Decide for Yourself

[GuyPerfect]

The following is an excerpt of a 2-hour conversation TonyV and I had on Skype last night. It seems that he's impervious to logic and has a very large bladder, so he can take this stuff longer than I can. The excerpt has not been modified in any way except for formatting purposes:

[10:00:16 PM] Guy Perfect: You said there was one prize.
[10:00:18 PM] TonyV: True.
[10:00:22 PM] Guy Perfect: And two doors.
[10:00:25 PM] TonyV: Yes.
[10:00:32 PM] Guy Perfect: Which means one of two doors has a prize.
[10:00:37 PM] TonyV: Yes.
[10:00:42 PM] Guy Perfect: Ta dah! 50%.
[10:00:46 PM] TonyV: Nope!

[Diellan]

Was this ins context of the Monty Hall problem?

[GuyPerfect]

It sure was. (-:

The scenario is basically that there's a prize behind one of three doors. You pick a door, and one of the remaining doors--one that did NOT have the prize--is eliminated. At this point, if you switch from the door you selected to the remaining door, you have a 2/3 chance of picking the one with the prize regardless of which of the three doors you picked first.

That point is not contested. By switching, you're basically gauging your likelihood of NOT guessing correctly on your first try, which is 2/3.

The question Tony kept asking, though, is after the first selection was made and two doors remained, "Which of the remaining doors is the better choice?" What this does is funnel down the scenario to the context of two doors, behind one of which is a prize. That means 1/2 chance for either door, and in fact you can flip a coin and be just as likely to pick the prize either way.

The 2/3 statistic comes from a scenario where you pick one of three doors, one of the wrong doors is eliminated, and then you switch to the remaining door.

The 1/2 statistic comes from a scenario where you pick one of two doors.

It's likely a matter of semantics that Tony didn't realize he was doing, but then I guess he was just so ardent in disagreeing with me that he quite literally stated that 1 in 2 is not 50%, so I just had to share it with the world. (-:

[GuyPerfect]

To put it another way...

"Pick one of these two doors."
Well, no matter how you slice it, it's going to be a 1/2 chance of getting it right.

"Pick two of these three doors, one of which will be removed after your first choice."
THERE we go. That's a situation where we can guarantee a 2/3 chance of success.

[Diellan]

Um, yeah, making your first choice when there are 3 doors is kind of necessary for it to be a Monty Hall problem. If all you do is take three doors and then rule out one of them, then you haven't done anything and it's a 50% chance for either door.

[GuyPerfect]

After some further discussion, we found that the source of the discrepancy came from a question along the lines of "Do you stay with door 1 or switch to door 3?" I interpreted it as "1 versus 3" and Tony was thinking "stay versus switch." And when it comes to who's right and wrong when two totally different things are being considered, it generally doesn't go well.

Now, Tony adamantly persists in that there is only one possible interpretation and continues to fancy me a buffoon, so we'll have to wait for his response in this thread to see his side of the situation.

Though I'm still curious about that 1 prize 2 doors != 50% thing. That quote up there was just one of many.

[TonyV]

Okay, here's the exact scenario that was given, directly quoted.  And yes, it's the Monty Hall problem.

TonyV: Well, here's the set up.  You're on Let's Make a Deal with Monty Hall.  He gives you a choice of three doors to pick.  Behind two of them are "goat" prizes--worthless stuff you don't want.  Behind one door is something really nice worth a lot. So you pick a door.  Go ahead, pick a door.  Which one do you want?  Door number one, door number two, or door number three?

Guy Perfect: Let's go with door number one.

TonyV: Okay, now Monty, like me, knows which door the prize is behind.  But like any good showman, he wants to build up the drama and suspense. So when you pick door number one, he says, "Okay, let's open up one of the other doors first.  How about we see what's behind...  Door number two!" Obviously, he's going to open up one of the goat doors. So now there are two doors left.  Monty says, "Okay Guy, I'll tell you what.  Are you still feeling good about door number one?  Because if you want, I'll let you change your mind and go with door number three.  Of course, if you like door number one, you are more than welcome to stick with that one." The question is, statistically speaking, if you want to win the nice prize, should you stay with door number one, or switch to door number three?  Or does it make a difference?

Guy Perfect: Well, my brain is telling me that since I still know nothing about doors 1 or 3, there may as well be an even 50/50 for it being the prize.

TonyV: The answer is, statistically speaking, if you stay with door number one, you only have a 1/3 chance of winning.  By switching to door number three, you increase your chances to 2/3. So if you want the prize, you should switch to door number three.

Guy Perfect: I don't believe you.

TonyV: No one ever does. Would you believe Marilyn vos Savant? The reason why basically boils down to one fact: Monty KNOWS which door the prize is behind. By opening up a "goat" door, that door's original probability (1/3) is shifted to any doors left that you didn't pick, but not yours, because you made your choice in ignorance.

Guy Perfect: I'm going to say that Marilyn vos Savant is full of hot air. The situation wouldn't have been any different if I'd picked door number 3.

Guy says, "The excerpt has not been modified in any way," but he doesn't mention the conversation that came immediately before it.  Here's a bit more context:

Guy Perfect: Two doors with equal chance of having a prize behind them comes down to a 1/2 chance of picking the right one.
TonyV: You're right...  But that's not the scenario that was presented. I never said it was.
Guy Perfect: The scenario was "You have two doors. Which is better?" You said THAT repeatedly.
TonyV: No, the scenario has ALWAYS been, you pick a door.  Monty opens a door that he KNOWS has a goat behind it.
Guy Perfect: And then you have two doors and pick one. Am I not correct?
TonyV: You are given a choice of sticking with your original door, or switching to the door that's left. I NEVER said, "two doors with equal chance..."
Guy Perfect: Yeah you did. 3 doors with 1 prize - 1 door = 2 doors with 1 prize. I know I'm a math genius, but this I think is easy enough for someone like you.
TonyV: True, but I NEVER said that those two doors have an EQUAL chance of having the prize.
Guy Perfect: You said there was one prize.
TonyV: True.
Guy Perfect: And two doors.
TonyV: Yes.
Guy Perfect: Which means one of two doors has a prize.
TonyV: Yes.
Guy Perfect: Ta dah! 50%.
TonyV: Nope!

And THAT is the full context.

Now, instead of sucking it up an saying, "Wow, that's pretty cool, I never would have thought of that!" like most people do who I present this problem to, he's trying to pretend like the original scenario I gave him was more along these lines:

TonyV: You walk out onto the stage and there are two doors, one of which has a prize behind it.  Pick one.
Guy Perfect: I'll take door number one.
TonyV: Okay, so do you want to keep that choice, or switch to door number two?
Guy Perfect: It doesn't matter, the odds are 50/50.
TonyV: No they're not!  If you switch, your odds of winning are 2/3!

Read the real exchanges again, because you can see that this is not the scenario given, and every time he tried to twist my words into saying something I didn't, I corrected him.

The thing is, this isn't the first time I've gotten people with this counter-intuitive example of probability doing weird things.  And almost every time I do and it becomes obvious that I'm right, the person I'm talking to tries to squirm out of it.  I've had people use the old, "But I like a goat, I'd rather have that!" (which is why I specifically said that "goat" prizes are worthless stuff you don't want).  The closest thing I've ever had to a valid argument was when someone pointed out that I didn't specifically mention that Monty knew which door the prize was behind, I only implied it when I said he opened up a "goat" door, so I've always made a special point of explicitly stating that Monty knows where the prize is.

So, there you go.  There's your "math genius" at work. 

So how about it, folks?  Can I get a, "Hell, yeah!  Tony's right!"

[TonyV]

Um, yeah, making your first choice when there are 3 doors is kind of necessary for it to be a Monty Hall problem. If all you do is take three doors and then rule out one of them, then you haven't done anything and it's a 50% chance for either door.

Read the full context above.  What Guy posted is deliberately misleading.  "Making your first choice when there are 3 doors" was always part of the scenario. 

[GuyPerfect]

What Guy posted is deliberately misleading.

I see what u did thar. There's a word for what u did thar, you know. (-:

Looking back on the conversation, it's fairly plain to see that it all goes back to the ambiguous sentence I mentioned earlier... The one about arbitrary doors (1 or 3) versus picking doors in the context of an earlier pick (stay or switch).

Knowing that now, it's pretty obvious that Tony and I had very different things in mind. Nonetheless, he continues to insist that I am wrong by his standard, regardless of how inapplicable that may be in this situation.

[8/2/2011 9:07:09 PM] Guy Perfect: My code would produce output something like this:
Doors: 3. Prizes: 1. Prizes per Door: 0.33333
Eliminating goat door.
Doors: 2. Prizes: 1. Prizes per Door: 0.5
Eliminating goat door.
Doors: 1. Prizes: 1. Prizes per Door: 1
[8/2/2011 9:07:16 PM] TonyV: Whatever it takes, man, write the program!  Prove me wrong!
[8/2/2011 9:07:38 PM] TonyV: I was thinking more something like this:
[8/2/2011 9:07:50 PM] Guy Perfect: That doesn't even require any randon numbers. Prizes per Door, by definition, is Prizes / Door
[8/2/2011 9:08:29 PM] TonyV: Prize door: 3
Contestant picks: 1
Monty opens: 2
Stay: LOSE, Switch: WIN
[8/2/2011 9:08:49 PM] TonyV: Prize door: 2
Contestant picks: 2
Monty opens: 3
Stay: WIN, Switch: LOSE
[8/2/2011 9:09:06 PM] TonyV: Keep track of all of the WINs and LOSEs, and add them up after a few thousand iterations.
[8/2/2011 9:09:31 PM] TonyV: The prize door and contestant door above are random.
[8/2/2011 9:09:50 PM] Guy Perfect: So basically what you're trying to figure out is "what is the likelihood of picking the correct door of 3 on the first try?"
[8/2/2011 9:10:31 PM] TonyV: No, it's exactly what I said above.  I want to know what is the probability of winning by sticking with your original door (which I contend is 1/3) versus winning by switching doors (which I contend is 2/3).
[8/2/2011 9:10:53 PM] Guy Perfect: Aka: picking the right one first.

For clarity, if you pick the right door first, you will lose when you switch. Monty's actions and prior knowledge notwithstanding, by switching after elimination, you turn your 2/3 chance of getting it wrong on the first pick into a 2/3 chance of getting it right on the second pick.

[8/2/2011 9:28:52 PM] TonyV: If you originally picked the right door and you switch, then you've lost.  If you originally picked a wrong door and switched, you win.
[8/2/2011 9:29:01 PM] Guy Perfect: So if you get the wrong one, then lose the other wrong one, and switch... You're going to win.
[8/2/2011 9:29:12 PM] Guy Perfect: If you get the right one, then lose one of the wrong ones, and switch... You're going to lose.
[8/2/2011 9:29:19 PM] Guy Perfect: So guess, lose, switch will yield a 2/3 chance of winning.
[8/2/2011 9:29:28 PM] TonyV: So you're changing your mind?
[8/2/2011 9:29:51 PM] Guy Perfect: No, because the question posed was only in the context of 2 remaining doors.

There's the discrepancy in a nutshell. Coulda saved an hour and a half with a little attention. (-:

[8/2/2011 9:35:46 PM] Guy Perfect: Say there are three doors and the prize is behind door 1.
[8/2/2011 9:35:52 PM] Guy Perfect: You pick door 1. What happens?
[8/2/2011 9:36:14 PM] TonyV: Monty will open door 2 or door 3, presumably at random.
[8/2/2011 9:36:20 PM] Guy Perfect: Are they equal?
[8/2/2011 9:36:28 PM] TonyV: Is what equal to what?
[8/2/2011 9:36:35 PM] Guy Perfect: I mean, door 3 may as well be the one in the middle, right?
[8/2/2011 9:36:46 PM] TonyV: It doesn't matter which door he opens, no.
[8/2/2011 9:37:03 PM] Guy Perfect: So how many doors are left?
[8/2/2011 9:37:11 PM] Guy Perfect: (this is part one of two)
[8/2/2011 9:37:15 PM] TonyV: Two doors are left.
[8/2/2011 9:37:23 PM] Guy Perfect: And how many have a prize?
[8/2/2011 9:37:39 PM] TonyV: One of them does.  If we're still following your example, the one you picked does--door one.
[8/2/2011 9:37:48 PM] Guy Perfect: Okay, now for a different situation.
[8/2/2011 9:37:58 PM] Guy Perfect: Door 1 still has the prize, but you pick door 3.
[8/2/2011 9:38:01 PM] Guy Perfect: What happens?
[8/2/2011 9:38:06 PM] TonyV: Monty opens door two.
[8/2/2011 9:38:16 PM] Guy Perfect: Which might as well be the one on the right, yes?
[8/2/2011 9:38:35 PM] TonyV: It doesn't matter where they're physically situated.
[8/2/2011 9:38:41 PM] Guy Perfect: Right.
[8/2/2011 9:38:42 PM] TonyV: But he MUST open door two.
[8/2/2011 9:38:45 PM] Guy Perfect: So, at this point, how many doors are left?
[8/2/2011 9:39:06 PM] TonyV: He can't open door one, because that's the one with the prize.  He can't open door three, because that's your door.  There are two doors left.
[8/2/2011 9:39:12 PM] Guy Perfect: And how many of them have a prize?
[8/2/2011 9:39:21 PM] TonyV: One--Door one.
[8/2/2011 9:39:25 PM] Guy Perfect: Right.
[8/2/2011 9:40:02 PM] Guy Perfect: So by your own admission... If you guess correctly first, one door is eliminated, and 1 of the remaining 2 doors has a prize. If you guess incorrectly first, one door is eliminated, and 1 of the remaining 2 doors has a prize.
[8/2/2011 9:40:17 PM] Guy Perfect: A -> C and B -> C. C = 1 of 2 doors has a prize.

Makes perfect sense if we take the second pick in isolation, no?

[Aggelakis]

Ya'll are geeks. :p

[GuyPerfect]

I represent that remark!

[TonyV]

Ya'll are geeks. :p

I don't disagree.  At least I'm the right one.  

And lest anyone misunderstand, this isn't like some kind of holy war or anything, no one is going to be banned or anything silly like that.  I respect people I argue with.

And really, I'm liking how Guy is posting even more evidence that I corrected him numerous times.

[GuyPerfect]

And lest anyone misunderstand, this isn't like some kind of holy war or anything [...]

Not to mention what a low move it would be to call out someone in public with citations from a private conversation. If I was actually angry with Tony, I have many far more effective ways to troll him. (-:

And really, I'm liking how Guy is posting even more evidence that I corrected him numerous times.

I was actually going more for the "Guy shows how the argument was over apples and oranges but Tony's too inflated to notice" effect. It's one of my many trolling ways. (-:

[Diellan]

TonyV: Well, here's the set up.  You're on Let's Make a Deal with Monty Hall.  He gives you a choice of three doors to pick.  Behind two of them are "goat" prizes--worthless stuff you don't want.  Behind one door is something really nice worth a lot. So you pick a door.  Go ahead, pick a door.  Which one do you want?  Door number one, door number two, or door number three?

Guy Perfect: Let's go with door number one.

TonyV: Okay, now Monty, like me, knows which door the prize is behind.  But like any good showman, he wants to build up the drama and suspense. So when you pick door number one, he says, "Okay, let's open up one of the other doors first.  How about we see what's behind...  Door number two!" Obviously, he's going to open up one of the goat doors. So now there are two doors left.  Monty says, "Okay Guy, I'll tell you what.  Are you still feeling good about door number one?  Because if you want, I'll let you change your mind and go with door number three.  Of course, if you like door number one, you are more than welcome to stick with that one." The question is, statistically speaking, if you want to win the nice prize, should you stay with door number one, or switch to door number three?  Or does it make a difference?

Guy Perfect: Well, my brain is telling me that since I still know nothing about doors 1 or 3, there may as well be an even 50/50 for it being the prize.

TonyV: The answer is, statistically speaking, if you stay with door number one, you only have a 1/3 chance of winning.  By switching to door number three, you increase your chances to 2/3. So if you want the prize, you should switch to door number three.

Guy Perfect: I don't believe you.

TonyV: No one ever does.  Would you believe Marilyn vos Savant? The reason why basically boils down to one fact: Monty KNOWS which door the prize is behind. By opening up a "goat" door, that door's original probability (1/3) is shifted to any doors left that you didn't pick, but not yours, because you made your choice in ignorance.

Guy Perfect: I'm going to say that Marilyn vos Savant is full of hot air. The situation wouldn't have been any different if I'd picked door number 3.

I was actually going more for the "Guy shows how the argument was over apples and oranges but Tony's too inflated to notice" effect. It's one of my many trolling ways. (-:

I'm sorry, but I think I have to side with Tony on this one. The initial quote had him state the Monty Hall problem precisely (you choose a door, he reveals a door, then asks you whether or not you should switch) and you fell for the common 50% fallacy. Everything after that just seems like trying to save face.

[GuyPerfect]

Everything after that just seems like trying to save face.

Even things like this?

[8/2/2011 8:46:39 PM] Guy Perfect: Then the contestant picks one of the remaining two at random.
[8/2/2011 8:46:46 PM] Guy Perfect: See, I'm all set to win this!
[8/2/2011 8:47:11 PM] TonyV: Right.  Run a few thousand trials of sticking with the contestant's original door versus switching to the other door.
[8/2/2011 8:47:48 PM] TonyV: I guarantee you that if you've coded it up right, you will get the contestant winning approximately 1/3 of the time by staying, and winning 2/3 of the time by switching.
[8/2/2011 8:48:15 PM] Guy Perfect: Relative to the original three, yes. I totally agree there.
[8/2/2011 8:48:39 PM] Guy Perfect: In context, statistics takes a wild shape.
[8/2/2011 8:48:53 PM] TonyV: Wait now, don't be placing qualifications on the claim.  If it's about 50/50, you are right.  If it's about 33/67, then I'm right.
[8/2/2011 8:49:18 PM] Guy Perfect: The question was "what's the probability of your door still having the prize after one of the goat doors was removed"?

[8/2/2011 9:21:25 PM] Guy Perfect: That's not the answer to the question. That's the answer to a different question.
<snip>
[8/2/2011 9:21:43 PM] Guy Perfect: Stay = Win -> You got it right on the first try. Aka, you had a 1/3 chance of guessing it right before any doors were eliminated
<snip>
[8/2/2011 9:22:14 PM] Guy Perfect: That's the answer to the question "What's the likelihood of picking the correct one of three doors?"

If you're going to chalk that up to "you've made a statistical fallacy" rather than "you've answered the wrong question," then you're just as bonkers as Tony.

This is the man you are agreeing with:

[8/2/2011 8:12:54 PM] TonyV: The answer is, statistically speaking, if you stay with door number one, you only have a 1/3 chance of winning.  By switching to door number three, you increase your chances to 2/3.
<snip>
[8/2/2011 8:14:27 PM] TonyV: The reason why basically boils down to one fact: Monty KNOWS which door the prize is behind.

Surely he didn't say Monty's knowledge alters reality? Maybe that's out of context... Let's try another:

[8/2/2011 8:59:03 PM] TonyV: If I toss a fair coin and it comes up heads 187 times in a row, what are the odds that it comes up heads on the next throw?  50/50.
[8/2/2011 8:59:20 PM] Guy Perfect: Yeah. Thanks for agreeing with me.
[8/2/2011 8:59:32 PM] Guy Perfect: So what's this nonsense about 1/3 to 2/3 for that two-sided coin?
[8/2/2011 8:59:48 PM] TonyV: Because like I said, knowledge makes the difference.
[8/2/2011 8:59:57 PM] TonyV: Monty KNOWS which door the prize is behind.
[8/2/2011 8:59:58 PM] Guy Perfect: *Facepalm*
[8/2/2011 9:00:13 PM] Guy Perfect: August 2, 2011: TonyV states that your knowledge will make one result of a coin toss 2/3

The all-uppercase "KNOWS" showed up a lot, directly asserting Monty Hall's amazing reality-bending abilities.

But perhaps that's all just a miscommunication. Tony's a smart guy, so he wouldn't say anything blatantly false, right? Let's take a look:

[8/2/2011 11:20:15 PM] Guy Perfect: You can use a coin to select between doors 1 and 3.
<snip>
[8/2/2011 11:20:27 PM] TonyV: You can, and you will lose most of the time.

Coins, as you know, have a 50% success rate. But it's just silly to say that a 1/2 chance is anything but fifty perce--oh... oh dear, that happened:

[10:00:16 PM] Guy Perfect: You said there was one prize.
[10:00:18 PM] TonyV: True.
[10:00:22 PM] Guy Perfect: And two doors.
[10:00:25 PM] TonyV: Yes.
[10:00:32 PM] Guy Perfect: Which means one of two doors has a prize.
[10:00:37 PM] TonyV: Yes.
[10:00:42 PM] Guy Perfect: Ta dah! 50%.
[10:00:46 PM] TonyV: Nope!

[Tazhyngarth]

I have to apologize for this, but I totally don't understand how staying with door 1 gives you a 1/3 chance but switching to door 3 ups it to 2/3.  2/3 doors remain, so both are 1/3 chance (if still factoring the eliminated door).....how is switching doors actually upping your chances?

I can see the whole thing with 50/50 versus not 50/50, but that the stay/switch chance difference, I'm lost to.

[GuyPerfect]

There are three doors, and one has a prize behind it. You pick a door, one of the other doors (one that does not have the prize behind it) is eliminated. You then switch from the door you picked first to the other remaining door.

Should you first pick the "right" door, then one of the "wrong" doors is removed, and switching will land you on the other "wrong" door.

Should you first pick one of the two "wrong" doors, then the other "wrong" door is removed, and switching will land you on the "right" door.

So to pick-and-switch with three starting doors, one of three things will happen:

Door 1 (Right) -> Eliminate -> Switch = Lose
Door 2 (Wrong) -> Eliminate -> Switch = Win
Door 3 (Wrong) -> Eliminate -> Switch = Win

Since you start with two wrong doors and switching from a wrong door will win, the likelihood of the pick-and-switch method winning is 2/3.

However, if you do Door -> Eliminate -> Flip a Coin, that's going to be 50/50.

[Tazhyngarth]

So to pick-and-switch with three starting doors, one of three things will happen:

Door 1 (Right) -> Eliminate -> Switch = Lose
Door 2 (Wrong) -> Eliminate -> Switch = Win
Door 3 (Wrong) -> Eliminate -> Switch = Win

Since you start with two wrong doors and switching from a wrong door will win, the likelihood of the pick-and-switch method winning is 2/3.

Now it makes sense, didn't mean for you to summarize the whole thing again, sorry about that.

[eabrace]

Well, let's see...

If it's a simple case of pick a door, see whether you win or lose, get a second chance if you lost, then you independently have a 1 in 3 chance of success and a 2 in 3 chance of failure on the first pick, and equal 1 in 2 chances of success or failure on the second pick.

However, since you only have to make a selection from two doors if you choose incorrectly the first time, the chances of winning on your second pick are contingent on an initial failure.

Since the prize could be behind any of the three doors, there's three possible scenarios where you win on your first try.  Since there are two possible ways to fail on your first try, each with one possible way of winning and one possible way of losing on your second try, there are two ways to succeed on a second pick and two ways to fail.

Therefore, there are fifteen total scenarios possible in the game, and your odds of winning are 9/15.

Looking at this another way:

If door A is the winning door, and you select door A the first time, you win.
If door A is the winning door and you select door B and then door A, you win.
If door A is the winning door and you select door B and then door C, you lose.
If door A is the winning door and you select door C and then door A, you win.
If door A is the winning door and you select door C and then door B, you lose.
If door B is the winning door and you select door A and then door B, you win.
If door B is the winning door and you select door A and then door C, you lose.
If door B is the winning door, and you select door B the first time, you win.
If door B is the winning door and you select door C and then door A, you lose.
If door B is the winning door and you select door C and then door B, you win.
If door C is the winning door and you select door A and then door B, you lose.
If door C is the winning door and you select door A and then door C, you win.
If door C is the winning door and you select door B and then door A, you lose.
If door C is the winning door and you select door B and then door C, you win.
If door C is the winning door, and you select door C the first time, you win.

BUT, that's not the scenario Tony was presenting if he was going for the Monty Hall scenario.

[Sekoia]

Well, let's see...

If it's a simple case of pick a door, see whether you win or lose, get a second chance if you lost, then you independently have a 1 in 3 chance of success and a 2 in 3 chance of failure on the first pick, and equal 1 in 2 chances of success or failure on the second pick.

However, since you only have to make a selection from two doors if you choose incorrectly the first time, the chances of winning on your second pick are contingent on an initial failure.

Since the prize could be behind any of the three doors, there's three possible scenarios where you win on your first try.  Since there are two possible ways to fail on your first try, each with one possible way of winning and one possible way of losing on your second try, there are two ways to succeed on a second pick and two ways to fail.

Therefore, there are fifteen total scenarios possible in the game, and your odds of winning are 9/15.

Looking at this another way:

If door A is the winning door, and you select door A the first time, you win.
If door A is the winning door and you select door B and then door A, you win.
If door A is the winning door and you select door B and then door C, you lose.
If door A is the winning door and you select door C and then door A, you win.
If door A is the winning door and you select door C and then door B, you lose.
If door B is the winning door and you select door A and then door B, you win.
If door B is the winning door and you select door A and then door C, you lose.
If door B is the winning door, and you select door B the first time, you win.
If door B is the winning door and you select door C and then door A, you lose.
If door B is the winning door and you select door C and then door B, you win.
If door C is the winning door and you select door A and then door B, you lose.
If door C is the winning door and you select door A and then door C, you win.
If door C is the winning door and you select door B and then door A, you lose.
If door C is the winning door and you select door B and then door C, you win.
If door C is the winning door, and you select door C the first time, you win.

I believe it would be a 2/3 chance of winning, not 9/15.

You have a 1 in 3 chance of winning on your first guess. You have a 1 in 2 chance of winning on your second guess.

Rephrased: Once you've made your first choice, you have a 1 in 3 chance for having a 100% chance of winning. You have a 2 in 3 chance for having a 50% chance of winning.

1/3 * 1 + 2/3 * 1/2 = 2/3

Because it's a two step process where the second is dependent on the first, your enumeration of the 15 scenarios does not accurately enumerate the odds.