And Behind Door No. 1, a Fatal Flaw

[JSngry]

No, you can guarantee that you've either chosen correctly or not.

Seems kinda...obvious, that one does...

How do you guarantee that you've chosen correctly? I'd like to have that talent!

No...

The guarantee is an all-encompassing one that you've either chosen correctly or not, not that you've chosen only one specifically.

[JSngry]

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.

i tend to call that number zero - matter of personality i guess

Well dude, unless something totally...unscripted goes down, you know you're either going to win or lose.

[JSngry]

Jim's 50/50 thing isn't related to any odds. He's just saying he has two choices - stay or change.

And that that choice will be either right or wrong. It will not be 67% right and 33% wrong. Those are the odds that it will be right or wrong, not the relative outcome of my desicison.

Thank you for translating this to Humanspeak for me!

[Swinging Swede]

What are the odds that the 67-33 odds will come to pass in a one-time only scenario with only two possible outcomes? 67-33? No, that's the odds for the choice itself being successful in theory, on an endless stream of Ultimate Outcomes, not the odds for the choice actually materializing in real time at any one time.

Once again it doesn't matter whether it is about one time or an endless stream of times, the odds are still the same.

This is not about the odds, it's about the odds of the odds. Can you guarantee that in a one shot game that the 67% outcome will be the one that materializes?

The outcome in a one shot game is always 100% or 0%. That doesn't prevent the odds from being 67% and 33% respectively.

This is not about the odds, it's about the odds of the odds.

OK, then I would say that the odds are 67-33, and the odds of the odds 100-0!

[JSngry]

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

Of course they are. That has never been a point of contention.

Well, you did actually say the following (see question in red):

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?

Exactly. Switching will always be the smart choice, but not always the successful one. And the object of the game is to be successful. Now, do you get to be successful by being smart? Yeah, sure, 2 out of 3 times over god knows how many chances. So you should take that route even when you only get one shot at Glory, and even if your selection has only two possible outcomes - win or lose, and even if you have no way of knowing which it's going to be, and even if fate has it lined up so that your time up is the 1 time in 3 that the smart choice is the wrong choice.

Unless your mojo is working, and/or unless you smell goat farts.

[JSngry]

This is not about the odds, it's about the odds of the odds.

This makes no sense.

In your world, maybe. In mine, it makes perfect sense, not as "science" but as "starting point". Guess we live in two different worlds (cue Freddie Roach...)

Good thing we're friends and not neighbors!

Edited April 11, 2008 by JSngry

[Swinging Swede]

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.

Actually you have a 100% chance of getting that 2 out of 3 chance, and you realize it by switching door.

So you're saying that you do not have the option of choosing not to switch?

Even when you have a 100% chance, you can choose not to take it. Monty himself, if he were to step in, could choose not to pick the door he knows that the car is behind, but pick the other door instead. I'm not sure why he would do it, but he certainly has the option.

[John L]

There is the so-called law of large numbers in statistics that says that if you play the switching strategy enough times in a row, the the number of times that you win the car will converge to 2 out of 3. If you only play a few times, there is a strong probability that you won't get the car two out of three times. But that is something different entirely than assessing the odds of a single event.

Well ok, you're talking prose, I'm mumbling poetry, but yeah, the earlier expressed notion that you should be surprised if you "smart" pick doesn't pan out the one and only time you get to use it is kinda...laughable to me.

Like I said, I've lost big betting on hands with "favorable odds" and won big by sometimes chasing after things that go against the odds. "Beating the odds" is not always "luck", ya' know. Some cats really do have an intuitive sense about which way the wind's blowing at any point in time, and I've lost money to some of them. But then again, there's been times when I've had that intuition working too, and I've won money from people who didn't. and on the whole, hey...I ain't poor, let me put it that way, and if even though that's in no way a result of gambling, it's also not a result of not gambling, if you know what i mean.

Odds, in a non-"theoretical" sense, are really just averages, which means that they don't always pan out, which means that sometimes, sometimes, counter-logical play is going to defeat logical play. I know that bugs people who like to think that the universe is a benign, non-fluid oasis of stasis, orderly place and if you just wait it all out you'll come out a winner, but...it is what it is, and what it is ain't no Sure Thing, ever.

OK, I think that I am beginning to get you. The notion is that something supernational might be at play. If you just finished listening to ESP and you feel charged, then your choice of an initial door may not have been random, but actually made for a real reason that you don't completely understand, etc.

Actually, mathematics doesn't go out the window in that case either. If you start out with a hunch that it is behind door number 1, then you can begin with different prior likelihood assessments (statisticians refer to them as Bayesian priors). For example, you can assign door number 1 50% and only 25% to doors number two or door number three, or something else that reflects your prior hunch.

But guess what? You should still switch! In that case, you should make an initial choice of door number 2 or 3 and then switch to door number 1 or the other door. That would make your subjective odds of winning become greater than 2/3. if Monty doesn't show a goat behind door number 1, your odds become 75%. If he does show a goat behind door number 1, they still become 50/50. So there!

Edited April 11, 2008 by John L

[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.

Actually you have a 100% chance of getting that 2 out of 3 chance, and you realize it by switching door.

So you're saying that you do not have the option of choosing not to switch?

Even when you have a 100% chance, you can choose not to take it. Monty himself, if he were to step in, could choose not to pick the door he knows that the car is behind, but pick the other door instead. I'm not sure why he would do it, but he certainly has the option.

Which is what I'm saying - that even though you know that one choice is "smarter" than the other one, you still have the chance not to make that choice.

You have a 50-50 chance (i.e. - you can choose or not choose to switch) to get that 2 out of 3 chance (which is what you have if you exercise your choice to switch).

Exactly what you're saying, just with numbers attached!

[JSngry]

There is the so-called law of large numbers in statistics that says that if you play the switching strategy enough times in a row, the the number of times that you win the car will converge to 2 out of 3. If you only play a few times, there is a strong probability that you won't get the car two out of three times. But that is something different entirely than assessing the odds of a single event.

Well ok, you're talking prose, I'm mumbling poetry, but yeah, the earlier expressed notion that you should be surprised if you "smart" pick doesn't pan out the one and only time you get to use it is kinda...laughable to me.

Like I said, I've lost big betting on hands with "favorable odds" and won big by sometimes chasing after things that go against the odds. "Beating the odds" is not always "luck", ya' know. Some cats really do have an intuitive sense about which way the wind's blowing at any point in time, and I've lost money to some of them. But then again, there's been times when I've had that intuition working too, and I've won money from people who didn't. and on the whole, hey...I ain't poor, let me put it that way, and if even though that's in no way a result of gambling, it's also not a result of not gambling, if you know what i mean.

Odds, in a non-"theoretical" sense, are really just averages, which means that they don't always pan out, which means that sometimes, sometimes, counter-logical play is going to defeat logical play. I know that bugs people who like to think that the universe is a benign, non-fluid oasis of stasis, orderly place and if you just wait it all out you'll come out a winner, but...it is what it is, and what it is ain't no Sure Thing, ever.

OK, I think that I am beginning to get you. The notion is that something supernational might be at play. If you just finished listening to ESP and you feel charged, then your choice of an initial door may not have been random, but actually made for a real reason that you don't completely understand, etc.

Actually, mathematics doesn't go out the window in that case either. If you start out with a hunch that it is behind door number 1, then you can begin with different prior likelihood assessments (statisticians refer to them as Bayesian priors). For example, you can assign door number 1 50% and only 25% to doors number two or door number three, or something else that reflects your prior hunch.

But guess what? You should still switch! In that case, you should make an initial choice of door number 2 or 3 and then switch to door number 1 or the other door. That would make your subjective odds of winning become greater than 2/3. if Monty doesn't show a goat behind door number 1, your odds become 75%. If he does show a goat behind door number 1, they still become 2/3. So there!

You expect me to take you seriously when you don't allow for goat farts?

[JSngry]

See, you mention goat farts and the next thing you know...

[Swinging Swede]

Jim's 50/50 thing isn't related to any odds. He's just saying he has two choices - stay or change.

OK, that explains a thing or two!

[Dan Gould]

the earlier expressed notion that you should be surprised if you "smart" pick doesn't pan out the one and only time you get to use it is kinda...laughable to me.

The idea I was getting at was that, using the simulator 100 times for each choice would lead to a recognition that switching results in a very large increase in the odds of winning (winning twice as often). Winning with one strategy at a rate of 2 out of 3 times while winning at a rate of 1 out of 3 times with the alternative strategy would lead me to have a strong expectation of winning when I switch. Two out of three times is winning a great deal - its the equivalent of a 108 win baseball season - so I would logically expect to win after switching.

Even though it leaves a not inconsequential chance of losing.

As for your feelings about mojo and intuitive sense of what is going to happen reminds me of what I said earlier - if you feel like you never have good initial instincts, you should always switch.

[RDK]

My right brain just cock-punched my left brain and now that dancing chick is moving counter-clockwise.

[JSngry]

the earlier expressed notion that you should be surprised if you "smart" pick doesn't pan out the one and only time you get to use it is kinda...laughable to me.

The idea I was getting at was that, using the simulator 100 times for each choice would lead to a recognition that switching results in a very large increase in the odds of winning (winning twice as often). Winning with one strategy at a rate of 2 out of 3 times while winning at a rate of 1 out of 3 times with the alternative strategy would lead me to have a strong expectation of winning when I switch. Two out of three times is winning a great deal - its the equivalent of a 108 win baseball season - so I would logically expect to win after switching.

Even though it leaves a not inconsequential chance of losing.

As for your feelings about mojo and intuitive sense of what is going to happen reminds me of what I said earlier - if you feel like you never have good initial instincts, you should always switch.

Yeah, I hear ya', but on Let's Make A Deal, all the simulator prep work in the world really doesn't come into play other than as a study in probability. When it comes time to actually play, the real question, the one that will definitely determine the outcome is simple - will this be one of the times that the 2/3 scenario goes down, or will it be one of the times that the 1/3 scenario happens?

Now, to figure out that one, you need to know...god knows what. Sure, your 2/3 choice is the only/best logical hedge against defeat, but where are you coming in on the "cosmic stage" that ultimately makes the final decision? Is it against every contestant who's ever played the game on Let's Make A Deal? Is it against every time you've played the game in your mind? Is it against all that plus all the times that everybody else has played the game in their mind? Is it in relation to all the people who will play the game in the future? Not only that, but how about in relation to all the people past present & future who have chosen/will choose correctly and/or incorrectly, where is the decision that you are about to make fall on that continuum of Let's Make A Deal-dom? 2-out-of-3, remember, it's statistically sound, but this stuff is never as "simple" as you think it's gonna be, just because nobody knows all they need to know to know where you stand until after you've already stood.

2 out of 3, sounds simple, but whose 2 and whose 3? Especially when you only got 1, and in all likelihood it's the only real 1 you'll ever get? Which is not to say that you don't make the smart call, because you do, it's all you got, but you still gotta hope like hell that this time, this one time, is gonna be your time, because it may not be, odds and all other hedging to the contrary, that sometimes all the odds in the world can be in your favor and you still end up fucked, you still gotta realize that it's still a gamble with nowhere near as "certain" an outcome in reality as it is in theory.

Which, actually, is right in line with the rest of life...

[The Magnificent Goldberg]

There'd be no fun in this at all if the outcome were certain.

Just like life.

No, wait, that outcome IS certain

MG

[The Magnificent Goldberg]

So I just went back to the original post.

Now we all, even or especially Jim, recognise that this is a statistical probability only, and when you're standing there in front of the door it's different, how do we feel about the monkeys and the M&Ms?

MG

[Tom Storer]

When it comes time to actually play, the real question, the one that will definitely determine the outcome is simple - will this be one of the times that the 2/3 scenario goes down, or will it be one of the times that the 1/3 scenario happens?

Jim, you're asked to make a choice between stay and switch.

Each choice can be assigned a probability, as exhaustively demonstrated in this thread:

Stay: 1/3

Switch: 2/3

You keep saying "what are the odds that my choice will be right?" and answering that by saying: "they must be 50/50 because there are only two possible outcomes."

But it isn't because there are two possible outcomes that the likelihood of each outcome is equal. By saying that there is a 50/50 chance because there are only two outcomes, that is what you are saying: that each outcome is as likely as the other. But that's not true. One outcome is twice as likely as the other.

There is no contradiction between saying "I'll either be right or wrong" and saying "I'm twice as likely to be right as to be wrong." In this case, both are true. You'll either get a car or a goat--there is no third possibility. And if you switch, you are twice as likely to get the car. Which doesn't mean you'll get the car, only that you are twice as likely to.

You also seem to think there is such a thing as "the odds on the odds." But there isn't.

OK, you've chosen to switch and are twice as likely to get the car. This you agree to. But then you've asked, as a further question, what are the odds that you will get the car? That this choice has been right?

You're comparing the question "What is the likelihood of the car being behind the door if I switch?" and the question "What is the likelihood that choosing to switch will get the car this time?"

Jim, it's the same question. And the same odds.

[JSngry]

Of course it is...

So ok, there's 100 people lined up to play this game, and you're #100. The first 67 of them switch, and they all win. Not likely, but it could happen. There's no "law" guaranteeing that it couldn't.

Sensing a trend, the next 32 also switch, and lo and behold, they win too! Again, unlikely as it is, it could happen.

So, it's your turn, Mr. 100. You know that switching is, in isolation, the smart move, but you also see right in front of you that the results have, in this immediate zone, outperformed the odds. At some point, not switching is going to have to result in a win.

So - where is that point, and how do you best predict it in order to win the car, which is, after all, the object of the game, not running an intellectual masturabatathon?

And - do you only look at the action going on in this room to figure that out, or do you look at everybody who's played the game in the past? Maybe at some point, five folks in a row didn't switch and won. How does that skew the point at which it's again going to be the winning choice? What if right there in the building you're in, there's another 20 studios full of people playing the same game, how does that affect the chances that your choice will perform according to the odds, not in theory, but in actuality? "The odds are always the same", I can hear everybody saying it. Well, yeah - it's still 2/3 vs 1/3. But unless there is a systematic pattern of every third person switching and losing (and that assumes that everybody will chose to switch, which ain't gonna happen either), then 2/3 vs 1/3 ain't necessarily gonna mean squat when your number comes up.

Mr. Litwack earlier mentioned the Law Of Large Numbers, which I first heard about in conjunction w/the MIT blackjack team. There were times when they took to the tables and lost big and long before the numbers came around in their favor, other times when the shit clicked from jump. Was there any predictor as to which it would be? No, of course not. And if they weren't deeply funded, there was no way to insure that on the nights when it took a long time for the odds to finally kick in that they wouldn't have run out of money first.

Now that's a game where if you got the time and money and system, you can guarantee winning results over the long haul. There, statistics are of comfort and practical use to you. Here, you got one shot to be right, and all the statistics can do for you is give you a little sense of faux-confidence that you've got a "good chance" to win the one time you play.

Well, maybe you do and maybe you don't. If you played in isolation, yeah. But you don't. The game's been played before and it will be played after, so you are not the only one to whom the odds are applying. Now, can anybody show me, not that the odds play out over time collectively, because it's obvious that they do, but rather that there is a predictor to individual distribution of same amongst people who precipitate this activity, i.e. - players?

"Small science", like "small religion" serves no real purpose other than to create a false sense of comfort rather than forcing one to confront the potentially horrifying, yet very real randomness of any given moment. Your everyday behavior may be the same either way, but at some point some shit could jump up and hit you and you're either gonna shit your pants and die, or else deal with it.

[Dan Gould]

Jim,

the odds are in no way affected by what has happened before. Period. Its equivalent to flipping a penny. The odds of it coming up heads is 50-50. Yet with those odds you can have a streak of X number of times that heads comes up. That doesn't increase the odds of tails coming up, even if in the long run, more tails will come up to even out the all those times it came up heads.

The odds of heads when flipping a fair coin is always 50-50.

The odds of winning by switching doors is always 67-33

[Dan Gould]

And - do you only look at the action going on in this room to figure that out, or do you look at everybody who's played the game in the past? Maybe at some point, five folks in a row didn't switch and won. How does that skew the point at which it's again going to be the winning choice? What if right there in the building you're in, there's another 20 studios full of people playing the same game, how does that affect the chances that your choice will perform according to the odds, not in theory, but in actuality? "The odds are always the same", I can hear everybody saying it. Well, yeah - it's still 2/3 vs 1/3. But unless there is a systematic pattern of every third person switching and losing (and that assumes that everybody will chose to switch, which ain't gonna happen either),

This would mean that the odds are not 67-33 but something else entirely

then 2/3 vs 1/3 ain't necessarily gonna mean squat when your number comes up.

Correct - because even if with a strategy of switching, you lose 1/3 of the time..

Let's summarize:

The odds are always 67-33 that you will win when you switch.

These odds are not effected in any manner by what has happened by other people playing the game.

Even when you switch, there is a substantial (1/3) chance that you will lose.

[Van Basten II]

I wish that i did not open the door that brought me to this thread.

[JSngry]

Jim,

the odds are in no way affected by what has happened before. Period. Its equivalent to flipping a penny. The odds of it coming up heads is 50-50. Yet with those odds you can have a streak of X number of times that heads comes up. That doesn't increase the odds of tails coming up, even if in the long run, more tails will come up to even out the all those times it came up heads.

The odds of heads when flipping a fair coin is always 50-50.

The odds of winning by switching doors is always 67-33

And - do you only look at the action going on in this room to figure that out, or do you look at everybody who's played the game in the past? Maybe at some point, five folks in a row didn't switch and won. How does that skew the point at which it's again going to be the winning choice? What if right there in the building you're in, there's another 20 studios full of people playing the same game, how does that affect the chances that your choice will perform according to the odds, not in theory, but in actuality? "The odds are always the same", I can hear everybody saying it. Well, yeah - it's still 2/3 vs 1/3. But unless there is a systematic pattern of every third person switching and losing (and that assumes that everybody will chose to switch, which ain't gonna happen either),

This would mean that the odds are not 67-33 but something else entirely

?

[JSngry]

Dan, you seem to be saying that the results stemming from the (constant) individual odds cannot be influenced by the (equally constant) collective results.

Seems to me that the only way for that to work would be if everybody who played played enough times to reach the "statistical truth". Unless and until that happens though, if the collective result = 2/3 and not every individual result = 2/3, then either some individuals are going to have to perform outside to odds, either positively or negatively to keep the collective even, or else something is an illusion.

In other words, if 60% of a state's residents always vote, and one county has a 45% turnout one year and 75% the next, some other counties are going to have to vary correspondingly to keep the 60% constant.

Similarly, if 100 people switch & win (or if 1 person switches and wins 100 times) then at some point the 1/3 outcome will have to happen either individually or collectively at a temporarily greater than 1/3 rate in order to uphold the "greater truth" of the overall 2/3 success rate. Right?

You said earlier that you won 80% of the time by always switching, right? Well, if you "retire" right now, you've outperformed the odds. In order for the odds to have true meaning (which I believe they do) doesn't there have to be underperformance somewhere at some time? Or can everybody win at an 80% rate and the odds still truly be 67% ?

Edited April 12, 2008 by JSngry

[Dan Gould]

Dan, you seem to be saying that the results stemming from the (constant) individual odds cannot be influenced by the (equally constant) collective results.

Seems to me that the only way for that to work would be if everybody who played played enough times to reach the "statistical truth". Unless and until that happens though, if the collective result = 2/3 and not every individual result = 2/3, then either some individuals are going to have to perform outside to odds, either positively or negatively to keep the collective even, or else something is an illusion.

In other words, if 60% of a state's residents always vote, and one county has a 45% turnout one year and 75% the next, some other counties are going to have to vary correspondingly to keep the 60% constant.

Right?

Yes but my point is that you can't use that information - say, 5 straight cases of switching coming up a winner - to predict the next iteration. The odds of a particular iteration are always the same, even as there can be what we might view as random variation in opposition to those odds.

Your example of the switching strategy resulting in:

win win lose win win lose win win lose ad infinitum is not a true example of a 67-33 set of odds.

These odds are 100, 100, 0, 100, 100, 0 etc.

which is why I called it "something else entirely".

Now the interesting thing is that yes, the sum of those probabilities is 67-33 but that doesn't describe the actual probability. At least, I don't think so. (For the first time in this thread I've talked myself into something I am not 100% certain of).