Kevin's Watch

Vraith wrote:

Z, show me the math that proves I'm just guessing and lucky to be right.

I provided an example, the coin toss logic applies precisely to this situation. And the rest of that describes how some gambler's [[the bad ones]] make the error.

Maybe you don't understand...there is pretty simple math that describes the situation...all the dispute and confusion isn't cuz the math is hard. So why is it disputed/confusing? Because it ISN't the math/logic itself.

Don't be a prick in my forum.

How am I being a prick? By having the temerity to disagree with the mod of this forum? By embarrassing you in pointing out your errors? Dude, I can't believe they give you any authority on this site whatsoever, you are absolutely the worst moderator here. Give you a little power, and you think it's ok to call people names, but no one else can!

I did not say that the math was hard. I just did not think that you had correctly identified the problem. They problem lies in finding the fallacy in how the original question is framed. And WF was correct in identifying the use of a single variable for two different things. How we formalize situations in the world with math IS a mathematical problem. Failing to see additional possibilities IS a logical problem.

Check out my link. This problem is more complex than it looks. There is quite a bit of literature that has been written about it over the years. I don't think your simplification (above) does it justice --- if I'm allowed to say that here, oh wise and powerful moderator of a forum on a barely read message board for a poorly selling has-been author! I tremble at your vast importance!

Vraith wrote:...there is pretty simple math that describes the situation...all the dispute and confusion isn't cuz the math is hard. So why is it disputed/confusing? Because it ISN't the math/logic itself.

Although strictly speaking true (kinda/sorta), the dispute results because of the persuasiveness of applying equally simple maths and logic... BUT ENTIRELY INCORRECTLY.

Sure it is dead simple to take the correct view that any game only ever contains X and 2X. Hang onto that and you'll never be bamboozled.

HOWEVER... there is a subtlety that needs to be realised. And that's how tempting it is, knowing for sure that the envelope opposite you contains either precisely twice or precisely half the amount within your own, to come up with an "X vs either X/2 or 2X" equation... that is simply non-applicable in this specific instance. But for all that it is just plain wrong, it is subtly wrong... it's ever so alluring.

And it is hard to spot WHY it is wrong. Coming up with the right equation/view is easy. But doing so fails to address pointing out why and where the wrong equation as detailed above is at fault. And as Zee has quite rightly said, ACHIEVING THAT is the tricky bit.

As I posted earlier, it only takes a small (but vital) tweak in the original set-up to actually make that "X vs either X/2 or 2X" equation both applicable and correct.

And if everyone's favourite flat-earther's maths degree is anything to set any store by, this is not as tritely simple as some are making it out to be - a fact further evidenced by the reams of web and other copy written on this very conundrum.

(Re that last, I think I'm committing the very unusual logical NON-fallacy of an appeal to flawed authority... that despite its expertise has got things plain wrong. Ergo it ain't dead simple )

PS looking at the lengthy Wikipedia page devoted to this conundrum, I don't even begin to understand half of the maths and/or probability equations written there - and the logical dissections and philosophical assertions are in a different league...

Really.... this is NOT "dead simple"..

TheFallen wrote:HOWEVER... there is a subtlety that needs to be realised. And that's how tempting it is, knowing for sure that the envelope opposite you contains either precisely twice or precisely half the amount within your own, to come up with an "X vs either X/2 or 2X" equation... that is simply non-applicable in this specific instance. But for all that it is just plain wrong, it is subtly wrong... it's ever so alluring.

Actually, I have checked on another math-related forum and my assessment checks out. If you wish to choose to remain stubborn about this, though...well, that is a shame. There is no paradox--it is a straightforward problem, which is the last I will say about it.

Quite the reverse, Hashi. I'm not being stubborn... I'm being fascinated. And better yet, I'm actually expecting my newfound understanding (bestowed on me courtesy of a few here) to be proven wrong.

I would absolutely love to see an understandable proof of what you've maintained, namely this:-

Hashi Lebwohl wrote:

TheFallen wrote:Okay, say I get my first envelope and open it, revealing a crisp $20 bill. I then know that the other envelope must contain either $10 or $40. It cannot contain anything else. Again on that basis, swapping would seem to be a very sensible policy?

Opening the envelope is a slightly different problem than the one originally posed. If you are allowed to open the envelope first and then switch you now have a version of the Monty Hall problem, in which case you definitely want to switch.

In your specific case, you are holding a $20 bill. The expected value of the other envelope is (.5)(10) + (.5)(40) = 25, which is higher than the value you are holding in your hand; therefore, it is in your best interest to switch.

I honestly cannot fathom any reason at all why opening my envelope makes any difference or in any way alters the conundrum as originally set up - and yet you consistently maintain that it does. I (thanks to others far more mathematically able than myself getting the penny to finally drop for me) have tried to explain why it doesn't - but I am yet to see any explanation as to why it does.

Here's my take once more. If I open my envelope and see a $20 bill, I STILL have no clue if I am in a game where the total cash is $30 (and I've randomly been given the larger amount to start with), or where the total cash is $60 (and I've randomly been given the smaller amount to start with). But I am exclusively in one or the other. I can't be in both. Therefore the only equation that can be correctly applied is that my game (and indeed any game under the stated set-up) merely contains X and 2X. Ergo if I am holding X and swap to 2X, I'll gain X. And if I am holding 2X and swap to X, I'll lose X. Parity is preserved and there is thus no point in swapping. And no extra info is gained by opening my envelope before deciding if there's any point in doing anything.

Surely for the love of God that is all that can apply? Surely this conundrum does not become Monty Hall if my envelope is opened? And surely - if you are correct in your assertion - if the envelope opposite me was the one opened to reveal $20 before I made my choice, your advice would then have to be "don't swap". How can that possibly make any sense? How can the still sealed envelope always be on balance "better"???

However please note that I would quite sincerely absolutely love it if, just when I think I've finally got something that's baffled me for a long while down pat, my admittedly not particularly mathematical mind gets blown again. So please, point me at an explanation - and hopefully one I can get my head around.

And that's a totally honest request, not me being mischievous (which you'll know I have a helpless habit for being... but categorically not in this instance).

TheFallen wrote:Here's my take once more. If I open my envelope and see a $20 bill, I STILL have no clue if I am in a game where the total cash is $30 (and I've randomly been given the larger amount to start with), or where the total cash is $60 (and I've randomly been given the smaller amount to start with). But I am exclusively in one or the other. I can't be in both. Therefore the only equation that can be correctly applied is that my game (and indeed any game under the stated set-up) merely contains X and 2X. Ergo if I am holding X and swap to 2X, I'll gain X. And if I am holding 2X and swap to X, I'll lose X. Parity is preserved and there is thus no point in swapping. And no extra info is gained by opening my envelope before deciding if there's any point in doing anything.

Here is the flaw in your reasoning--you are never holding 2x; instead, you are ever only holding x, whether you open the envelope or not. The other envelope thus contains either .5x or 2x. The moment you go from "if I am holding x" to "if I am holding 2x" you are talking about two different problems.

Regardless of whether you open the envelope or not, the amount of money in your envelope is x. There is a 50% chance the other envelope contains .5x and a 50% chance the other envelope contains 2x. The expected value of that envelope is:

(0.5)(0.5x) + (0.5)(2x) = 0.25x + 1x = 1.25x

this value is greater than the x you already have, so you should definitely switch. Opening the envelope only lets you put a value on x but it does not change the logic involved.

...and so much for me not responding. Oh, well.

Unfortunately the logic and maths you've applied above is:-

a. what I also mistakenly first thought with the exact same error in modelling.

and

b. what literally everyone else in this thread has thoroughly debunked with full proof as being fallacious - and I also now understand why it is fallacious.

I can't say any more on this either Hashi... but you're literally on your own here.

Hashi, I agree that your logic seems straightforward and, on the surface, plausible. But the same exact logic would still apply after switching, which would compel you to switch again. How is that not a paradox? If the recommended course of action would have you reverse this decision as soon as you make it, then why was it logical to take it in the first place?

From the Wikipedia page, this the simplest solution:

The total amount in both envelopes is a constant c=3x, with x in one envelope and 2x in the other.

If you select the envelope with x first you gain the amount x by swapping. If you select the envelope with 2x first you lose the amount x by swapping. So you gain on average G=1/2(x)+1/2(-x)=1/2(x-x)=0 by swapping.

That sort of says what Vraith says above (though showing the math) about the same probability applying to either envelope. However, it still doesn't address how the mistake is made, as here:

The famous mystification is evoked by secretly mixing up of two completely different circumstances and situations, giving wrong results.

The so-called "paradox" presents two already appointed and already locked envelopes, where one envelope is already locked with twice the amount of the other already locked envelope. Whereas step 6 boldly claims:

"Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2." - Although, in the given situation, that claim can never be applicable to "any A" nor to "any average A".

This claim is never correct for the situation presented, this claim applies to the "Nalebuff asymmetric variant" only (see below).

In the situation presented, the other envelope cannot "generally" contain 2A, but can contain 2A only in the very specific instance where envelope A, by chance, actually contains the smaller amount of Total/3, but nowhere else.

And the other envelope cannot "generally" contain A/2, but can contain A/2 only in the very specific instance where envelope A, by chance, actually contains 2*Total/3, but nowhere else.


So the difference between the two already appointed and locked envelopes is always Total/3. No "average amount A" can ever form any initial basis for any "expected value", as this does not get to the heart of the problem.

I've bolded the important part. Your error, Hashi, is in taking an average for something which isn't generally possible, but only possible for specific cases.

TheFallen wrote:I can't say any more on this either Hashi... but you're literally on your own here.

I would rather be correct and on my own, then.

The original problem:

TheFallen wrote:There are two sealed envelopes and one contains twice the amount of money than the other. That's all you know.

You are given one of the sealed envelopes.

You are told that you can either keep the envelope you've got... or exchange for the other sealed envelope. Is there anything you can do in order to maximise potential gain?

Your envelope contains x. The other envelope must contain either .5x, in which case yours in the envelope which contains twice as much as the other, or the other envelope contains 2x, in which case it is the one containing twice as much. There are no alternate universes or "what if" scenarios--you have x in hand and the other envelope is either 0.5x or 2x.

Now, once you are handed the envelope you have an amount, which we call x. Since the other envelope contains either 0.5x or 2x I refer back to my earlier solution--the expected value of the other envelope is 1.25x, so to maximize value you switch. The amount you stand to gain is 1.25x - x = 0.25x.

If you *open* the envelope and you see a $20, then the other envelope could contain $10 or $40. You have a 50% chance of losing $10 or a 50% chance of gaining $20, which becomes (0.5)(-10) + (0.5)(20) = -5 + 10 = $5. This is *exactly* what I predict--you gained the expected amount of 0.25(20) = $5.

The error you made--and continue to make--is when you switch from "if my envelope contains x" then say "if my envelope contains 2x"--those are two different problems. It may seem as if they are the same problem but they are not.

I am afraid that I cannot state the solution any more simply than that.

Zarathustra wrote:Hashi, I agree that your logic seems straightforward and, on the surface, plausible. But the same exact logic would still apply after switching, which would compel you to switch again. How is that not a paradox? If the recommended course of action would have you reverse this decision as soon as you make it, then why was it logical to take it in the first place?

The initial problem did not say "you may switch as many times as you want". If you may switch as many times as you want then the expected value of the envelope not in your hand is always 1.25x since the envelope you contain is x. This is a slightly different problem, in which case it won't matter whether you switch or not--the decision tree as to whether or not to begin switching lies on a guess of "I have x in hand" or "I have 2x" in hand.

My caution still applies: you cannot begin guessing "I have x in hand" then later on second-guess yourself by thinking "but what if I have 2x in hand?" because those are different problems. You must choose a starting presumption and work from there. There is a paradox only if you start mixing up the presumptions.

Hashi Lebwohl wrote:If you *open* the envelope and you see a $20, then the other envelope could contain $10 or $40. [/color]

No, it can't. It most definitely contains one or the other, already. That is fixed prior to opening the envelopes. That means one is a fact and the other is impossible (your ignorance of which-is-which doesn't change that). It is not like flipping a coin; there is no 50/50 chance of it being something else when it's already something.

This may not seem to matter on the first round, but we're talking about averages (which is what the notion of probability rests upon). That's where you're tripping things up. Let's say you play the game repeatedly, and the person stuffing the envelopes keeps putting in 10s and 20s. There will NEVER be a time when, upon getting the 20, the other envelope could possibly contain 40. Every time you get that 20, and you switch, you're losing. Each time, you're either gaining 10 or losing 10, relative to what you hold prior to switching, so this gain/loss evens out. But you are acting like every time you get that 20, you have a 50/50 chance of gaining, when in fact it's impossible (for the scenario I've given: 10s and 20s only). Your chances of gaining by switching, once you initially pick the 20, are in fact zero.

You are treating a situation which has definite values as if it is indefinite. You are letting your use of variables trick you. "X" represents your ignorance of what's in the envelopes, but that doesn't mean what's in the envelopes is variable. Your ignorance of a known, predetermined reality isn't the same thing as the probability of that reality occurring.

You have a 50% chance of losing $10 or a 50% chance of gaining $20, which becomes (0.5)(-10) + (0.5)(20) = -5 + 10 = $5. This is *exactly* what I predict--you gained the expected amount of 0.25(20) = $5.

This requires multiple rounds (i.e. "on average"), because if you play it once, you can can never gain 5 (in this scenario: 10/20/40).* You'll either lose 10 or gain 20, as you say. But if you play multiple rounds in order to see that average result, and the envelope stuffer never puts in 40--always 10/20--you'll never gain 20. His choice to stuff the envelopes was never said to be random. You can't assume that. But you are, by assuming more variability than exists in a definite situation.

*[Edit: actually, I'm not even sure how you would gain 5 on average, in a game which has only 10, 20, or 40. Can you describe such a scenario? Who is making change, here? How does this ever add up to 5??]

Zarathustra wrote:Let's say you play the game repeatedly, and the person stuffing the envelopes keeps putting in 10s and 20s. There will NEVER be a time when, upon getting the 20, the other envelope could possibly contain 40. Every time you get that 20, and you switch, you're losing.

This is another reason the rest of you are wrong--you keep changing the problem. First "you may switch envelopes" turned into "you may switch envelopes any number of times" and now, all of a sudden, the person handing you the envelopes is changing the amount of money in them.

When the rest of you pick one, specific problem and stick with it without changing the rules in the middle of the discussion, please let me know.

Hashi Lebwohl wrote:

Zarathustra wrote:Let's say you play the game repeatedly, and the person stuffing the envelopes keeps putting in 10s and 20s. There will NEVER be a time when, upon getting the 20, the other envelope could possibly contain 40. Every time you get that 20, and you switch, you're losing.

This is another reason the rest of you are wrong--you keep changing the problem. First "you may switch envelopes" turned into "you may switch envelopes any number of times" and now, all of a sudden, the person handing you the envelopes is changing the amount of money in them.

When the rest of you pick one, specific problem and stick with it without changing the rules in the middle of the discussion, please let me know.

I didn't say the person is changing the amount of money in them. I said he's putting 10s and 20s in them each time, and yet your argument depends on the possibility that he might put in $40 in one envelope.

Nor did I say you can switch any number of times. I said that your logic would tell you that it would be the right choice to switch, even after you've just switched (regardless whether that's allowed or not). It's the logic I'm debating, not the initial conditions. If your logic would hypothetically lead you to reverse your decision--and it would!--then there was something wrong with your first decision and your logic.

And none of that has anything to do with getting $5 out of a 10/20/40 scenario. I'm not trying to be difficult. I'm honestly puzzled.

Has anyone considered running this as an experiment? It would be exceedingly easy to do. I think I'll take a stab at it.

I can already predict, with a stack of 10s and 20s, I'll never come out with 1.25 of either denomination.

Okay, let me try this one last way (and yes, it's a reprise of what's been said before but still...)

Hashi Lebwohl wrote:
The original problem:

TheFallen wrote:There are two sealed envelopes and one contains twice the amount of money than the other. That's all you know.

You are given one of the sealed envelopes.

You are told that you can either keep the envelope you've got... or exchange for the other sealed envelope. Is there anything you can do in order to maximise potential gain?

That is an exact description of the starting conditions of the problem. You know no more and no less than as stated above. No semantic trickery of any kind.

Okay Hashi, bear with me.

Would you agree with the following statement?

The total (which we will call T) of the cash within the two envelopes must in every case consist of 3X, where X is the smaller amount of cash residing in one of the envelopes - and 2X is the larger amount of cash residing in one of the envelopes.

So T = 3X and of course therefore X = T/3.

If you do not agree with that statement, please point out where it is wrong?

So, presuming you do agree with that statement, when you are initially given an envelope at random, you are either given one containing X (or T/3 if you'd rather) or you are given one containing 2X (or 2T/3, if you'd rather).

So... the only two possibilities - with a 50/50 chance of either - are:-

1. that you are initially given an envelope that contains X. If so, the other envelope must contain 2X, so if you swap, you will gain X (you started with X and you exchanged for 2X).

or

2. that you are initially given an envelope that contains 2X. If so, the other envelope must contain X, so if you swap, you will lose X (you started with 2X and you exchanged for X).

On that basis, if you see nothing faulty in the above mathematical modelling, you will agree that there is no benefit in switching. It's a pure 50/50 between ending up with X or 2X, no matter whether you stick or swap.

Again, please feel free to point out where any flaw in the above lies.

If you cannot find any flaw in the above reasoning/maths, then you KNOW that your own mathematical modelling (where you recommend switching) MUST be wrong.

Now... if I slightly but crucially change the initial set-up, your modelling would then become correct. Here's a description of a set-up where you would be correct.

To make Hashi right, it was wrote: Revised initial conditions to make Hashi's assertion correct

A fixed amount of cash is placed in an envelope marked A. A fair coin flip is then used to decide whether to put twice that amount of money in an envelope marked B, or half that amount of money in an envelope marked B. You are then handed the envelope marked A and asked if you want to switch.

In the above revised case, your reasoning would be absolutely correct - the odds of increasing your gain are definitely in your favour if you switch - exactly as per the equation ... 0.5(X/2) + 0.5(2X) = 1.25X ... that you've stated several times.

BUT that is NOT the problem as originally laid out.

And if you examine the difference between the two statements of initial conditions, you may well see where your reasoning is faulty regarding the problem as originally posed.

Oh... and opening your envelope before deciding what to do in EITHER scenario as detailed above does not make the slightest bit of difference.

Vraith, what were you saying about this being simple?

Zarathustra wrote:I can already predict, with a stack of 10s and 20s, I'll never come out with 1.25 of either denomination.

erm...that is not how "expected value" works. The "expected value" when you roll a d6 is 3.5 but you will never actually roll a 3.5.

Zarathustra wrote: By having the temerity to disagree with the mod of this forum? By embarrassing you in pointing out your errors?

You can disagree with me all you want.
And I'm not embarassed by errors...well, except when I say "hey, steve, how ya doin?" and the person is actually dave.
Other than that, I consider correction a plus.

[[fuck it...stuff deleted, cuz this ain't gonna be the tank-tone or devolving place. You said something prickish [as did WF further upthread, and I'd have told him the same if I'd seen it in a reasonable amount of time, and TF pointedly moved on from it]. It's hard in the tank to avoid that shit sometimes. It's easy here. Take the easy way.]]


TF: About saying it's easy...
I suspect there are still cross purposes/differences in satisfying answers.
It looks to me like the odds/stats of the possible outcomes are precisely 50/50.
But because of the [extraneous to the probabilities] "more better" VALUES of the players things go fuckity.
Suppose, for whatever reason, the setup is identical, but the "better" outcome was the LOWER value? [x and 2x are the number of fingers the mob boss at this facility cuts off]. As far as I can tell, the higher and lower outcomes occur at precisely the same rate. Half the time.
The math doesn't change. The strategy doesn't change. Only your value/preferences are different.

TheFallen wrote:BUT that is NOT the problem as originally laid out.

Is the presumption that I misquoted you?

Anyway, now I see how we have arrived at different perspectives on this problem and the disconnect is a simple one: you care about how much money is in the initial envelope you are given whereas I do not.

You are trying to figure out a strategy based on whether the envelope you are given contains x or 2x. If that is the initial presumption then you do wind up with the "optimal" strategy being 50/50. The problem with that reasoning is that the envelope you are given does not contain x or 2x; rather, it contains .5(x + 2x) = 1.5x, which will be exactly how much is in the other envelope...until you open them. You also care about whether you are losing money; I do not.

The way I look at the problem is this: I am given an envelope which contains m financial units; probability tells us that the other envelope contains .5(.5m + 2m)--since it could be half as much or twice as much--or 1.25m. This means my strategy is simple--always switch.

V, I honestly have no idea what you're talking about re: "prick." Maybe if you could tell me which of my words/sentences were the offending ones--instead of just hurling insults/sexual pejoratives at me---I could better adapt my language to your demonstrably high standards.

Anyway, I don't think either you or Hashi is fully appreciating that TF showed precisely how Hashi's reasoning (and thus the reasoning in the original problem) would be CORRECT. It is a small but crucial variant. And by crucial, I mean that it demonstrates by example precisely how the reasoning in the original problem (and by extention , Hashi's reasoning) is wrong. Making that one change means that switching IS the correct strategy, for the mathematical reasons Hashi listed.

In case you skimmed over it: the change is having the second envelope determined by a coin toss, thus making it random and variable, rather than fixed, while having the results of the coin toss determine either 2x or 1/2x, thus keeping it within the original parameters.

Now, the task for you, Vraith, is to explain--upon your "pyschological" explanation--how this is the case. Honestly, I don't think you can explain the difference with your theory, and this inability is why I think your explanation misses the point and over simplifies the issues at hand.

Hashi Lebwohl wrote:

TheFallen wrote:BUT that is NOT the problem as originally laid out.

Is the presumption that I misquoted you?

No. The fact is that there is a fault in your reasoning re the problem as originally laid out.

Hashi Lebwohl wrote:Anyway, now I see how we have arrived at different perspectives on this problem and the disconnect is a simple one: you care about how much money is in the initial envelope you are given whereas I do not.

Again incorrect, I am afraid. This is not about "caring". It's about coming up with a valid mathematical model that describes the game properly and accurately.

Hashi Lebwohl wrote:You are trying to figure out a strategy based on whether the envelope you are given contains x or 2x. If that is the initial presumption then you do wind up with the "optimal" strategy being 50/50.

Yes... except that is the ONLY VALID way in which to model the problem. There is no "choice" in how to correctly model it.

Hashi Lebwohl wrote:The way I look at the problem is this: I am given an envelope which contains m financial units; probability tells us that the other envelope contains .5(.5m + 2m)--since it could be half as much or twice as much--or 1.25m. This means my strategy is simple--always switch.

But looking at the problem in that way - and thus coming up with that equation - is incorrect. That equation CANNOT be validly applied to this situation.

Okay let's try this. Say that one of the envelopes is green and the other red. Everything else is exactly the same as per the originally posed problem.

I give you the green envelope to start with - and you assert that swapping envelopes is an advantageous strategy. I note this down, stop that game. THEN WITHOUT CHANGING A DAMN THING (so using the exact same envelopes with the exact same amounts in) I start a new game with you, this time handing you the red envelope to start with. If your modelling of the problem is correct, you will necessarily again maintain that swapping is advantageous.

Okay so on this second occasion I let you swap, so you pass me the red envelope and I give you the green. BUT this time I continue the game and ask you (having already swapped once) again whether you'd like to stick or swap... Again by your reasoning, you would have to say that it's advantageous to swap... so I let you, and then AGAIN offer you the same "stick or swap" choice...

That's the reductio ad absurdum that we've all tried to demonstrate. Your modelling leads to infinite swapping (if a chance to swap is offered every time you come into possession of an envelope) - and that's clearly nonsensical. Therefore there MUST be an error in your modelling.

Zarathustra wrote:In case you skimmed over it: the change is having the second envelope determined by a coin toss, thus making it random and variable, rather than fixed, while having the results of the coin toss determine either 2x or 1/2x, thus keeping it within the original parameters.

Zee, you missed an (I believe) crucial thing about the revised set-up and the changes needed to make Hashi right... as per the bolding below.

Revised Set-up wrote:Revised initial conditions to make Hashi's assertion correct

A fixed amount of cash is placed in an envelope marked A. A fair coin flip is then used to decide whether to put twice that amount of money in an envelope marked B, or half that amount of money in an envelope marked B. You are then handed the envelope marked A and asked if you want to switch.