And Behind Door No. 1, a Fatal Flaw

[Aggie87]

I got it. Thanks.

Door A (my initial pick) = 33% chance of car

Doors B + C (the ones I didn't pick) = 67% chance of car

When Monty opens one of the other doors, it still means my door has a 33% chance, and the remaining door has the 67%.

I understood that logic from the beginning, I just wasn't sure I was accepting of it.

Thanks Swede & Dan (& Sen. Rockefeller).

[Dan Gould]

I got it. Thanks.

Door A (my initial pick) = 33% chance of car

Doors B + C (the ones I didn't pick) = 67% chance of car

When Monty opens one of the other doors, it still means my door has a 33% chance, and the remaining door has the 67%.

I understood that logic from the beginning, I just wasn't sure I was accepting of it.

Thanks Swede & Dan (& Sen. Rockefeller).

We'll split the Fellowship money.

[JSngry]

But what if you had the car in the first place & lost it by "playing the odds"? Wouldn't you feel like a dumbass or something?

[RDK]

But what if you had the car in the first place & lost it by "playing the odds"? Wouldn't you feel like a dumbass or something?

Hell, you're already wearing a Bozo costume or something - how much more of a dumbass could you feel?

[RDK]

This is really mind-fucking me. I know the numbers add up, but it's so counterintuitive to me that it's giving me a headache.

[Jim Alfredson]

Makes sense to me.

[JSngry]

Ok, here's the deal - yeah, ok, over the long haul, over multiple shots, the 1-in-3 vs 2-in-3 thing plays out.

But...

In real life, you got only one shot to be right. And the question is - Is this one of the times when the odds work for me or against me? And at that one moment, you got a 50-50 chance of it being so, since there's only two answers to that question - yes or no.

Hey, maybe it's not scientific. But it works for me.

[Aggie87]

Ok, here's the deal - yeah, ok, over the long haul, over multiple shots, the 1-in-3 vs 2-in-3 thing plays out.

But...

In real life, you got only one shot to be right. And the question is - Is this one of the times when the odds work for me or against me? And at that one moment, you got a 50-50 chance of it being so, since there's only two answers to that question - yes or no.

Hey, maybe it's not scientific. But it works for me.

That's not correct, but it's what I was trying to do too, LOL.

It's only 50/50 for the guy who walks up at the point where there are two doors left, and he is allowed to pick one of the two doors.

For the person who was there when there were 3 doors, his likelihood of winning the car is still better if he switches - by a 2 to 1 margin.

It's counterintuitive for sure.

[John L]

I haven't run the simulation, Dan - can't access it from this computer. I also understand the arguments that have been shared here, and they make sense.

I'm still saying that Monty's opening of a door with a goat behind it leaves 2 closed doors, one with a car behind it. It seems immaterial which door you chose first, KNOWING that you will again have to choose one of the two doors that are left. The data at that point in time is that you have two closed doors, and a car behind one of them - and that's it. That's what you have to go by AT THAT POINT IN TIME.

I'm not arguing that you're wrong. I just am not convinced by the logic, whether it's true or not.

The basic point to make the intuition jive with the reality is the following: Yes, you know before hand that Monty is going to show you a door with a goat. But you don't know which door. When you see which door, that reveals valuable information. Furthermore, which door Monty picks also depends, with some probability, on the first door that you choose.

So your initial choice of a door gives a chance of winning of 1/3 if your strategy is to stick with it and not use any of the revealed information.

In this case, you also know that with probability 2/3, it is behind one of the two other doors.

After updating your information following the revelation of a door that has a goat (one of the two that you didn't choose), you can now place a 2/3 probability that the car is behind the other door. You initially placed a 2/3 probability that it was behind one of the two. Now you know which one of the two to give the 2/3 probability.

Edited April 10, 2008 by John L

[Jazzmoose]

All I know is that this thread is useless without the word "zonk"...

[JSngry]

Ok, here's the deal - yeah, ok, over the long haul, over multiple shots, the 1-in-3 vs 2-in-3 thing plays out.

But...

In real life, you got only one shot to be right. And the question is - Is this one of the times when the odds work for me or against me? And at that one moment, you got a 50-50 chance of it being so, since there's only two answers to that question - yes or no.

Hey, maybe it's not scientific. But it works for me.

That's not correct, but it's what I was trying to do too, LOL.

It's only 50/50 for the guy who walks up at the point where there are two doors left, and he is allowed to pick one of the two doors.

For the person who was there when there were 3 doors, his likelihood of winning the car is still better if he switches - by a 2 to 1 margin.

It's counterintuitive for sure.

The more I think about it, the more I think it is correct, seriously, because the question here is not "Which door is the car behind?", or "Over the long haul, is it a good idea to change picks?", it's now "Ok, I've made a decision - say, to change my pick - that is going to be right 2-out-of-3 times, Is this going to be one of those 2 times?"

There's only two answers here, yes or no. When it's no longer about the overall, long-term success rate but instead about the immediate, one-time possibility of good move vs bad move, that's a subtly but truly different criterion. Long term, yeah, you got 2-out-of-3. But the nature of the game is that you got one shot, period. No do-overs or multiple choices.

Since you only get one shot, you don't have a chance to play it out long-term to see how it averages. So what are the odds that the odds will come through the one and only time you call upon them? Gotta be 50-50, because the only possibilities are they will or they won't. "They should" or "they probably will" are not options, because those are not results, those are hopes. Even if they're based on statistically sound reasoning, they're still simply hopes. Sounder hopes to be sure, but still, ultimately, just hopes. They're in no way a result.

So yeah, it's no doubt smart to make the change, but it's not smart to think that doing so automatically doubles your chances or anything unless you're playing repeatedly over time, which you don't do, not in the "real-world" scenario of being a contestant on Let's Make A Deal. You get one shot, period. You play the odds sure, but it's wither going to work or not, and then it's over, period.

It's a gamble either way, and although in theory one choice is better than the other over the long haul, anybody who gambles will tell you that the only way the odds really work out as they should is over time, never, ever in a right-here-right-now moment. There ain't no such thing, and as somebody who's had 4 kings beaten by a straight flush (at no small expense, btw), I'm here to tell you that the moment is where you win or lose and over time is by how much. A subtle but very real distinction.

[Dan Gould]

Your odds do in fact double by making the switch. Period. There is no other way to describe the odds and be mathematically correct.

You are right or wrong in the instant that Monty reveals the door you've chosen, but by switching, your odds have measurably improved, from 1/3 to 2/3.

Just because 2/3 is not 3/3 or 100% does not mean that you should not switch. And it definitely does not mean that your odds at that moment are 50/50.

[Aggie87]

As far as I can tell, your logic is still the same as it would be if you were the person who walked up to the game show with only two doors remaining, and got to pick one of them. Then it's a 50/50 chance.

If you were the person standing there the whole time, there's STILL a 66% chance that the car is behind the other door.

The other folks here have done alot better job of explaining it already, than I could.

[JSngry]

Put another way, 2/3 is the measurement of what will usually win. 50/50 is the measurement of that "usuality" occurring any one time in isolation/absence of any second (or more) chances.

[Aggie87]

Put another way, 2/3 is the measurement of what will usually win. 50/50 is the measurement of that "usuality" occurring any one time in isolation/absence of any second (or more) chances.

I'm not sure usuality is a word (it's a fuquitous one if it is! ), but the measurement would still be 66/33 in favor of switching for that one isolated decision.

[Niko]

so if you have a disease that, say, kills 90% of those who have it within one year; and then you get a possibility to exchange this disease against one that only kills 2% within that same year, you wouldn't do it?

(you may argue that this type of thing doesn't happen too often)

when people tell me i shouldn't be afraid of flying because the odds that something will happen are ... (very good number), i always say it doesn't impress me because i will fly only once and when the disaster happens i won't care how unlikely it was... but in that goat thing where you can really improve the odds its stupid not to do so, (again, that's like taking a motorcycle ride although you don't enjoy it if you could have walked instead)

[JSngry]

Somebody explain to me how a question - Did I make the right choice for this one occasion? - that can only have two answers - yes or no - translates into anything other than 50-50 odds.

Again, remember that we're not measuring the probability of the choice itself being smart, we're measuring the chances of the certainty that it will be correct in the one and only time we put it to the test.

These are not the same things.

"Improving your odds" does not equate with "guaranteeing a win" or "preventing a loss".

And again - in the real-world scenario of the game show, you only get one shot to be right. So the best you can do is "improve your odds", which still leaves you with only one of two outcomes.

I've seen motherfuckers go broke "playing the odds", and I've seen 'em get rich (for a little while anyway ) going against them. Ultimately, no matter how well you position yourself (and make no mistake, I am always in favor of doing so), it always, always comes down to this time as to whether or not you got it right. And when there's only one "this time", hey...

[JSngry]

Put another way, 2/3 is the measurement of what will usually win. 50/50 is the measurement of that "usuality" occurring any one time in isolation/absence of any second (or more) chances.

I'm not sure usuality is a word (it's a fuquitous one if it is! ), but the measurement would still be 66/33 in favor of switching for that one isolated decision.

AHA!

EXACTLY!

Yes, in favor of switching - not in favor of it actually working one time and one time only.

Yu're trying to fit a 2-out-of-3 peg (probability of outcomes) into a 1-out-of-2 hole (definitely possible outcomes). They won't fit, nor should they, because they're not the same thing.

[Niko]

Somebody explain to me how a question - Did I make the right choice for this one occasion? - that can only have two answers - yes or no - translates into anything other than 50-50 odds.

i vaguely get what you mean; would you say, playing a lottery (together with a million others and only one of us gets a price) is a 50-50 situation for each of us, because each of us either wins or loses... ?

[JSngry]

...but in that goat thing where you can really improve the odds its stupid not to do so...

Indeed.

But that's not my point. My point is that the seeming "discredited" "50-50 logic" is not entirely without basis. It's not relevant to what the smart choice is, but it is relevant to whether or not the smart choice will be the winning choice when there's only one chance to play.

It's probably abstract beyond redemption but there is a distinction between the two criteria, and it kinda bugs me when "logic" is trumpeted as such without the parameters being as clearly defined as they probably :g :g should be.

[Aggie87]

If a car is placed behind one of three doors, randomly, and if you were offered a choice of what's behind (Door A) or what's behind (Door B+Door C), you'd pick (Door B+DoorC) wouldn't you? Just about every time?

It's a 50/50 choice (i.e. either the one door or the set of two doors), but the odds of winning the car by picking the two door set is greater. By a 2 to 1 margin.

[JSngry]

Somebody explain to me how a question - Did I make the right choice for this one occasion? - that can only have two answers - yes or no - translates into anything other than 50-50 odds.

i vaguely get what you mean; would you say, playing a lottery (together with a million others and only one of us gets a price) is a 50-50 situation for each of us, because each of us either wins or loses... ?

Although that's in no way comparable to the singular experience of the game show in question...no I would not say that. Because in the lottery, it's a matter of "if I win, then XYZ number of people must lose". On the game show, the question is simply "will I win or will I lose". There are no other competitors.

[JSngry]

If a car is placed behind one of three doors, randomly, and if you were offered a choice of what's behind (Door A) or what's behind (Door B+Door C), you'd pick (Door B+DoorC) wouldn't you? Just about every time?

It's a 50/50 choice (i.e. either the one door or the set of two doors), but the odds of winning the car by picking the two door set is greater. By a 2 to 1 margin.

That's the odds of winning the car. What are the odds that you made the right choice to do so?

[Swinging Swede]

And at that one moment, you got a 50-50 chance of it being so, since there's only two answers to that question - yes or no.

Gotta be 50-50, because the only possibilities are they will or they won't.

According to the two quotes above you seem to be under the common misperception that if there are two choices, then it must be a 50/50 chance. That's just not so. If that were the case, then Monty would also have a 50/50 chance, even if he actually knows which door the car is behind.

So yeah, it's no doubt smart to make the change, but it's not smart to think that doing so automatically doubles your chances or anything unless you're playing repeatedly over time, which you don't do, not in the "real-world" scenario of being a contestant on Let's Make A Deal. You get one shot, period.

Switching doors does automatically double your chances. Keeping the door gives you a 33 % chance, switching to the other door gives you a 67 % chance. That's doubling your chances.

You also seem to make a distinction between playing the game once and multiple times, in that you accept that over time it would be a 67 % chance. But playing the game 100 times and winning 67 times is the same as having a 67 % chance if you do it once. Those 100 times are just 100 single shots counted together.

Apologies if I misunderstood your argument.

Edited April 11, 2008 by Swinging Swede

[DukeCity]

OK, I get how it works (just barely), but I keep thinking about some comic I heard one time who said: "Whenver I travel, I always bring a bomb onto the airplane. I know that I'm not going to set mine off, and what are the odds of there being two guys with bombs on the same plane?"