but a bottom line. Furthermore, the first choice IS known by the host and the stage assistant who elect to "help" you make a decision. Perhaps the idea is to prolong the segment, therefore the decision to reveal one of the "goat" doors that was not chosen. Or perhaps, the idea is to persuade you to change your choice because the first choice was correct.

IF the first choice was INCORRECT, the host could have ended this segment immediately by revealing the chosen door and sending the hapless contestant home with a goat and a headache. Granted, the host could also have revealed the other goat and left us on pins and needles of anticipation for another brief moment, but we just don't know which strategy he has employed. Can we look at the clock?

IF the first choice was CORRECT, the host could have attempted to lure the contestant into changing his choice by distracting him with an incorrect door. Doing so would serve the dual purpose of prolonging the airtime for this particular segment, heightening the anticipation of the audience, increasing the ratings of the show, and, perhaps, enticing the contestant into sparing the advertisers the expense of a new car.

SOooooo taking into account the unknowns and using a personal interpretation of the biases involved, I would be inclined to answer (would switching improve odds?) NO. In fact, switching might only DECREASE the odds.

In both scenarios, an incorrect choice is eliminated. The difference, of course, is the bias involved in the first scenario which does not exist in the second. In the second, we deal purely with statistical probablitites. In the first, we deal with the bias of the host and his ability to manipulate information supplied, and our own best efforts to divine his intentions. The second is purely mathematics, the first invovles [EDIT: involves] human nature.

 

[VNugget]

Doh, forget what I said; I misread the first paragraph. I thought that he opened YOUR door and showed you a wrong answer, and then gave you a second chance.

I take back what I said. It would be pointless to change your choice in either case. The analogy holds up.

 

[Snoopy58]

Tony,

Let me remove a little bit of the psychology from the Gameshow question, and put it back into the realm of probability/statistics:

You (the contestant) has seen the show on TV enough to know that, when this game is played, EVERY contestant is shown the same thing: you picked door X, so I'll open door Y, what could it be? It's a ... (dramatic pause) IT'S A GOAT! Whew! Sigh of relief, I haven't lost the car yet. Now, do you want to switch to door Z, or stay with door X?

Substitute 1,2, or 3 as applicable for the X,Y, and Z, but the host ALWAYS opens a door you didn't choose & reveals a goat. That's what he does. So when he reveals the goat for you, it's no great surprise (except for the viewers who don't watch the show often enough to have known what was coming next). You also know that he NEVER fails to offer the choice, NEVER reveals the car at this point, ALWAYS opens a door you DIDN'T choose, and has, in fact, offered the choice both to players who'd picked the car initially, and to those who hadn't. He reveals a goat & offers an option to switch to EVERY contestant who plays this game. The host doesn't care if you get a car or a goat -- the prizes are donated, and editing takes care of the time element for the show. (Did I dismiss all the human nature issues & get this back into the realm of prob/stat? It's intended as a prob/stat question, not one of psychology, really!)

Now, you're on the stage with that long thin microphone in front of you: do you want to switch from the door you picked to the other door on stage? As a mathmatical choice!

Snoopy

 

[VNugget]

1/2 chance in any case.

And I'll stick to that

 

[TonyC]

Tony,

Let me remove a little bit of the psychology from the Gameshow question, and put it back into the realm of probability/statistics:

You (the contestant) has seen the show on TV enough to know that, when this game is played, EVERY contestant is shown the same thing: you picked door X, so I'll open door Y, what could it be? It's a ... (dramatic pause) IT'S A GOAT! Whew! Sigh of relief, I haven't lost the car yet. Now, do you want to switch to door Z, or stay with door X?

Substitute 1,2, or 3 as applicable for the X,Y, and Z, but the host ALWAYS opens a door you didn't choose & reveals a goat. That's what he does. So when he reveals the goat for you, it's no great surprise (except for the viewers who don't watch the show often enough to have known what was coming next). You also know that he NEVER fails to offer the choice, NEVER reveals the car at this point, ALWAYS opens a door you DIDN'T choose, and has, in fact, offered the choice both to players who'd picked the car initially, and to those who hadn't. He reveals a goat & offers an option to switch to EVERY contestant who plays this game. The host doesn't care if you get a car or a goat -- the prizes are donated, and editing takes care of the time element for the show. (Did I dismiss all the human nature issues & get this back into the realm of prob/stat? It's intended as a prob/stat question, not one of psychology, really!)

Now, you're on the stage with that long thin microphone in front of you: do you want to switch from the door you picked to the other door on stage? As a mathmatical choice!

Snoopy

Cool... remove the psychology, make it pure math - - I can dig it.

I'll stick with curtain A - - changing to B will have no effect on the odds, just like the second example you gave (#17 on the test).

 

[TonyC]

... and by the way

If that's the way the game is ALWAYS played, then the odds were 1 out of 2 all along.

 

[Snoopy58]

Tony, you lost me with your last comment.

If you pick door 1, the car could be behind 1, or 2, or 3. Your odds are 1 in 3. What he shows you doesn't improve your odds if you don't switch, and you've said that you don't believe switching improves your odds. So they're still 1 in 3.

Now, let me suggest something to you that may change your mind about the value of switching.

Suppose that, instead of 3 curtains, the gameshow has 100 mailboxes. Inside of one mailbox are the keys to the new car, while the other 99 have a length of rope to slip around the goat's neck. If you pick the mailbox with the keys, you get the car, otherwise, the goat.

Now, you walk out on stage (never having seen this game before, but knowing that the gameshow has a reputation for being PURELY MATH-BASED & no mind-games) & you're asked which mailbox you want to choose. You pick a favorite number, 42, and hope for the best. Looking at all the mailboxes, you feel like your odds aren't very good.

Now, however, the host starts opening mailboxes one at a time: #1, has a rope; #2, a rope; #3, a rope, and so on down the line. He gets to #42, and doesn't open it, since you have picked it, but keeps on going, until he reaches #58. He also leaves that one closed as well, but opens #59 (another rope), and keeps going. A couple minutes later, you're staring at 98 open mailboxes, all with ropes inside them, and only mailboxes #42 (the one you chose) and #58 (the other one he skipped) unopened. The host assures you that behind those 2 remaining curtains, I mean mailboxes, are one set of keys, and one rope.

He now asks you if you'd like to switch your pick, and take what's in mailbox #58 instead. Of course, if you were right in your 1 in 100 guess initially, switching would mean losing the car. However, if you had originally picked one of the 99 ropes, switching will get you the car.

What will you do? Is there value in switching your selection?

 

[TonyC]

Probabilities

Tony, you lost me with your last comment.

If you pick door 1, the car could be behind 1, or 2, or 3. Your odds are 1 in 3. What he shows you doesn't improve your odds if you don't switch, and you've said that you don't believe switching improves your odds. So they're still 1 in 3.

Sorry, didn't mean to lose you. I'll try to explain.

What you described is a situation where everybody knows from the start that we will ultimately be looking at 2 doors (or curtains) as one of the original three will always be opened revealing a goat. Consequently, we know that the REAL choice will be between 2 doors, even though there initially appears to be 3. Since the revealing of the first goat is simply drama, and we know it will always occur, we have a 50% probability that the door we choose will have a car, and a 50% probability that the remaining door (after the drama exercise) will have the car. No matter which we choose, the odds are 1 in 2. The third door is simply noise to confuse the issue.

Think of it this way. Look at a single door. What could they put behind that door? Well, we have a car -- that's one thing. And they have a goat - - that makes 2. And they have another goat - - that makes three. So we have a 1 in three chance that the door will have a car, right? Wrong. There can't be a third goat, because we already know that the third goat will be sacrificed to drama when one door is opened. Consequently, there are only 2 possibilities behind the door we choose - - car and goat. So the odds are 1 in 2. Don't think of it as the number of doors. Think of it as the number of possible results when the door is opened.

Now, let me suggest something to you that may change your mind about the value of switching.

Suppose that, instead of 3 curtains, the gameshow has 100 mailboxes. Inside of one mailbox are the keys to the new car, while the other 99 have a length of rope to slip around the goat's neck. If you pick the mailbox with the keys, you get the car, otherwise, the goat.

Now, you walk out on stage (never having seen this game before, but knowing that the gameshow has a reputation for being PURELY MATH-BASED & no mind-games) & you're asked which mailbox you want to choose. You pick a favorite number, 42, and hope for the best. Looking at all the mailboxes, you feel like your odds aren't very good.

Now, however, the host starts opening mailboxes one at a time: #1, has a rope; #2, a rope; #3, a rope, and so on down the line. He gets to #42, and doesn't open it, since you have picked it, but keeps on going, until he reaches #58. He also leaves that one closed as well, but opens #59 (another rope), and keeps going. A couple minutes later, you're staring at 98 open mailboxes, all with ropes inside them, and only mailboxes #42 (the one you chose) and #58 (the other one he skipped) unopened. The host assures you that behind those 2 remaining curtains, I mean mailboxes, are one set of keys, and one rope.

He now asks you if you'd like to switch your pick, and take what's in mailbox #58 instead. Of course, if you were right in your 1 in 100 guess initially, switching would mean losing the car. However, if you had originally picked one of the 99 ropes, switching will get you the car.

What will you do? Is there value in switching your selection?

Interesting scenario, but it boils down to the same thing. 3 or 100, or a million, if we know that ultimately we will ALWAYS be looking at 2 choices, with 50% probability of success, then our odds are the same: 1 in 2. If all but 2 of the incorrect choices will ALWAYS be removed before the final choice, then we always have 1 in 2 odds, regardless of how tempting it may be to switch. Before the ceremonial revealing of 98 boxes, #42 and #58 had an equal chance of being the correct box. After the revealing they still have an equal chance. Switching the choice doesn't change that.

Again, consider the possible outcomes of opening the mailbox. Since we know that 98 of the ropes will be revealed to increase the drama, we also know that there are only 2 outcomes of opening THIS mailbox - - one set of keys, and one rope.

 

[Snoopy58]

Tony, I'm going to disagree with you here, and to do so, I'm going to change the original question in a semantic way: behind the 3 curtains, there are a car, a goat, and a sheep. After the contestant makes his choice, one curtain is opened, revealing an animal. Next you're asked if you want to switch.

The possible setups are as follows (C=car, S=sheep, G= goat, and the number represents the door that prize is behind).

C1 S2 G3 or, expressed differently, 1=C, 2=S, 3=G
C1 S3 G2 or, expressed differently, 1=C, 2=G, 3=S
C2 S1 G3 or, expressed differently, 1=S, 2=C, 3=G
C2 S3 G1 or, expressed differently, 1=G, 2=C, 3=S
C3 S1 G2 or, expressed differently, 1=S, 2=G, 3=C
C3 S2 G1 or, expressed differently, 1=G, 2=S, 3=C

Now, can we agree that with a random setup, all 6 of the above are equally likely, and are the only possible distributions of prizes?

Now, the game begins. You pick, say, door #1.

IF YOU DON'T SWITCH: 2 of the 6 times you will win, 4 of the 6 times you will get a farm animal (whether you can RECOGNIZE the difference between a sheep & a goat in immaterial -- with either one, you lose the car). Your odds of winning are 1 in 3. If you're not going to switch, the whole "here's door # __, do you want to switch, blah blah blah" is only drama that doesn't effect the outcome, as you point out.
Same odds if you pick any other door.

IF YOU WILL SWITCH, the process becomes more interesting, because we have to consider what is revealed.

C1 S2 G3 Pick door 1, door 2 (S) revealed, switch to 3, and lose
or, Pick door 1, door 3 (G) revealed, switch to 2, and lose
C1 S3 G2 Pick door 1, door 2 (G) revealed, switch to 3, and lose
or, Pick door 1, door 3 revealed, switch to 2, and lose
C2 S1 G3 Pick door 1, door 3 (G) revealed, switch to 2, and win
C2 S3 G1 Pick door 1, door 3 (S) revealed, switch to 2, and win
C3 S1 G2 Pick door 1, door 2 (G) revealed, switch to 3, and win
C3 S2 G1 Pick door 1, door 2 (S) revealed, switch to 3, and win

Looks like there are 4 ways to win, and 4 ways to lose, yes? But those 8 possibilities don't occur equally often! Of the 6 equally likely initial setups, if you switch, 2 of the setups will result in a loss (i.e. the 2 times out of 6 that you initially picked the door with the car behind it, even though each one can play out different ways depending on which animal the host reveals), but the other 4 times you will WIN if you switch (i.e. the host has no option which animal to reveal).

Again, the math will work the same if you pick door #2 initially, or door #3. Don't switch, you have your initial 1 in 3 chance of winning. Switch, and you have 2/3 chance of winning.

Similarly, in the 100 mailboxes game, the chance that your #42 (or whatever box you pick) had the keys in it was 1 in 100. If there had been no "reveal some that you didn't pick & ask if you want to switch" and they had just opened the box you picked, your odds of winning are 1 in 100. The fact that the host KNOWS which mailbox to skip over is pretty strong evidence that the ONLY ONE of 99 that he could have opened but didn't, probably contains the keys! Of course, in the 1 time in 100 that you made the lucky guess initially, he can skip over anything, and if you switch you'll lose. But, the rest of the time, switching will mean you win.

There's a joke about "all odds are 50/50 -- either it happens or it doesn't." Cute, but not true, since not all outcomes aren't necessarily equally likely... if there are 99 ways to get the rope and only 1 way to get the keys, the fact that there are only 2 outcomes doesn't make the chances that that's what you'll get 1 in 2!

Yes, this puzzle is a mindbender, but switching is a good thing, BECAUSE the host is giving you information in his "reveal the goat (or sheep)" act that you can act on when you make the switch, but which you DIDN'T HAVE with your first choice.

Cheers!

 

[TonyC]

Probability

Yes, this puzzle is a mindbender, but switching is a good thing, BECAUSE the host is giving you information in his "reveal the goat (or sheep)" act that you can act on when you make the switch, but which you DIDN'T HAVE with your first choice.

Cheers!

The tricky thing is, though, the host is not giving us any new information. We knew all along that he would reveal an incorrect guess. No news.

It seems to me the math you've used is right on the money when used to describe two random, INDEPENDENT events. In the two scenarios you described, though, the revelation of wrong choices is neither random nor independent. If so, there would always be the possibility of the host opening one of the doors you didn't choose and revealing the car (or the keys).

Before you ever make a choice you KNOW that hands will be waved and curtains/doors will be opened and you will see but 2 doors remaining, one with the correct choice, and one with an incorrect choice. No matter what the setup, no matter what your initial choice, you will always have but 2 choices. No news.

That's my opinion...

 

[dash8driver]

when you are first presented the choice between the 3 curtains, you have a 1 in 3 chance of getting it correct. but when he reveals one and asks if you want to switch, you have a new choice with new odds. the act of changing your curtain wont increase your odds... you have a new decision to make, and that is now between 2 curtains. if you chose to switch you have a 50% chance of being right, if you chose to stay you have a 50% chance of being right.

the 3 curtain choice becomes moot as you now have a new decision...either choice you make is a decision that carries equal odds.

though its not the same this reminds me of the "stats change odds" concept of gambling. if you flip a coin 9 times and it comes up heads, what will it be when you flip it a 10th time. well, statistically it should be tails, becuase of the outcomes over time are supposed to be roughly equal... however, at the time of the flip, a brand new flip with brand new options...the odds remain 50/50. for some reason, an ex of mine used to make money at the roulette table using this technique.. "well, its been (black, even, high, red, odd, low) 5 times in a row, its bound to be (insert opposite here) next.

 

[Snoopy58]

The odds in the original game are NEVER 50/50 for you!

At the outset, you have a 1/3 chance of picking the correct door. If the contestant makes his pick & never gets any more information or an offer to switch but simply is given what's behind the door he picked, the contestant will win 1 time in 3.

Now, if he sees what is behind one of the doors that he didn't pick but doesn't alter his choice, nothing has changed. His odds are still 1 in 3.

HOWEVER: after one door has been opened revealing a goat, there are now 2 options: one wins, the other loses. Holding on to the original choice is still a 1/3 chance, so WHAT ARE THE ODDS THAT THE OTHER DOOR IS A WINNER? 2 in 3!!!!!

Once more, here are the possibilities:

Car behind door 1; contestant (initially) picks 1; switch = LOSE (door 2 or 3 revealed, doesn't matter)
Car behind door 2; contestant picks 1; switch = WIN (door 3 revealed)
Car behind door 3; contestant picks 1; switch = WIN (door 2 revealed)
Car behind door 1; contestant picks 2; switch = WIN (door 3 revealed)
Car behind door 2; contestant picks 2; switch = LOSE (door 1 or 3 revealed, matters not)
Car behind door 3; contestant picks 2; switch = WIN (door 1 revealed)
Car behind door 1; contestant picks 3; switch = WIN (door 2 revealed)
Car behind door 2; contestant picks 3; switch = WIN (door 1 revealed)
Car behind door 3; contestant picks 3; switch = LOSE (door 1 or 2 revealed, machts nichts)

Those 9 situations are the only ways the game can play out: the car is behind one of 3 doors, and the contestant picks one of 3 doors initially. Those 2 events are independent of each other, so each of the 9 outcomes is equally likely. In 3 of the 9 outcomes, switching will be a losing choice, and staying will be a winning choice. In the other 6, staying will be a losing choice, and switching will be a winning choice. The odds of winning after switching are 2/3. Questions?

Tony, the host is absolutely giving us new information. He is revealing (i.e. eliminating) a losing door! The fact that we know that he WILL tell us something isn't the same thing at all as knowing WHAT he will tell us! Knowing that "some stocks will go up on Monday" is far, far less useful than knowing WHICH stocks will go up! Knowing that, in the "100 mailboxes" version, the host will eliminate 98 wrong choices, isn't AT ALL the same thing as knowing WHICH 98 are wrong choices!

Consider that 100 mailboxes game -- at the outset (not knowing what will happen, but only that the game you're playing is about math & not mindgames) you have a 1 in 100 chance of picking the mailbox with the keys... staying with that choice that you initially made is NEVER BETTER than a 1 in 100 deal! After the host opens up 98 of the mailboxes, unless you got the 1 in 100 by a lucky guess, you're almost certainly wrong about your pick, but now the host has given you a world of information... in 1 time in 100, staying with your choice is a winning option; the other 99 times... it would be better to switch! The host isn't opening mailboxes randomly -- he KNOWS which one has the keys, and hence which one he CAN'T open!

Tony, as to your "random" and "independent" thoughts, where the car/keys are is random, as is the contestant's initial guess, and the two are independent. What the host reveals is NOT AT ALL random, nor independent... it is utterly determined by the above two variables! (If he WERE opening doors at random, there would be plenty of times he'd reveal the car & there'd be no opportunity to offer a switch... a situation I specifically excluded {"you've seen the show enough to know..."} in my first response to you, and also in my response to VNugget before that.)

"You have but two choices" is correct. That is not at all the same as saying that your two choices are equally likely to produce a winner!

 

[dash8driver]

edited out my response because i have to rethink my response before i put it out there.

 

[TonyC]

At the outset, you have a 1/3 chance of picking the correct door.

If there were a possibility of the game ending after the first choice, this would be true. However, we know that there will ALWAYS be a "Round 2," an event that is dependent on the first choice.

HOWEVER: after one door has been opened revealing a goat, there are now 2 options: one wins, the other loses. Holding on to the original choice is still a 1/3 chance, so WHAT ARE THE ODDS THAT THE OTHER DOOR IS A WINNER? 2 in 3!!!!!

When there are 2 choices before you, and only two possible outcomes of either choice, how can either have a probability of success of 2/3?!?!

Let's refer to the concept dash8driver mentioned. What are the odds of 10 consecutive coin flips resulting in 10 consecutive "heads"? There's a mathematical way of computing the odds = > (½ * ½ * ½ * ½ * ½ * ½ * ½ * ½ * ½ * ½) or 1 in 1024 (That's the probablity of a single event raised to the tenth power.) So, let's consider that we've just witnessed 9 consecutive "heads". WOW. What incredible odds have just been beaten! (½)^9 or 1 in 512. Incredible. What are the odds that this next coin flip will be "heads"? ½ to the tenth power? NO. It's simply ½. The single coin flip is an independent event, and bears the odds of the single coin flip.

The odds of 10 consecutive, random, independent events resulting in 10 consecutive "heads" is 1 in 1024. The probablity of each individual coin flip, each random, independent event, is always 1 in 2.

Once more, here are the possibilities:

Car behind door 1; contestant (initially) picks 1; switch = LOSE (door 2 or 3 revealed, doesn't matter)
[this is 2 situations and should be counted as such]
Car behind door 2; contestant picks 1; switch = WIN (door 3 revealed)
Car behind door 3; contestant picks 1; switch = WIN (door 2 revealed)
Car behind door 1; contestant picks 2; switch = WIN (door 3 revealed)
Car behind door 2; contestant picks 2; switch = LOSE (door 1 or 3 revealed, matters not)
[again, this is 2 situations and should be counted as such]
Car behind door 3; contestant picks 2; switch = WIN (door 1 revealed)
Car behind door 1; contestant picks 3; switch = WIN (door 2 revealed)
Car behind door 2; contestant picks 3; switch = WIN (door 1 revealed)
Car behind door 3; contestant picks 3; switch = LOSE (door 1 or 2 revealed, machts nichts)
[once again, this is 2 situations and should be counted as such]


Those 9 [12] situations are the only ways the game can play out: the car is behind one of 3 doors, and the contestant picks one of 3 doors initially. Those 2 events are independent of each other, ...

First, there are 12 situations, and only half of them result in a win.

The second problem is they're NOT random, not independent of each other. The first event (the initial choice) dictates the second event. Otherwise, the host would have the option of revealing the contestant's initial choice, revealing either a correct choice or an incorrect choice. Also, the host might reveal the car behind a door that has not been chosen. You've excluded those two options in this game, therefore the second event is NOT independent of the first. It is ENTIRELY dependent. It is NOT random.

Here's how 2 INDEPENDENT, RANDOM events might play out:

CAR 1, Pick 1, Reveal 1, AUTOMATIC = WIN
CAR 1, Pick 1, Reveal 2, NO SWITCH = WIN
CAR 1, Pick 1, Reveal 3, NO SWITCH = WIN
CAR 1, Pick 2, Reveal 1, AUTOMATIC = LOOSE
CAR 1, Pick 2, Reveal 2, AUTOMATIC = LOOSE
CAR 1, Pick 2, Reveal 3, NO SWITCH = LOOSE
CAR 1, Pick 3, Reveal 1, AUTOMATIC = LOOSE
CAR 1, Pick 3, Reveal 2, NO SWITCH = LOOSE
CAR 1, Pick 3, Reveal 3, AUTOMATIC = LOOSE
CAR 2, Pick 1, Reveal 1, AUTOMATIC = LOOSE
CAR 2, Pick 1, Reveal 2, AUTOMATIC = LOOSE
CAR 2, Pick 1, Reveal 3, NO SWITCH = LOOSE
CAR 2, Pick 2, Reveal 1, NO SWITCH = WIN
CAR 2, Pick 2, Reveal 2, AUTOMATIC = WIN
CAR 2, Pick 2, Reveal 3, NO SWITCH = WIN
CAR 2, Pick 3, Reveal 1, NO SWITCH = LOOSE
CAR 2, Pick 3, Reveal 2, AUTOMATIC = LOOSE
CAR 2, Pick 3, Reveal 3, AUTOMATIC = LOOSE
CAR 3, Pick 1, Reveal 1, AUTOMATIC = LOOSE
CAR 3, Pick 1, Reveal 2, NO SWITCH = LOOSE
CAR 3, Pick 1, Reveal 3, AUTOMATIC = LOOSE
CAR 3, Pick 2, Reveal 1, NO SWITCH = LOOSE
CAR 3, Pick 2, Reveal 2, AUTOMATIC = LOOSE
CAR 3, Pick 2, Reveal 3, AUTOMATIC = LOOSE
CAR 3, Pick 3, Reveal 1, NO SWITCH = WIN
CAR 3, Pick 3, Reveal 2, NO SWITCH = WIN
CAR 3, Pick 3, Reveal 3, AUTOMATIC = WIN

In the above scenarios where a choice is made, a RANDOM revelation occurs, and no switch is made (if it's even possible at that point). The odds of ending up with a car are 9 in 27, or 1 in 3. And we haven't even yet looked at the scenarios where a decision is made to switch.

CAR 1, Pick 1, Reveal 1, AUTOMATIC = WIN
CAR 1, Pick 1, Reveal 2, SWITCH = LOOSE
CAR 1, Pick 1, Reveal 3, SWITCH = LOOSE
CAR 1, Pick 2, Reveal 1, AUTOMATIC = LOOSE
CAR 1, Pick 2, Reveal 2, AUTOMATIC = LOOSE
CAR 1, Pick 2, Reveal 3, SWITCH = WIN
CAR 1, Pick 3, Reveal 1, AUTOMATIC = LOOSE
CAR 1, Pick 3, Reveal 2, SWITCH = WIN
CAR 1, Pick 3, Reveal 3, AUTOMATIC = LOOSE
CAR 2, Pick 1, Reveal 1, AUTOMATIC = LOOSE
CAR 2, Pick 1, Reveal 2, AUTOMATIC = LOOSE
CAR 2, Pick 1, Reveal 3, SWITCH = WIN
CAR 2, Pick 2, Reveal 1, SWITCH = LOOSE
CAR 2, Pick 2, Reveal 2, AUTOMATIC = WIN
CAR 2, Pick 2, Reveal 3, SWITCH = LOOSE
CAR 2, Pick 3, Reveal 1, SWITCH = WIN
CAR 2, Pick 3, Reveal 2, AUTOMATIC = LOOSE
CAR 2, Pick 3, Reveal 3, AUTOMATIC = LOOSE
CAR 3, Pick 1, Reveal 1, AUTOMATIC = LOOSE
CAR 3, Pick 1, Reveal 2, SWITCH = WIN
CAR 3, Pick 1, Reveal 3, AUTOMATIC = LOOSE
CAR 3, Pick 2, Reveal 1, SWITCH = WIN
CAR 3, Pick 2, Reveal 2, AUTOMATIC = LOOSE
CAR 3, Pick 2, Reveal 3, AUTOMATIC = LOOSE
CAR 3, Pick 3, Reveal 1, SWITCH = LOOSE
CAR 3, Pick 3, Reveal 2, SWITCH = LOOSE
CAR 3, Pick 3, Reveal 3, AUTOMATIC = WIN

Of the 12 cases where a switch was even possible (the outcome wasn't automatically determined by the first "reveal") 6 result in winning, and 6 result in losing. Of all the cases, 9 of the 27 result in a WIN, or 1 in 3. BUT, notice that 15 of the sequences listed are duplicates of the "NO SWITCH" list, and so cannot be counted as separate outcomes.

If we combine the "NO SWITCH" and "SWITCH" lists, we find that there are a total of 39 possible random sequences (27+12). Of those, 15 result in an AUTOMATIC outcome, 12 involve NO SWITCH, and 12 involve a SWITCH.
Of the 15 AUTOMATIC outcomes, the odds of winning are 3 in 15, or 1 in 5.
In the NO SWITCH sequences, the odds are 6 in 12, or 1 in 2.
In the SWITCH sequences, the odds are the same, 6 in 12, or 1 in 2.

IN THIS SCENARIO, where the opening of a door after the initial choice is completely random, the odds of choosing the correct door THE FIRST TIME are simply 1 in 3. IF the sequence progresses to a second choice, the odds of winning after switching are 6 in 12, or 1 in 2. Likewise, the odds of winning after NOT switching are 6 in 12, or 1 in 2. Nothing is gained by switching.

--- TO BE CONTINUED ---