And Behind Door No. 1, a Fatal Flaw

[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... ?

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.

as i said, i think i partly understand and can appreciate you argument, people tend to take statistics to seriously... but then

i think it is strange to make such a huge difference between a lottery where one person (out of a million) must win and, say, a random generator which gives you the price with a probability of one in a million... does the guy in Oklahoma who ends up with the money in the first case really make that much of a difference (turning a 50-50 opportunity into a one in a million thing)...

[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.

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

The odds of having made the right choice by picking Door A would be 33%. And the odds of having made the right choice by picking Door B+C would be 67%.

Even though there are only two choices.

[Dan Gould]

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.

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.

Aggie is correct. The odds are not 50/50 in the "isolated" example of an individual with one chance to win. His odds are 2/3 when he switches, 1/3 when he doesn't. Period. His odds increase when he switches, and its not a long-term thing. The odds are better if you switch.

Maybe if someone could get all the old tapes of Let's Make a Deal and collated the number of times that a contestant switched and the number of times that he stuck with his first choice. I guarantee that like the simulator that was posted earlier, 2/3 of all people who switch won, and 1/3 of those who didn't won.

What part of that can't you get? The odds improve when you switch.

[Aggie87]

I think basically what Jim's saying is that you have two choices - switch or don't switch. He's saying that you have an either/or decision, which isn't the same thing as a 50/50 chance of winning the car.

[JSngry]

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 % if you do it once. Those 100 times are just 100 single shots counted together.

Apologies if I misunderstood your argument.

I think you do. I'm not talking about whether or not it's a smart choice to make the switch, obviously it is. But once you do, there's still only two possible outcomes. Not "choices" or "chances", but definite outcomes. And no matter how well you position yourself, unless you make a 100% foolproof decision (which in this case is impossible), you're subject to the possibility that your decision, no matter how smart it is, will not pan out this time. And this time is all you get.

Measure the probability that switching will be a good idea, and yeah, hey, there it is. You should win 2 of 3 times. But how does that play out when you only got one time? Are you gonna 67% win and 33% lose the one time you play? No way. You will either 100% win or 100% lose, and although your odds by changing favor you winning, they do not guarantee it.

Now, what about somebody who doesn't switch but wins anyway? How does that work? Is it "luck"? Or is it that the one time that this person played was a time that the favorable odds did not apply? And again, if we're only playing one time, then it's either all win or all lose. So if the favorable odds are not going to apply every time and we only have one incident with which to work, what other possible odds are there than 50/50 that the smart choice will be successful this one time? Not that it should be, but that it will be?

[Aggie87]

Outcome A = winning the car

Outcome B= losing the car

You can still win the car with the 33% likelihood (by keeping with Door A), or lose the car even with the 67% likelihood of winning (by swithing to Door B+C). It's just a probability.

You still had two choices. But it wasn't a 50/50 thing. There's no way to guarantee the win (unless you're Monty and know which door has the car).

[Dan Gould]

So if the favorable odds are not going to apply every time and we only have one incident with which to work, what other possible odds are there than 50/50 that the smart choice will be successful this one time? Not that it should be, but that it will be?

The favorable odds will apply every time. The odds are not 50-50 for your one chance. They are 2/3 if you switch, 1/3 if you do not.

How about if you go try the simulator that I posted? Follow the two different strategies 20 times each. I guarantee that you will win the car more often if you switch, less often if you don't. If you can't grasp that this applies for each single attempt, then I don't know how to make you understand that there is a 2/3 chance each time you switch.

Try it this way:

When you go through the simulator, repeat the words "wow, I won!" every time that you win after switching, every time you win after sticking. You will be exclaiming the magic phrase so many more times by switching, something has got to click inside your head.

[JSngry]

What part of that can't you get?

The part that says that the "smart" choice will not always be the right choice, and how can you tell for sure when it will or won't be?

You can't, that's how. So 1 out of 3 times, the "dumb" choice will be the right choice.

Now, how can you know when to make the smart or the dumb choice in order to make the right choice? You can't, that's how.

So even though 2 out of 3 times the smart choice will be the right choice, and 1 out of 3 times the dumb choice will be the right choice, whatever choice you make may or may not be the right choice for the one time you play. You have a 2 out of 3 chance of being right, but only a 50-50 chance that you'll get that 2 out of 3 chance.

What part of that can't you get?

[Aggie87]

You have a 2 out of 3 chance of being right, but only a 50-50 chance that you'll get that 2 out of 3 chance.

I don't think that's true.

[JSngry]

The favorable odds will apply every time.

Indeed, the will apply every time.

But they will not come through every time.

Remind me to play poker with you some time...

[Aggie87]

The favorable odds will apply every time.

Indeed, the will apply every time.

But they will not come through every time.

Remind me to play poker with you some time...

That's very true. Odds can be in your favor and you can still lose, or odds can be against you and you can still win.

That's why so many people play the lottery. Not worth throwing away my money.

[Dan Gould]

What part of that can't you get?

The part that says that the "smart" choice will not always be the right choice, and how can you tell for sure when it will or won't be?

You can't, that's how. So 1 out of 3 times, the "dumb" choice will be the right choice.

Now, how can you know when to make the smart or the dumb choice in order to make the right choice? You can't, that's how.

So even though 2 out of 3 times the smart choice will be the right choice, and 1 out of 3 times the dumb choice will be the right choice, whatever choice you make may or may not be the right choice for the one time you play. You have a 2 out of 3 chance of being right, but only a 50-50 chance that you'll get that 2 out of 3 chance.

What part of that can't you get?

The part that reminds me that you seemingly have no grasp of statistics or probability theory.

The chance is not 50-50 under any circumstances, except the one Aggie mentioned where a new contestant is introduced to the situation after the first goat has been revealed. And I don't have any idea how to interpret "only a 50-50 chance the you'll get that 2 out of 3 chance". There is exactly a 2/3 chance that you will "get that 2/3 chance". 2/3rds of the time that I switch, I will win. One-third of the time that I don't switch, I will win.

Repeat after me:

There are no guarantees in games of chance.

BUT, under these circumstances, switching is the only intelligent decision.

Now, go play the simulation. You think the odds are 50-50, so for every time you change doors, notice how much more often you win. Take 10 minutes and do it 100 times each. Absorb the results you observe. Then pretend there really is a car behind the door, and ask yourself if you should switch or not.

Then play the game.

You may win or you may lose, but if you've absorbed how the odds change by switching, you should not be viewing the outcome as though there are even odds of winning. If you switch, and lose, you should be legitimately surprised at the outcome.

[Dan Gould]

The favorable odds will apply every time.

Indeed, the will apply every time.

But they will not come through every time.

Remind me to play poker with you some time...

This is not something I have trouble grasping.

[JSngry]

You have a 2 out of 3 chance of being right, but only a 50-50 chance that you'll get that 2 out of 3 chance.

I don't think that's true.

Well, here we go back to what everybody's been saying all along - you got two choices, right or wrong. 50-50.

Y'all are arguing "wisdom" of choices, and all I'm saying is that, not so fast, sometimes, 1 in 3 times, the smart choice will be wrong. and you really don't know which one it's gonna be until it goes down.

So it stands to reason that at some point, making the dumb choice will be making the right choice. So when, then do you make it?

Easy - when you decide what to do. And there you got two choices. Switch or don't. One will be right, one will be wrong, and although one will be right more often than the other, you have no way of knowing which will be which or when, so you make one of two choices.

And that gives you a 50-50 chance at getting the 2 out of 3 chance.

[JSngry]

What part of that can't you get?

The part that says that the "smart" choice will not always be the right choice, and how can you tell for sure when it will or won't be?

You can't, that's how. So 1 out of 3 times, the "dumb" choice will be the right choice.

Now, how can you know when to make the smart or the dumb choice in order to make the right choice? You can't, that's how.

So even though 2 out of 3 times the smart choice will be the right choice, and 1 out of 3 times the dumb choice will be the right choice, whatever choice you make may or may not be the right choice for the one time you play. You have a 2 out of 3 chance of being right, but only a 50-50 chance that you'll get that 2 out of 3 chance.

What part of that can't you get?

Repeat after me:

There are no guarantees in games of chance.

BUT, under these circumstances, switching is the only intelligent decision.

Now, go play the simulation. You think the odds are 50-50, so for every time you change doors, notice how much more often you win. Take 10 minutes and do it 100 times each. Absorb the results you observe. Then pretend there really is a car behind the door, and ask yourself if you should switch or not.

Then play the game.

You may win or you may lose, but if you've absorbed how the odds change by switching, you should not be viewing the outcome as though there are even odds of winning. If you switch, and lose, you should be legitimately surprised at the outcome.

We're on the same page up until that last sentence.

If you got a one shot, you oughta know that even if the odds are 99-1, you might get the 1.

Pissed off, yeah. Surprised, no.

The way most amateur gamblers fuck up is by not walking away while (if) they're ahead, by giving the odds a chance to catch up with them. And they always do.

But a smart motherfucker quits while ahead, stashes the money away, and lets the odds catch up with him some other time, with some other seed money, and then walks away before they catch up with him too much.

People who have statistics as their friends should really watch their backs...

[Aggie87]

You have a 2 out of 3 chance of being right, but only a 50-50 chance that you'll get that 2 out of 3 chance.

I don't think that's true.

Well, here we go back to what everybody's been saying all along - you got two choices, right or wrong. 50-50.

I still think it's "two choices, right (switch) or wrong (stay), 66-33." And I was one of the few that started out kinda thinking like you are, I think.

Y'all are arguing "wisdom" of choices, and all I'm saying is that, not so fast, sometimes, 1 in 3 times, the smart choice will be wrong. and you really don't know which one it's gonna be until it goes down.

So it stands to reason that at some point, making the dumb choice will be making the right choice. So when, then do you make it?

Making the right choice is switching, 66% of the time. Making the right choice is staying with your original door, 33% of the time.

Easy - when you decide what to do. And there you got two choices. Switch or don't. One will be right, one will be wrong, and although one will be right more often than the other, you have no way of knowing which will be which or when, so you make one of two choices.

And that gives you a 50-50 chance at getting the 2 out of 3 chance.

I still say you have a 66-33 chance at getting the 2 out of 3 chance.

[John L]

You have a 2 out of 3 chance of being right, but only a 50-50 chance that you'll get that 2 out of 3 chance.

I don't think that's true.

Well, here we go back to what everybody's been saying all along - you got two choices, right or wrong. 50-50.

Y'all are arguing "wisdom" of choices, and all I'm saying is that, not so fast, sometimes, 1 in 3 times, the smart choice will be wrong. and you really don't know which one it's gonna be until it goes down.

So it stands to reason that at some point, making the dumb choice will be making the right choice. So when, then do you make it?

Easy - when you decide what to do. And there you got two choices. Switch or don't. One will be right, one will be wrong, and although one will be right more often than the other, you have no way of knowing which will be which or when, so you make one of two choices.

And that gives you a 50-50 chance at getting the 2 out of 3 chance.

Jim: Everything sounds OK to me until you cite 50-50. Like everyone else is saying it's actually 1/3 , 2/3. Sure, you might just feel lucky by picking the 1/3 option, and you just might get lucky. In fact, with probability 1/3, you will be lucky. But that doesn't contradict the point here: if the objective is to pick a strategy that will increase you chances of winning, then the switching strategy is the one to choose. But maybe you just have your mojo working or something.

[JSngry]

I still say you have a 66-33 chance at getting the 2 out of 3 chance.

Then that means you have three choices to make - switch, don't switch, and....?

[aparxa]

The only way to have 50/50 occurs if YOU decide to open one of the two remaining doors (and not Monty) and that door reveals a goat.

[JSngry]

In fact, with probability 1/3, you will be lucky.

How is a statistical inevitability, even a less likely one, "luck"?

[JSngry]

The only way to have 50/50 occurs if YOU decide to open one of the two remaining doors (and not Monty) and that door reveals a goat.

Ok, one...more...time...

50-50 as I'm using it does not refer to the probability that switching will be successful. That is clearly 2/3.

It (50-50) refers to what you can guarantee as far as your choice being the right one for any given time, before the results are revealed.

[Aggie87]

I still say you have a 66-33 chance at getting the 2 out of 3 chance.

Then that means you have three choices to make - switch, don't switch, and....?

No it doesn't. the 66-33 is still dealing with only two alternatives - switching or staying.

There are four separate scenarios though, one of which WILL play out:

1 - stay with door A and be correct (33% likelihood of this happening)

2 - stay with door A and be wrong (67% likelihood of this happening)

3 - switch to door B+C and be correct (67% likelihood of this happening)

4 - switch to door B+C and be wrong (33% likelihood of this happening)

[JSngry]

Nope.

Switching or not switching are the only choices you have. The results/outcomes only come into play after the choice is made.

You have four possible outcomes, but they result from only two choices.

[Swinging Swede]

I'm not talking about whether or not it's a smart choice to make the switch, obviously it is. But once you do, there's still only two possible outcomes. Not "choices" or "chances", but definite outcomes. And no matter how well you position yourself, unless you make a 100% foolproof decision (which in this case is impossible), you're subject to the possibility that your decision, no matter how smart it is, will not pan out this time.

Sure. In 33% of the cases this scenario is repeated the person who does this chocie once will realize, after Monty opens the door, that it didn't pan out, and in 67% of the cases the person who does the choice once will realize that it did pan out.

And this time is all you get.

How many times you get doesn't matter for the statistical probability. If it's a 67% chance if you do it 1000 times, then there's also a 67% chance if you do it once.

Measure the probability that switching will be a good idea, and yeah, hey, there it is. You should win 2 of 3 times. But how does that play out when you only got one time? Are you gonna 67% win and 33% lose the one time you play? No way. You will either 100% win or 100% lose, and although your odds by changing favor you winning, they do not guarantee it.

Of course 67% does not guarantee anything, but it is still a lot better than 50%. What will happen is that you will 100% win in 67% of the cases, and 100% lose in 33% of the cases, if you want to express it that way.

Now, what about somebody who doesn't switch but wins anyway? How does that work? Is it "luck"? Or is it that the one time that this person played was a time that the favorable odds did not apply?

Well, a 33% chance is still a decent chance. It's just not as good as a 50% chance or a 67% chance. But people who keep their originally chosen doors will fairly often win a car. To be more specific, it will happen in one third of the cases!

And again, if we're only playing one time, then it's either all win or all lose. So if the favorable odds are not going to apply every time

They do apply every time. 67% is not 100% and 33% is not 0%, so of course it will happen that people who switch doors will lose and people who keep the door will win. That doesn't mean that the odds don't apply.

and we only have one incident with which to work, what other possible odds are there than 50/50 that the smart choice will be successful this one time? Not that it should be, but that it will be?

The only correct odds that the smart choice will be successful (i.e. winning the car) are 67/33, not 50/50.

If the odds somehow were 50/50 for one incident, then it would be 50/50 for 1000 incidents too, since they only consist of 1000 single incidents counted together. I can't understand how you can think that it's a 50% chance for one single incident, but a 67 % chance for 100 or 1000 of them. That's just illogical.

[JSngry]

I still say you have a 66-33 chance at getting the 2 out of 3 chance.

Then that means you have three choices to make - switch, don't switch, and....?

No it doesn't. the 66-33 is still dealing with only two alternatives - switching or staying.

There are four separate scenarios though, one of which WILL play out:

1 - stay with door A and be correct (33% likelihood of this happening)

2 - stay with door A and be wrong (67% likelihood of this happening)

3 - switch to door B+C and be correct (67% likelihood of this happening)

4 - switch to door B+C and be wrong (33% likelihood of this happening)

Even if you choose to look it it like this, you still got 2 out of 4 choices that result in 2/3 scenarios. Even that's 50/50, although it's now a matter of which odds you end up with (2/3 or 1/3, ach value-neutral) rather than whether you've selected the best winning strategy, which is of course to switch..