## 3 door problem

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SpitValve
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### 3 door problem

A classic one often taught in undergrad, not sure if I remember it being done here though.

You are on a high-stakes game-show. For the final round, you are shown three doors. Two of the doors have nothing behind them. The other door has a boxload of kittens. All the doors appear identical. You are told to select one door to attempt to obtain to the boxload of kittens.

You choose one door - but it is not opened yet. Then the host opens one of the other two doors, revealing that it has no kittens behind it.

There are now two closed doors, one of which contains a box of kittens. The host gives you a chance to change your choice.

Assuming you want to maximise your odds of getting the kittens, should you stay with your original door or switch to the other closed door? Does it make a difference?

Tractor
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More importantly, if you don't want a box full of kittens, can you pick the known empty one?

solution? wrote:Really, it shouldn't matter, right? You already increased your odds from 33% to 50%. So at this point your odds are equal on either door.
9 x 6 = 42

Note: Randall kicks ass.

parsonsb
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always switch it gives you better odds of winning since there are three different possiblilities. either you picked empty door 1 initially and switching would get you the kittens, or you picked empty door 2 initially and switching would get you the kittens, or you picked the kittens initially and switching would lose it for you. so by always switching you would improve your odds from 1/3 to 2/3

3.14159265...
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This is a famous one, wiki has it too. I think it was up on this site though not sure.
You pick the door you didn't pick.
The door you picked (say door A) had 1/3 chances of having girl (don't like kittens)
The doors together had 2/3. (say door B and C)
When the host eliminated on of those two doors (Say door C), the chances of the girl being behind the door you didn't pick (door B) is still 2/3 . Your door (door A) just has the 1/3 chance.
So depending on whether you want the girl or have a gf you pick appropriately.

Edit: Damn you parsonb, for once I thought I had the solution
"The best times in life are the ones when you can genuinely add a "Bwa" to your "ha""- Chris Hastings

parsonsb
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i's sorry

and its parsonsb

no one ever gets the s

SpitValve
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Yeah didn't think it'd last for long, it's pretty famous...

At first everybody thinks it's Tractor's solution, and I personally took a look of convincing before I understood the right one (i.e. parsonsb's and Pi's solution)...

3.14159265...
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I just assumed it was something like personb, like Person A and person B.
I am curious now, wut does it mean?
"The best times in life are the ones when you can genuinely add a "Bwa" to your "ha""- Chris Hastings

tendays
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By the way, this problem was already discussed a long time ago (except that it wasn't kittens but a car instead - the solution still works, doesn't it?? ) :

Wikipedia's writeup

EDIT: restored original contents. Seriously guys, WHO edited my post? (or was my account hijacked? - given that we can't go on https)
Last edited by tendays on Wed Mar 14, 2007 12:20 am UTC, edited 1 time in total.

tendays
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Hey - did a mod edit my post? That's not what I wrote...

bbctol
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This is a fairly famous problem, but it's always good to use on first-comers to probability. When in doubt, always use the slow-but-always-works outcome tree!

parsonsb
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parsons is my last name b is my first initial (stands for bryan)

Sygnon
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I recently had to explain this to a small group, I have found that when explaining the problem you can change the numbers and imagine that you start with 100 doors, select 1, open 98, This immediately keyed people in on what was happening.

Anyway, orthogonal to the thread but i thought it was appropriate.

skeptical scientist
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Sygnon wrote:I recently had to explain this to a small group, I have found that when explaining the problem you can change the numbers and imagine that you start with 100 doors, select 1, open 98, This immediately keyed people in on what was happening.

Anyway, orthogonal to the thread but i thought it was appropriate.

I once asked this as a riddle on a long car trip, and unsurprisingly nobody got the answer. However, after I revealed the correct answer and explained it, they kept arguing and wouldn't believe I had the correct answer (in fact, the debate lasted several days, even after I gave internet references to the solution, and performed the experiment with some cards). I've never given it as a riddle since.
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gmalivuk
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While we're at it, what's 0.9999999... again?

I can at least understand resisting the Monty Hall problem, because it's (initially) counterintuitive. But it's still annoying when people are that dense.

Yakk
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The problem with this question is that unstated assumptions change the answer.

Pattern A: Suppose the door opening guy has the following algorithm:
If you pick the door with the kittens, open a door and show it to be empty.
If you pick a door without the kittens, don't open a door.

Or, pattern B: the host only opens a blank door if you picked a door that was already blank.

In Pattern A, your chance when swapping is 0% and 100% when sticking.
In Pattern B, your chance when swapping is 100% and 0% when sticking.

The host can probablistically choose between Pattern A and Pattern B to generate any chance in between.

...

Now when you are on the show, you observe exactly what the problem describes. Yet changing doors when the host opens an empty one makes you win with any probability the host wants.

For the Monty Kitten proof to work, assumptions about how the door opening algorithm works must be made. When people state the problem, they basically never mention these assumptions, yet the answer is completely dependant on them.

...

You can fix this by specifying the exact algorithm the host follows. "The host always picks and opens a door that is empty, and you know this", or "The host always opens a door at random. This time you got an empty door."

But that kind of information can't be glazed over and ignored.

Toam
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Yakk wrote:You can fix this by specifying the exact algorithm the host follows. "The host always picks and opens a door that is empty, and you know this", or "The host always opens a door at random. This time you got an empty door."

An entirely valid point except that the idea of the "puzzle" is that it is entirely random and the latter algorithm is implied.

tendays
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Yakk wrote:"The host always opens a door at random. This time you got an empty door."

I think you get the "50% chances of winning if you switch" with that scenario.

Case 1 with 1/3 chances: You pick the kittens
Case 2 with 2/3 chances: You don't pick the kittens.

Case 2a (probability 1/3): the host doesn't pick the kittens.
Case 2b (probability 1/3): the host picks the kittens

In case 1 you lose if you switch (what the host chose doesn't matter).
In case 2a you win if you switch.
In case 2b the game is void because the host "accidentally" revealed the kittens - maybe you restart the game or see it as a draw - anyway whether you should switch or not is no longer a valid question in this case YET this case may happen, invalidating the reasoning done in the original problem.

Conclusion: if the host did not reveal the kittens (probability 2/3) you win 50% of the cases if you switch ( (1/3)/(2/3) ) (and also 50% of the cases if you don't - i.e. whether you switch or not doesn't change your chances of winning)

elminster
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Edit: Completely rewritten.

I don't like this question. It should be 50/50, the only reason why it may not be is if you study it from a particular point within the game tree, in a particular manner, that doesn't represent the question which was asked.
We all know each statistical process is individual and random when looking at the scope of what it involves.

Ie, regardless of weather you flip a coin a million times on heads in a row, it still has a 50/50 chance the next time. If you link together their probabilities by saying you will get X amount of heads, it becomes a different question.

At this point in the game show, regardless of what could have happened, or what may happen. We only know for 100% certain that:
1) There are 3 doors.
2) 1 Door has kittens behind it
3) 1 Door does not have kittens behind it.

Hence, in this particular statistical state, he happened to pick the one without kittens. So you cant physically change the fact that there isnt any kittens there after it has happened.
Last edited by elminster on Mon Mar 19, 2007 12:31 am UTC, edited 2 times in total.

EvanED
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Sygnon wrote:I recently had to explain this to a small group, I have found that when explaining the problem you can change the numbers and imagine that you start with 100 doors, select 1, open 98, This immediately keyed people in on what was happening.

Anyway, orthogonal to the thread but i thought it was appropriate.

I never found that argument compelling. I don't see how it works with 100 doors any better than 3 myself.

The only explanation that I've found actually *satisfying* (as opposed to just a mechanical proof or empirical demonstration that proves or shows what the right answer is that doesn't help with the intuition) is that if you choose the right door in the first place (1/3 chance) you lose if you switch, and if you choose the wrong door in the first place (2/3 chance) you always win if you switch, so in some sense you're picking the other *two* doors if you switch.

phlip
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NB: for the following, I am referring to the "You pick a door, the host then always opens one of the two other doors, which is always a losing door, and you're always given the opportunity to switch" problem, which is the classic Monty Hall problem.

The first step when you're trying to convince someone that an intuitive answer is wrong should be to disprove the intuition. Showing the proof of the correct answer should be second.

In this case: the intuition is "there are two doors, therefore 50/50". The maths says "If there are n indistinguishable options, the chances are 1/n"... but the doors aren't indistinguishable... one of them the host couldn't open because you've chosen it... the other the host could open, but elected not to... possibly at random, but possibly because it's the winning door. So the doors are distinct, and a such, they're not necessarily 50/50.

To make the 1/3 result more clear, imagine that the doors are able to be physically moved around (I often use an analogy with Deal or No Deal here, and the briefcases). The door that you choose, you take with you to your side of the room, the two you don't choose I keep with me. The chance that you have the prize is 1/3, the chance that I have the prize is 2/3... this much is clear, since the doors are still indistinguishable (the only distinguishing feature is that you picked the one that you did, and you made the choice without any information).
Now, I open one of my doors and show it's the loser. This doesn't give any new information with respect to which side of the room the prize is on, since you know that one of my doors is, indeed, a loser.
So the chance you have the prize is still 1/3, and the chance I have it is 2/3. In the 2/3 chances that I have the prize, however, it is always the door I haven't opened, so there's a 2/3 chance that the prize is behind my closed door - you'll have a 2/3 chance if winning if you switch.

It is as EvanED says... if you choose wrong in the first place, which you have a 2/3 chance of doing, then switch, you will always win.

However, Yakk's gripe is very valid, hence the note at the top of my post.

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`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}`
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evilbeanfiend
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skeptical scientist wrote:I once asked this as a riddle on a long car trip, and unsurprisingly nobody got the answer. However, after I revealed the correct answer and explained it, they kept arguing and wouldn't believe I had the correct answer (in fact, the debate lasted several days....

that's the point you start playing it for money until they twig

elminster
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Say for example, we have a different quiz.
There are 3 cups, 2 are upside down, 1 is the right way up. You are told one of the 3 cups contains a ball. If you know for a fact that the right way up cup does not, and will not contain the ball, even if it could, it wont for any particular asking of this simulation. The probability would be 50/50.
It represents the same probabilistic state as the game show though.

The only reason why it would be 1/3 and 2/3 is on the assumption that he "could" have picked the winning one. Its only really the solution to the Nth cases. But using probability of Nth case does not apply when you are already part way through a probability tree for a particular instance of a probability that contains more than 1 step.
Nth case probabilities cant be used to "predict" outcomes in reality. Just because you got heads on the last flip of the coin, the Nth case says it should be tails, but that is irrelevant to what's actually happening.

Edit: also im probably wrong somewhere, even after reading the wiki on it 3 times. This is just a confusion by looking at the second probabilistic state from the first probabilistic state. ie, assuming the chances at the first state are still true, even though were in the second state.

Patashu
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Here's how I see it:

There are three possible cases after your first turn: Two where you got it wrong, one where you got it right. You don't know which it is yet, but this is important

Now the host shows you a selection which is wrong. For the first two cases, only one other selection is wrong and thus if you switch you'll get it right. For the last one you already have it so if you switch you get it wrong. But as you can see, not switching gives you a 1/3 chance of getting it right while switching gives you a 2/3 chance.

Dashiva
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Cranking up the number of doors usually works for me when people need convincing.
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phlip
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elminster wrote:There are 3 cups, 2 are upside down, 1 is the right way up. You are told one of the 3 cups contains a ball. If you know for a fact that the right way up cup does not, and will not contain the ball, even if it could, it wont for any particular asking of this simulation. The probability would be 50/50.
It represents the same probabilistic state as the game show though.

Not quite... in this example, the incorrect cup is turned over before you make your first selection, but in the Monty Hall problem, the incorrect door is opened after your first selection.

The two cases are distinct, since in the Monty Hall problem, when you choose a door, the host can't open the door you chose... so you're restricting the host's choices, so the probabilities change.

In your problem with the cups, the probability is 50/50, because the two facedown cups are indistinguishable. In the Monty Hall problem the two closed doors are distinct, so it's not necessarily 50/50.

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`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}`
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Icaruse
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It's rather easy to prove that the answer with math if you make define a distribution:

Assume a uniform discrete distribution between which of the doors (1, 2, or 3) the kittens are behind (a fairly resonable assumtion, agreed?) :
f(x) = 1/3, x = 1,2,3.

Define the set A as the set where x = 3, so the size of A is 1
Define the set B is the set where x != 2, so the size of B is 2
The size of (A int B), or the intersection of the two sets, would be 1 because the element x = 3 is present in both.

P(A|B) = P(A int B) / P(B)
... = {|(A int B)| / |S|} / {|B| / |S|}
... = |(A int B)| / |B|
... = 1 / 2

Same goes for P(C|B) where C is the set x = 1

P(A|B) would be the probability that the kittens are behind door # 3 given that they are not behind door # 2. This is the only thing you know: that the kittens are not behind door # 2. Say that x = 1 is the door that you picked initially and x = 2 is the door that the host showed you. It doesn't matter if this is the case, since you can define a set where that is a case (for, say, y) and do the above with that set.

Edit: However, the question, as posed, also gives that the door is different from the one you picked, so there's an extra calculation since you can measure which door the host picks in terms of wether it has kittens behind it as a uniform probability (or it's probably good to assume it is rather than base it off the other facts since the host is rather external to the picking of the correct door).

Eventually it'll lead down to 2/3. The graph at wikipedia was the best tool in making me see it intuitively.
Last edited by Icaruse on Tue Mar 20, 2007 3:41 am UTC, edited 3 times in total.

Patashu
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I think I just found an even easier way to explain it.

You pick Door X. There is a 1/3 chance that the prize is behind it.
The host opens Door Y. It does not have the prize.
Door Z therefore has the prize whenever Door X does not, or 2/3 of the time.

Icaruse
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You forgot to condition your probability to the fact that you now know that it is not Y.

Otherwise you have that P(Z) = 1/3 and P(X) = 1/3 leaving a contradiction that the sum of all probabilities = 1

phlip
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Hey guys, Airplane on a Treadmill is still on the first page of the forum... let's stick to one question where people are arguing cross-purposes based on differing interpretations of the question at a time, shall we?

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`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}`
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Icaruse
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The Wikipedia write-up has that you know the host to always open a closed door that you didn't pick, then you can indeed calculate the conditional probability to have a 2/3 chance for switching. That requires that the host never opens the door you pick, weigting the probability favorably that he narrows it down for you. So I guess this is, indeed, a problem of phrasing and interpretation.

(sorry)

demeteloaf
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EvanED wrote:
Sygnon wrote:I recently had to explain this to a small group, I have found that when explaining the problem you can change the numbers and imagine that you start with 100 doors, select 1, open 98, This immediately keyed people in on what was happening.

Anyway, orthogonal to the thread but i thought it was appropriate.

I never found that argument compelling. I don't see how it works with 100 doors any better than 3 myself.

say there are 1 million doors and you guess that the car is behind door #4. The host goes, "ok, now i'll open all the doors except door #532,592." and there's no car behind any of them.

at this point, which one are you going to pick, your initial guess 4, or the one that the host left open?

Ren
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Hey, hurray, it's the Monty Hall problem! I learned about it first in this book called "The Curious Incident of the Dog in the Night-time."
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cmacis
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That book is a good read. However, all of probability is wrong.

Any event that happens always had probability 1 as no other event could have happened. Any other event has probability 0 as it didn't take its chance to happen.

So the probability of you winning in a real situation is 1 or 0 depending on so many factors.
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Uh, cmacis, are you piss-taking, or are you actually posting in support of actualism?

Because under that philosophy, there's really no point in doing any kind of science, ever. Which means that, even if actualism is true, it's useless for actually figuring anything out.
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cmacis
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I just don't like probability. Or statistics.
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QED is Latin for small empty box.

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yellowtamarin
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I can't see any posts that have put it as simply as I see it:

(The scenario where we know a door without kittens will be opened after we choose.)

The probability is 50/50, as we are always only choosing between two doors, one with and one without the kittens. There is no third door to consider because it is always removed from the equation and we know this.

As for the other scenario where we don't know that a losing door will be opened, there is a neat pictorial represenation of this in the book The Curious Incident of the Dog in the Night-time. Although it is used to illustrate the first scenario, which makes it incorrect.
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SpitValve
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yellowtamarin wrote:I can't see any posts that have put it as simply as I see it:

(The scenario where we know a door without kittens will be opened after we choose.)

The probability is 50/50, as we are always only choosing between two doors, one with and one without the kittens. There is no third door to consider because it is always removed from the equation and we know this.

As for the other scenario where we don't know that a losing door will be opened, there is a neat pictorial represenation of this in the book The Curious Incident of the Dog in the Night-time. Although it is used to illustrate the first scenario, which makes it incorrect.

Nope

Yes, we know the 3rd door doesn't have kittens. That doesn't mean there's a 50/50 chance between the other two doors. That would be assuming the doors were identical - and that is not true. You have been given more information about the door you didn't chose than the door you chose - that is, if one of the doors you didn't chose had kittens behind it, then that's the door that wasn't opened for you.

But that's confusing. Here's my simple version:

You choose one door. You have a 1/3 chance of that door being correct and a 2/3 chance one of the other doors is correct.

One of the other doors is revealed to not contain kittens.

There is still a 2/3 chance one of the two doors you haven't chosen contains kittens. Of course, there's a 0 chance that the door that was revealled contains kittens. So there's a 2/3 chance the remaining door contains kittens.

So if you stay, you win 1/3. If you change, you win 2/3.

But it took me about half an hour of arguing to figure it out Wikipedia has a good article on it apparently.

yellowtamarin
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You choose one door. You have a 1/3 chance of that door being correct and a 2/3 chance one of the other doors is correct.

One of the other doors is revealed to not contain kittens.

Your explanation is true if the chooser doesn't know that a losing door is going to be opened after they choose. But if they already know this is going to happen, they are essentially only chosing between 2 doors. The quote above can be swapped around the other way and will look like this:

"One of the doors is revealed to not contain kittens.

You choose one door. You have a 1/2 chance of that door being correct and a 1/2 chance the other door is correct."

You can only swap it around because you already know a door is going to be removed after you choose.
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yellowtamarin
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Sorry SpitValve for not quoting you properly, I haven't done a post in a forum for about 8 years.

I've read through the Wikipedia explanation, including the part that tries to tell me where I'm "going wrong" and I still disagree

If I were the contestant on the quiz show, and I had watched it many times before so I knew what was going to happen, I would be choosing between a door with a car or a door with a goat. One of the goat doors is going to be removed so it is irrelevant.

Wikipedia says you can't ignore the past, meaning you can't change the odds after one door has been opened. I agree, but I say the odds were always 50/50.

BTW I actually originally thought the chance was 1/3 v 2/3. I didn't think it was 50/50 until I thought about it more (a lot).
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Lothar
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Yellow, run simulations. Do a thousand runs sticking with your door, then a thousand switching to the other door. Now pretend you don't know that the host is going to open one of the other doors (though he always does) and repeat the experiment. Do you actually expect different results?
Last edited by Lothar on Wed Mar 28, 2007 9:25 am UTC, edited 1 time in total.
Always program as if the person who will be maintaining your program is a violent psychopath that knows where you live.

If you're not part of the solution, you're part of the precipitate.

1+1=3 for large values of 1.