PARADE!

[boomzilla]

@TDWTF123 said:

Fucksake, are you a complete moron or just not reading? No-one's arguing about how to solve either problem. The question is which one the problem actually is. Why fucking three cards?

The problem, as commonly stated, uses three doors. So for a directly analogous experiment, you would use 3 cards. That's the minimum number needed. You can adjust the number cards higher if you like (NB: it requires upping the number of cards the host reveals by an equal amount, too), but why make things more complicated than they need to be when working with physical objects like that?

You cannot possibly demonstrate anything about MHP with only 2 cards. If you only have 2 cards, like you keep insisting, then you've changed the problem and the answer. If you only have 2 cards, how do you get a expected values of 1/3 and 2/3 for the two strategies (THAT would be question begging if we hadn't already derived the answer multiple times in this thread, but we did, so no begging here).

[Hmmmm]

@TDWTF123 said:

Filed under: Where's Ronald? He's not this dumb.

I suspect he is keeping well out of this thread because of that very fact.

[Quinnum]

@boomzilla said:

The continued asininity on TDWTF123's part (and politeness on mine) says pure trolling.

Certainly this is becoming the new SpectateSwamp thread.

[flabdablet]

@TDWTF123 said:

What we are discussing, once again, is not how to solve those two problems, because anyone who's not innumerate can see how to do that. We're discussing which of those two problems the MHP actually represents.

You have two problems? Everybody else here only sees one. Did you perhaps try to solve the MHP with a regexp?

[PJH]

@Hmmmm said:

@TDWTF123 said:

Filed under: Where's Ronald? He's not this dumb.

I suspect he is keeping well out of this thread because of that very fact.

He's not the only one.

[flabdablet]

@boomzilla said:

The problem, as commonly stated, uses three doors.

Suppose you're on a web forum, and you're given the choice of three threads: behind one thread is an idiot; behind the others, trolls. You pick a thread, say No. 1, and the moderator, who knows who's behind the threads, opens another thread, say No. 3, which has a troll. He then says to you, "Do you want to reply in thread No. 2?" Is it to your advantage to switch your choice?

[Hmmmm]

 @flabdablet said:

@boomzilla said:

The problem, as commonly stated, uses three doors.

Suppose you're on a web forum, and you're given the choice of three threads: behind one thread is an idiot; behind the others, trolls. You pick a thread, say No. 1, and the moderator, who knows who's behind the threads, opens another thread, say No. 3, which has a troll. He then says to you, "Do you want to reply in thread No. 2?" Is it to your advantage to switch your choice?

That really depends on whether you prefer arguing with idiots or trolls (assuming you can tell the difference and there is actually a difference to tell).

[dhromed]

@flabdablet said:

Is it to your advantage to switch your choice?

 

You drive a hard bargain, Mr. Hall.

[boomzilla]

@flabdablet said:

@boomzilla said:

The problem, as commonly stated, uses three doors.

Suppose you're on a web forum, and you're given the choice of three threads: behind one thread is an idiot; behind the others, trolls. You pick a thread, say No. 1, and the moderator, who knows who's behind the threads, opens another thread, say No. 3, which has a troll. He then says to you, "Do you want to reply in thread No. 2?" Is it to your advantage to switch your choice?

Accepting such a false choice is not the way to a 6 digit post count on TDWTF, I can tell you.

[dhromed]

@boomzilla said:

So for a directly analogous experiment, you would use 3 cards. That's the minimum number needed.

 

Hell, you don't even need playing cards, actually. It'll take too long. Just do the following:

- copypaste this bit 30 times: [_] [_] [_] (Each group is a game show. 30 should be statistically significant and the whole thing doable in ~10 minutes)
- add a column where you mark wins. You'll be switching all the time. The non-switch is the complement, obviously.

For each game show:
-- load the prize into a door by rolling a dice (or hit up random.org), and put a dot in it. (It's okay if you know the winning door, because you don't control the dice. It won't affect the outcome)
-- pick the first door every time. (This is to prevent human selection bias. Your randomness is shit. The dice guarantees random placement anyway, and your own picking can't add any randomness)
-- cross out the next empty door. Monty has opened it and that door is unavailable for you.
-- switch to the next available door.
-- mark down whether your final door has the prize

Due to a miracle proper statistics, the marks you put down will show ~2/3 wins.

I am purposely holding back my hideous javascript because people will just argue about how the implementation is wrong and I don't want a protracted multidimensional argument, especially with poor live goats waiting behind doors.

Excuses: I've rewritten this post 30 times, so there's a 2/3 chance there are problems with the algorithmic MHP.

[dhromed]

@TDWTF123 said:

Why fucking three cards? [...] explain why all three doors are relevant.

 

Your first choice, Monty's choice, and your switch-choice are each dependent on the previous choice.

Monty can't pick your door.

You can't switch to the door that Monty picked.

[TDWTF123]

@flabdablet said:

@TDWTF123 said:

What we are discussing, once again, is not how to solve those two problems, because anyone who's not innumerate can see how to do that. We're discussing which of those two problems the MHP actually represents.

You have two problems? Everybody else here only sees one.

You seem determined to prove that you've been responding without reading a single one of my posts. I can't imagine how you've missed such a blatant point so spectacularly otherwise.

As I said at the start, and at least a dozen times since, the point of the MHP is to trick people into thinking it's a more complicated question than it really is. People who don't think the probability is two-thirds do not necessarily have trouble with conditional probability: they've merely seen through the trick and noticed that there are actually only two relevant doors and so it isn't a conditional probability problem.

[TDWTF123]

@dhromed said:

Your first choice, Monty's choice, and your switch-choice are each dependent on the previous choice.

OK, good, now we're getting somewhere. The dependency matters because...?

[boomzilla]

@TDWTF123 said:

As I said at the start, and at least a dozen times since, the point of the MHP is to trick people into thinking it's a more complicated question than it really is. People who don't think the probability is two-thirds do not necessarily have trouble with conditional probability: they've merely seen through the trick and noticed that there are actually only two relevant doors and so it isn't a conditional probability problem.

The "trick" is in understanding the conditions that give the values for the strategies. There are, of course, three relevant doors. Two are losers, one is a winner. What does the word "relevant" mean in your head?

[flabdablet]

@TDWTF123 said:

People who don't think the probability is two-thirds do not necessarily have trouble with conditional probability: they've merely seen through the trick and noticed that there are actually only two relevant doors and so it isn't a conditional probability problem.

Not at all. People who think the probability is 1/2 think they've seen a shortcut to the answer when they haven't, because in fact the entire procedure is relevant.

That's the trick.

[flabdablet]

@TDWTF123 said:

The dependency matters because...?

It makes your final choice a choice between an outcome with probability 1/3 of winning a car (don't switch) and another with probability 2/3 of winning a car (switch).

[TDWTF123]

@flabdablet said:

@TDWTF123 said:

People who don't think the probability is two-thirds do not necessarily have trouble with conditional probability: they've merely seen through the trick and noticed that there are actually only two relevant doors and so it isn't a conditional probability problem.

Not at all. People who think the probability is 1/2 think they've seen a shortcut to the answer when they haven't, because in fact the entire procedure is relevant.

That's the trick.

No. The trick is in getting people who think they know maths to think what you just said by presenting an answer that is 'obviously' wrong for the reason you gave - but then it turns out that is the answer anyway because it's a degenerate case.

But if you disagree, feel free to state why rather than just saying so.

[TDWTF123]

@flabdablet said:

@TDWTF123 said:

The dependency matters because...?

It makes your final choice a choice between an outcome with probability 1/3 of winning a car (don't switch) and another with probability 2/3 of winning a car (switch).

Was that intended to be a joke, or are you really that dumb? Either way, total failure on your part.

[dhromed]

@TDWTF123 said:

The dependency matters because...?

 

Because if you're holding on to the door, you can't re-roll the doors. Its contents, goat or car, are now fixed. After Monty opens his door, you're not in a fresh random scenario.

If, and only if,  the doors re-roll after Monty clears out one of them, do you end up with 50/50. Because then the door was indeed irrelavant. But they don't re-roll. So you stay with the original three door situation.

[dhromed]

@TDWTF123 said:

@flabdablet said:

@TDWTF123 said:

The dependency matters because...?

It makes your final choice a choice between an outcome with probability 1/3 of winning a car (don't switch) and another with probability 2/3 of winning a car (switch).

Was that intended to be a joke, or are you really that dumb? Either way, total failure on your part.

 

It's how the MHP works, though.

Have you tried my experiment yet?

Because if the third card is irrelevant, or if there are any shenanigans in the problem, the execution of the experiment will show this once you are either unable to proceed, or once you've added up the final tally of the odds.

Sometimes when I write some code, I create a variable that turns out to be entirely irrelevant later on, but in my initial idea of the program, it wasn't obvious. So eventually I remove the variable, but that's only after I've clearly established that the variable is unused after writing the code, something that I was unable to do merely by thinking about the code.

So.

Experiment.

Experiment. Experiment. Experiment. Experiment. Experiment. 

Experiment. 

Experiment. Experiment. Experiment. 

Experiment. 

[TDWTF123]

@dhromed said:

After Monty opens his door, you're not in a fresh random scenario.

I never suggested you are.
@dhromed said:

@TDWTF123 said:

The dependency matters because...?

 

Because if you're holding on to the door, you can't re-roll the doors.

Any other reason, since you've imagined the necessity to re-roll(?) something?

Hint: what differentiates these three doors from the fire-exit, or Monty Hall's dressing room door, for example?

[Lorne Kates]

@TDWTF123 said:

As I've repeatedly stated, the trick is that you're told there are three doors, but only two actually matter.

 

<font size="26">AND YOU DO NOT KNOW WHICH TWO MATTER UNTIL AFTER YOU PICK YOURS!</font>

[TDWTF123]

@dhromed said:

Because if the third card is irrelevant, or if there are any shenanigans in the problem, the execution of the experiment will show this once you are either unable to proceed, or once you've added up the final tally of the odds.

No, that's just plain wrong. However you have interpreted the starting conditions, all you're doing is proving that a certain result follows from that interpretation.

Plainly, if you set it up with only two cards to start with, as you should, the outcome is different.

@dhromed said:

Have you tried my experiment yet?

No, because you didn't even manage to describe the MHP as you see it. Plainly you have no idea just how different your algorithm is to the MHP as you see it, or you wouldn't have written it, but I'm afraid you badly screwed it up. The whole point is the knowledge asymmetries, and you've altered them whilst removing Monty's freedom of action (such as it was).

[TDWTF123]

@Lorne Kates said:

@TDWTF123 said:

As I've repeatedly stated, the trick is that you're told there are three doors, but only two actually matter.

 

<font size="26">AND YOU DO NOT KNOW WHICH TWO MATTER UNTIL AFTER YOU PICK YOURS!</font>

If it's that obvious, you won't have any trouble explaining why that is, will you?

I get the impression this thread's full of people who actually don't understand the maths behind conditional probability, which is why you're all shouting and blustering rather than clearly working through the chain of logic.

[Faxmachinen]

@TDWTF123 said:

People who don't think the probability is two-thirds do not necessarily have trouble with conditional probability: they've merely seen through the trick and noticed that there are actually only two relevant doors and so it isn't a conditional probability problem.

In case you're too dimwitted to figure it out, the original 1/3 per door comes from an assumption. Since the process of reducing the three doors into two is well known (i.e. in the problem description), you cannot make a similar assumption (i.e. 1/2) for the two doors.

@TDWTF123 said:

Admittedly I don't know how to do maths,

Maybe you should learn it.

[dhromed]

@TDWTF123 said:

@dhromed said:

Because if the third card is irrelevant, or if there are any shenanigans in the problem, the execution of the experiment will show this once you are either unable to proceed, or once you've added up the final tally of the odds.

No, that's just plain wrong.

 

You've just countered the scientific method. Good job.

@TDWTF123 said:

all you're doing is proving that a certain result follows from that interpretation

My "interpretation" is word-for-word literal.

[error]

<html><body style='color:#000000; background:#ffffff; '>
function rand( minInclusive, maxInclusive ) {
    return Math.floor( Math.random() * maxInclusive - minInclusive + 1 ) + minInclusive;
}

function montyHall( switchDoors ) {
    // Suppose you're on a game show, and you're given the choice of three doors:
    // behind one door is a car; behind the others, goats
    var doors = { 1: 'goat', 2: 'goat', 3: 'goat' };
    doors[ rand( 1, 3 ) ] = 'car';
    
    // You pick a door, say No. 1,
    var myDoor = 1; // the first door
    
    // and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat.
    var revealedDoor;
    do {
        revealedDoor = rand( 1, 3 ); // pick a random door
    } while(
        revealedDoor === myDoor // if we picked the same door as the contestant
    ||  doors[ revealedDoor ] !== 'goat' // or we didn't reveal a goat, then pick another
    );
    
    // He then says to you, "Do you want to pick door No. 2?"
    var switched = false; 
    if( switchDoors ) {
        for( var i = 1; i <= 3; ++i ) { // find the door...
            if( i !== myDoor // ... that wasn't the contestant's first pick ...
            &&  i !== revealedDoor // ... and wasn't the one the host revealed ...
            ) {
                myDoor = i; // ... and switch to that door
                switched = true;
                break;
            }
        }
    } 
    
    if( switched !== switchDoors ) {
        throw 'Something went wrong.'; // make sure our preference was honored
    }
    
    // Is it to your advantage to switch your choice?
    return doors[ myDoor ] === 'car';
}

function testRng( iterations ) {
    var i;
    var results = { 1: 0, 2: 0, 3: 0 };
    for( i = 0; i < iterations; ++i ) {
        ++results[ rand( 1, 3 ) ];
    }
    console.log( 'Tested RNG for ' + iterations + ' iterations:' );
    for( i in results ) if( results.hasOwnProperty( i ) ) {
        console.log( i + ': ' + ( results[ i ] / iterations ) );
    }
}

function simulate( switchDoors, iterations ) {
    var total = 0;
    for( var i = 0; i < iterations; ++i ) {
        total += montyHall( switchDoors ) ? 1 : 0;
    }
    console.log( 'Simulated for ' + iterations + ' iterations:' );
    console.log( 'Probability of winning after ' + ( switchDoors ? '' : 'not ' ) + 'switching: ' + ( total / iterations ) );

}


var iterations = 100000;
testRng( iterations );
simulate( false, iterations );
simulate( true, iterations );


Example Output:

Tested RNG for 100000 iterations:
1: 0.33086
2: 0.33343
3: 0.33571
Simulated for 100000 iterations:
Probability of winning after not switching: 0.33227
Simulated for 100000 iterations:
Probability of winning after switching: 0.66659

[dhromed]

@TDWTF123 said:

I get the impression this thread's full of people who actually don't understand the maths behind conditional probability, which is why you're all shouting and blustering rather than clearly working through the chain of logic.

 

Do the experiment. Don't reinterpret the first sentence by changing three into two, just use the three doors as described, and see what happens.

[boomzilla]

@TDWTF123 said:

Hint: what differentiates these three doors from the fire-exit, or Monty Hall's dressing room door, for example?

Because one of those three doors (mentioned in the problem) is the winning door. Maybe one of them actually is the fire exit, but that isn't relevant, because maybe the prize is out back behind it. These 3 doors are the doors that determine whether you are a winner or a loser. The other doors are not relevant (you seem to have that right).

But, for the sake of trolling, which two doors are the doors that you believe are relevant? I don't recall a straight answer about that. Perhaps I'll start one upping you and asserting that oneno doors are relevant! Beat that!

[boomzilla]

@TDWTF123 said:

I get the impression this thread's full of people who actually don't understand the maths behind conditional probability,

Were you looking in the mirror when posting this?

[Hmmmm]

 @TDWTF123 said:

Hint: what differentiates these three doors from the fire-exit, or Monty Hall's dressing room door, for example?

The fact that the problem description strongly implies (I would say "says" but you would argue that it doesn't say that and then I would have to argue that any other interpretation wouldn't be considered a "fair" game in the context of a game show and then the argument would shift into whether game shows actually are "fair" and, while that might be more amusing than the current discussion, it would be a whole other bag of WTF) that those three doors have a 1 in 3 chance of having a car behind them and you only get the option of picking one of those three anyway.

 

[TDWTF123]

@dhromed said:

@TDWTF123 said:

@dhromed said:

Because if the third card is irrelevant, or if there are any shenanigans in the problem, the execution of the experiment will show this once you are either unable to proceed, or once you've added up the final tally of the odds.

No, that's just plain wrong.

 

You've just countered the scientific method. Good job.

Seriously? What the hell do you think you're proving experimentally there? What's your null hypothesis? Because it sure as hell isn't what I've said.

@dhromed said:

@TDWTF123 said:

all you're doing is proving that a certain result follows from that interpretation

My "interpretation" is word-for-word literal.

What the hell does that matter? You're still only proving what you've already assumed. It's entirely circular. If I start with two cards, that also doesn't prove anything.

[dhromed]

@TDWTF123 said:

I never suggested you are.

 

You are forced to, since it's the only way to get 50/50.

@TDWTF123 said:

re-roll(?)

Sorry, I thought the term was clear. Re-randomize is what I mean.

@TDWTF123 said:

Hint: what differentiates these three doors from the fire-exit, or Monty Hall's dressing room door, for example?

Because the prize is guaranteed to not be placed behind one of those.

 

 

[TDWTF123]

@dhromed said:

@TDWTF123 said:

I get the impression this thread's full of people who actually don't understand the maths behind conditional probability, which is why you're all shouting and blustering rather than clearly working through the chain of logic.

 

Do the experiment. Don't reinterpret the first sentence by changing three into two, just use the three doors as described, and see what happens.

Are you a complete dimmock? Obviously, two-thirds of the time switching will win. That's simply going to follow from your starting conditions. The question is whether those starting conditions are correct, and you can't prove or disprove that experimentally.

[Lorne Kates]

@TDWTF123 said:

If it's that obvious, you won't have any trouble explaining why that is, will you?

 

You can lead an idiot to water...

You are an idiot because you can't do the math yourself. You are an idiot because you can't see that the host eliminates a losing door but *DOES NOT MAKE A DECISION*, and thus does not branch a probability tree. You are an idiot because you can't do simple branching or use fucking Excel.

The motherfucking answer, you idiot: https://docs.google.com/spreadsheet/ccc?key=0AvvZs8EfItY_dHRTbVBzbkpOWkt5V1haSzZKX0Q4TFE&usp=sharing

When you don't switch, there are 6 possible outcomes, only 2 of which result in you winning.

When you DO switch, <font size="60">there are only 3 possible outcomes,</font> 2 of which result in you winning.

You idiot.

[TDWTF123]

@dhromed said:

@TDWTF123 said:

I never suggested you are.

 

You are forced to, since it's the only way to get 50/50.

Fucking hell, are you really this thick? That's only true once you've already assumed your starting conditions are correct. Now please stop assuming and start demonstrating.
@dhromed said:

@TDWTF123 said:

Hint: what differentiates these three doors from the fire-exit, or Monty Hall's dressing room door, for example?

Because the prize is guaranteed to not be placed behind one of those.

Are you doing this on purpose? I can lead you to the answer, but I can't understand it for you. Why does it matter that the prize is guaranteed not to be behind any particular door?

[TDWTF123]

@Hmmmm said:

 @TDWTF123 said:

Hint: what differentiates these three doors from the fire-exit, or Monty Hall's dressing room door, for example?

The fact that the problem description strongly implies (I would say "says" but you would argue that it doesn't say that and then I would have to argue that any other interpretation wouldn't be considered a "fair" game in the context of a game show and then the argument would shift into whether game shows actually are "fair" and, while that might be more amusing than the current discussion, it would be a whole other bag of WTF) that those three doors have a 1 in 3 chance of having a car behind them and you only get the option of picking one of those three anyway.

 

No, this isn't an argument about practicalities. If you want to be that literal, the problem never states anywhere that the doors are opaque. I'm happy to assume that much, though :)

[dhromed]

@TDWTF123 said:

What the hell do you think you're proving experimentally there?

 

That a literal execution of the problem text will result in 2/3 win chance for switching.

@TDWTF123 said:

all you're doing is proving that a certain result follows from that interpretation

@TDWTF123 said:

@dhromed said:

My "interpretation" is word-for-word literal.

What the hell does that matter?

Because the first step in demonstrating a problem with in a description is by taking it literally at first, and then ending up at a contradiction or similarly nonsensical result.

@TDWTF123 said:

If I start with two cards, that also doesn't prove anything.

It demonstrates that you're re-interpreting the first sentence of the problem in a non-literal way, in advance.

 

[TDWTF123]

@Lorne Kates said:

the host eliminates a losing door but DOES NOT MAKE A DECISION

How can you see that, unlike the others, and still have trouble with what I'm saying? It's an obvious corollary of what you've just agreed with.@Lorne Kates said:

When you don't switch, there are 6 possible outcomes, only 2 of which result in you winning.

When you DO switch, <font size="60">there are only 3 possible outcomes,</font> 2 of which result in you winning.

Nonsense. That's only true if there are three doors. But as you've just said, Monty has no choice: there are only two relevant doors. In which case, there are only four possible outcomes, two of which result in you winning, and two in you losing.

[flabdablet]

There is no need for complicated simulations. Just count the outcomes of all the equally likely cases.

1. You pick door A. Monty knows car is behind A and opens B or C. Keeping your initial door wins; switching loses.

2. You pick door A. Monty knows car is behind B and must open C. Keeping your initial door loses; switching wins.

3. You pick door A. Monty knows car is behind C and must open B. Keeping your initial door loses; switching wins.

4. You pick door B. Monty knows car is behind A and must open C. Keeping your initial door loses; switching wins.

5. You pick door B. Monty knows car is behind B and opens A or C. Keeping your initial door wins; switching loses.

6. You pick door B. Monty knows car is behind C and must open A. Keeping your initial door loses; switching wins.

7. You pick door C. Monty knows car is behind A and must open B. Keeping your initial door loses; switching wins.

8. You pick door C. Monty knows car is behind B and must open A. Keeping your initial door loses; switching wins.

9. You pick door C. Monty knows car is behind C and opens A or B. Keeping your initial door wins; switching loses.

In cases 1, 5 and 9 where Monty's choice of door to open is unforced, his choice can't affect your chance of winning.This is why there are only nine equally likely cases, not twelve. In six of those nine cases, switching wins. Therefore switching is to your advantage. QED.

[TDWTF123]

@dhromed said:

@TDWTF123 said:

What the hell do you think you're proving experimentally there?

 

That a literal execution of the problem text will result in 2/3 win chance for switching.

@TDWTF123 said:

all you're doing is proving that a certain result follows from that interpretation

@TDWTF123 said:

@dhromed said:

My "interpretation" is word-for-word literal.

What the hell does that matter?

Because the first step in demonstrating a problem with in a description is by taking it literally at first, and then ending up at a contradiction or similarly nonsensical result.

@TDWTF123 said:

If I start with two cards, that also doesn't prove anything.

It demonstrates that you're re-interpreting the first sentence of the problem in a non-literal way, in advance.

 

I thought we were getting somewhere before, but you're right back round the circle again. Once more, what we're arguing about is the correct interpretation. All your experiment proves is that the two-thirds answer follows from your interpretation, which is the only thing you and I agreed on to start with.

[dhromed]

@TDWTF123 said:

Why does it matter that the prize is guaranteed not to be behind any particular door?

Because the contestants already know that the prize is not behind them, so they don't factor in the problem.

You could start with five doors, if you like, and have Monty open two of them and reveal they're empty, and then you make your pick from the remaining three.

 

 

[Lorne Kates]

 In fact, I'll do you one better.

I will set up a website that simulates the MHP. I will hook it up to Paypal. I will let TDWTF play it. I will give him even payouts-- be bets $1, I will pay him $1 if he wins-- BUT ONLY if he agrees to never switch.

After all, this is a 50/50 game according to him, so even money is fair.

If he's right, at the end of, say, $10,000 worth of play, we should be even.  Since it's a coin-flip game, I believe $10k of play should be enough to hit the long term, and we'll be within 1 standard dev. Neither of us will have really lost anything.

Or if I'm right, which I obviously am not-- he'll be out around $3K and an idiot will have been rightfully parted from his money.

 

[dhromed]

@TDWTF123 said:

Once more, what we're arguing about is the correct interpretation. All your experiment proves is that the two-thirds answer follows from your interpretation,

 

Why are you opposed to the literal interpretation?

[flabdablet]

@Lorne Kates said:

You idiot.

I picked troll to start with and I'm going to stick with that thanks Monty.

[Lorne Kates]

@TDWTF123 said:

Nonsense. That's only true if there are three doors. But as you've just said, Monty has no choice: there are only two relevant doors. In which case, there are only four possible outcomes, two of which result in you winning, and two in you losing.

 

Show me an Excel, you idiot. Or just keep repeating your idiotic idocy like an idiot. The burden of proof is on you, the idiot. The rest of the world can do math-- and has done math. You're just being an idiot.

[Lorne Kates]

@flabdablet said:

@Lorne Kates said:

You idiot.

I picked troll to start with and I'm going to stick with that thanks Monty.

Your avatar is scary.

[flabdablet]

@dhromed said:

Why are you opposed to the literal interpretation?

The devil may quote Scripture for his own purpose.

[flabdablet]

@Lorne Kates said:

Your avatar is scary.

Sorry, I'll change it.

[dhromed]

@TDWTF123 said:

Nonsense. That's only true if there are three doors. But as you've just said, Monty has no choice: there are only two relevant doors. In which case, there are only four possible outcomes, two of which result in you winning, and two in you losing.

 

lol

I'm going home. Enjoy your goat.