Philosophy Now Forum

Hi dionisos,

First, I have to ask you to NOT insult. I don't insult you for your view and this only makes me less concerned to care to deal with you. You are also here anonymously while I am not. I would actually prefer that if you want to be fair to your authority, you'd have to reveal your own identity. Without doing so, while I can infer your background knowledge, you risk nothing here by speaking one way or the other. If you opt to remain anonymous, then respect that my own presence here to speak makes only me accountable and deserve better treatment.

Second, note how the fact that I'm known and you are hidden relates to this puzzle. The question which appears to be contradictory is whether KNOWING of the host's reveal MATTERs or does NOT MATTER. It appears to be BOTH and why it is reasonably questioned as a paradox (contradiction).

Third, I am aware of a need for the details that include using a logic and a given methodology to measure the 'weights' of the values (the probabilities) of each thing. Default to assuming I understand at least intuitively to the problem. That is, I don't DENY that one may gain an advantage for their knowing and is why I argued before the fact that increased incidents (or multiple games) increases one's odds overall. But the other issue is to whether one gains by + 0, + 1/6, or 1/3? The first 'gain' is one which recognizes no difference between knowing or not knowing [+0]; The second case is the 'gain' when one begins with 1/3 but increases only to 1/2. [1/2 - 1/3 = +1/6]. The third case is the standard one you support: a 1/3 to 2/3 is a gain of +1/3.

Please recognize that I already KNOW the traditional argument of which you appear to believe that I disagree with. What you DO disagree with me is that you believe that there are no other possible perceptions which are valid. I will definitively prove this and at least thank you for adding the motivation for me to even TRY to prove my point as I WILL be able to prove that:

Having knowledge of the host's reveal leads to two distinct possible conclusions. One is that by placing VALUE in the first part of the game applies to the final result. And the second is that placing VALUE in the first part of the game is removed or canceled. The first option is the apparent 'gain' by switching. The second option is split in two: (a) the perception of the gain as REAL [+1/6], and (b) the perception of the gain as NON-REAL [+ 0].

I will be completely formalizing this using (1) a multi-valued logical extension of Boolean logic, as per Peano (or Fuzzy Logic, if necessary), (2) Propositional Calculus, and (3) Math (accepting Algebra, Probability and/or Calculus, if necessary).

I need time to prepare and present this. However, for my efforts and conclusiveness of the proof, I also deserve credit for this if I am able to prove this. Thus I also need the time to determine how to assure my credit. I've already given enough hints at this in this thread but my whole effort could easily become moot if this site has the advantage to either co-opt my effort and/or delete this thread. This has already occurred here as I've witnessed.

Do you accept this? I am not against you either way and will still be doing this on my own regardless. But you can act as a 'witness' and add value by your criticisms as you already have been doing. Just don't forget that you're anonymity requires you to remain fair OR I have no reason to trust you here.

Scott Mayers.

First, I have to ask you to NOT insult. I don't insult you for your view and this only makes me less concerned to care to deal with you.

I have no time now, i will answer tomorrow to the rest of the message, but i wanted to say i don’t want to insult you, and if you felt insulted i am sorry.
My goal was to be honest about what i believed, and i believe you confuse yourself. (and i know that i could be wrong on this).
Take into account my english is bad, and maybe something i said didn’t have the meaning i wanted, my last message was neutral in my perceptive.

Motivations for my concern
(1) Knowing that and how people choose to gamble leads to real problems, I have a personal motivation to dislodge the mystery apparent to justify the perspective of the gambler as well as to those who utilize the behaviors to take advantage of them. When one believes that they improve their odds by acting in some way in games of chance, it can be literally illusive with respect to the ones risking themselves by continuing to gamble. Thus I suspect such puzzles' apparent paradoxes only favor the ones who run these games only.

(2) I also have an interest to show how using contradiction (or apparent paradoxes) can be resolved by including the concern within logic to recognize that contradictions are NOT to be simply represented as something to STOP with. That is, if we come across a contradiction that matters in reality to resolve (paradoxes), we must be motivated to repair this condition rather than ignore going further. This is the basis for other theories that I have which is necessary to understand and so these puzzles act as special cases that can demonstrate how and why I am correct.

Two Solutions?
Here I show what each of the two main problem solutions are.

Let's presume the defaulted opinion of the Monty Hall Problem.

Argument for the 2/3 solution:
At first we have 1/3 chance to win car.
Argument for the 1/2 solution:
Continuing from the above game, assume that in the reveal, we discover that this is Goat 1. Below we thus have the case where Goat 2 remains as a contingent possibility. Now imagine that I only pretend to Switch in my head. In this case, I then want to question whether it is wise to RE-Switch back to the original options and what these odds are?
So only upon imagining that I switch in my head, I then learn that to return back to my original choice, I also gain a 2/3 advantage to return and still get the Car. This means that since both options are 2/3 to Stay and win a car as well as to Switch and win the car. This demonstrates even using the traditional model to argue for this, that it reduces to (2/3) = (2/3) or 50/50 odds to either stay or switch.

And this means that our actual odds are still 1/2. This is because in actuality, once you begin to play the second round, in reality, the game is reset and become 1/2 no matter what.

Note that because the odds are even, that the 2/3 appears to be valid because of that (2/3) = (2/3). This suggests that other perspectives can be realized, such as 1/3. The 1:1 relationship assures this.

Conclusion: There is thus NO advantage to optimize by actually Switching nor Staying. But if one is suckered into believing this, it only encourages people to gamble and lose money in these type of games. This is why I find it harmful and should be challenged.

Scott Mayers wrote: Conclusion: There is thus NO advantage to optimize by actually Switching nor Staying.

Scott. Your hypothesis meets the gold standard for a legitimate scientific theory because it yields a prediction which can be empirically tested. It has in fact been empirically tested millions of times and falsified on every occasion. Why have you never bothered to test it for yourself? All you need is three cups, a coin, a test subject and about ten minutes of your time.

Obvious Leo wrote:

Scott Mayers wrote: Conclusion: There is thus NO advantage to optimize by actually Switching nor Staying.

Scott. Your hypothesis meets the gold standard for a legitimate scientific theory because it yields a prediction which can be empirically tested. It has in fact been empirically tested millions of times and falsified on every occasion. Why have you never bothered to test it for yourself? All you need is three cups, a coin, a test subject and about ten minutes of your time.

I don't know what you're saying? Are you agreeing or disagreeing with my conclusion? And I have done the literal experiments and know my solution works. dionisos accepts the two out of four possibilities above but questioned the weights. In two of the cases where one picks the car, he wanted me to demonstrate the probabilities respecting this is 1/6 each and why he believed you can only consider the two events as 1/3. So I demonstrated this above showing first why the original 2/3 solution 'works'. Then I follow it up to show that it is also 2/3 to return to win the car. It is the fact that the second part of the game fools those believing that the first 1/3 probability still counts when you begin the next part of the game. It cancels this 1/3 and makes it 2/3 if you stayed too. But this means that since both options are true, this is no different than interpreting 1/2 as the real probability to win the car in the game as a whole.

Scott Mayers wrote: I don't know what you're saying? Are you agreeing or disagreeing with my conclusion?

I'm saying your conclusion is testable.

Scott Mayers wrote:And I have done the literal experiments and know my solution works.

How many guesses to switch or stay did you use and what were the results. Most experts say that 50 guesses of each is statistically significant so I may have exaggerated somewhat with my estimate of ten minutes to run the experiment. It's been a while since I did it but half an hour might be a bit more realistic.

Obvious Leo wrote:

Scott Mayers wrote: I don't know what you're saying? Are you agreeing or disagreeing with my conclusion?

I'm saying your conclusion is testable.

Scott Mayers wrote:And I have done the literal experiments and know my solution works.

How many guesses to switch or stay did you use and what were the results. Most experts say that 50 guesses of each is statistically significant so I may have exaggerated somewhat with my estimate of ten minutes to run the experiment. It's been a while since I did it but half an hour might be a bit more realistic.

You can't test it alone other than to the logic. The reason why is because if you do it yourself, you act as both the 'guest' as well as the 'host'. This is because the host tosses a coin in his head and doesn't actually show you his toss. If he ONLY tossed the coin when you have the Car because he knows where it is, the simple fact of him tossing a coin would tell you that the car is in your original selection.

He could opt to hide this by always choosing to toss a coin for the other two remaining cases. But if he did, he'd still have to hide the result of his toss because if you saw it, he couldn't cheat and so would have to expose the car in those two cases. So the only way he could do this is if he fakes tossing a coin in those cases and pretending to accept the value of it. Thus in these two cases, he is acting deceptive (= unfair) even if he doesn't have to fake it when he has real options to reveal a goat.

You cannot possibly do this experiment you're thinking without at least two people. But is is also not necessary. My proof above is NOT a theory, it is a theorem and so doesn't even NEED any experiment. If you are unaware, a "theorem" is an absolute proof. A "theory" is not unless one could prove it using logic alone.

This is a valid Monty Hall experiment which anybody can do.

You are the game host and anybody can be the game subject. It doesn't matter what you tell your subject about the experiment.

1. Place three cups upside-down on the table and hide a coin under one of them while your subject looks away.

2. The subject then guesses under which cup the coin is hidden. You know immediately if he is right or not because you hid the bloody thing so you also know which of the other two cups definitely has no coin under it.

3. If your subject guessed correctly you can then turn over either of the other other two cups to reveal nothing beneath it. However if your subject guessed incorrectly then you have no choice in the matter. You cannot turn over the cup concealing the coin but MUST turn over the one with nothing under it. (N.B. this is the entire point of the story)

4. You then invite your subject to either switch his guess or stick with his original choice but strictly speaking this step is unnecessary. All you need to do is record the result if he had decided to switch and compare it with the result if he had decided to stay. To speed your experiment up you can simply do this as an "as if" scenario because you can then test both the switch and stay guesses at the same time.

5. Repeat the above at least 50 times.

I apologise for the manner in which I've dealt with you on this question, Scott, and I implore you as a friend to perform this experiment. When my eldest son was in high school his entire maths class had to perform this task as a homework assignment and every single result confirmed that switching guesses doubles the chances of correctly guessing the location of the coin. However it was only when the results were collated for the entire class that the odds actually came out at almost exactly 1/3 for staying and 2/3 for switching. I'm sure you don't need anybody to explain to you why this would be so.

There's a little by-play which goes with this story about my eldest son because he chose me as his subject. After he explained what we were doing I just told him I was going to switch guesses every single time regardless. He just gave me that sneering look that adolescent boys are so good at and said "you know this one don't you, you cunning old bugger?" I just winked at him and then we played the game.

Scott Mayers, thanks for trying something more clear.
But i don’t agree, and it is still not the kind for formalized answer i wanted. (I will let Obvious Leo try to convince you on it.)

I still wait for the definition of "probability". I said you i will not go outside of a step by step method.

dionisos wrote:(I will let Obvious Leo try to convince you on it.)

On balance I'd have to say that I've not made a lot of progress thus far but I'm a stubborn bastard.

Obvious Leo wrote:This is a valid Monty Hall experiment which anybody can do.

You are the game host and anybody can be the game subject. It doesn't matter what you tell your subject about the experiment.

1. Place three cups upside-down on the table and hide a coin under one of them while your subject looks away.

2. The subject then guesses under which cup the coin is hidden. You know immediately if he is right or not because you hid the bloody thing so you also know which of the other two cups definitely has no coin under it.

3. If your subject guessed correctly you can then turn over either of the other other two cups to reveal nothing beneath it. However if your subject guessed incorrectly then you have no choice in the matter. You cannot turn over the cup concealing the coin but MUST turn over the one with nothing under it. (N.B. this is the entire point of the story)

4. You then invite your subject to either switch his guess or stick with his original choice but strictly speaking this step is unnecessary. All you need to do is record the result if he had decided to switch and compare it with the result if he had decided to stay. To speed your experiment up you can simply do this as an "as if" scenario because you can then test both the switch and stay guesses at the same time.

5. Repeat the above at least 50 times.

I apologise for the manner in which I've dealt with you on this question, Scott, and I implore you as a friend to perform this experiment. When my eldest son was in high school his entire maths class had to perform this task as a homework assignment and every single result confirmed that switching guesses doubles the chances of correctly guessing the location of the coin. However it was only when the results were collated for the entire class that the odds actually came out at almost exactly 1/3 for staying and 2/3 for switching. I'm sure you don't need anybody to explain to you why this would be so.

The bolded part is where the illusion occurs. The host has two cases at once where he flips a coin in his head. But these two cases are granted as 1/2 x 1/3 of the original = 1/6 each. But because these values are both 'goats', you multiply them by (0)1/6 each. This becomes 0/6 + 0/6 in that case which is equivalent to 0 + 0 = 0. This leaves the other two cases, 1/3 + 1/3 = 2/3.

The trick is in the 0 + 0 = 0. People who perceive the right side of this equation use this when they add up the totals:

(0 + 1 + 1)/3 = 2/3.

The '3' here is the number of elements of the top part of the fraction in the brackets.

But if you include the left side, this becomes

(0 + 0 + 1 + 1)/4 = 1/2,

where the four is the number of relevant terms in the denominator.

Those '0's above represent the two cases where the Guest unknowingly picked the car in the first case so the host has freedom to randomly pick either '0' and the '1's are the cases where the cars are always a win by switching. The problem is that since the Guest does NOT know the cases where he has the car on the first pick [He is unaware], he can't determine when the host is forced to reveal or free to choose. As such, you have to treat each case as being fair.
1) [Hidden car D1][Guest guesses D1] ---> [Host flips a coin H = D2 T = D1] ---> [1/2 time goat 1 (H)]
2) ..................................................................................... ---> [1/2 time goat 2 (T)]
3) [Hidden car D1][Guest guesses D2] ---> [......Host forced to show.......] ---> [1 time Car]
4) [Hidden car D1][Guest guesses D3] ---> [......Host forced to show.......] ---> [1 time Car]

From the Guest perspective, each of the 1/2 times are unable to determine. So they either have to interpret this as:
1) [Hidden car D1][Guest guesses D1] ---> [Host flips a coin H = D2 T = D1] ---> [1 time goat 1 (H)]
2) ..................................................................................... ---> [1 time goat 2 (T)]
3) [Hidden car D1][Guest guesses D2] ---> [......Host forced to show.......] ---> [1 time Car]
4) [Hidden car D1][Guest guesses D3] ---> [......Host forced to show.......] ---> [1 time Car]

OR.... the Guest has to count the events where the host doesn't show you the goat in (3) and (4) as null but equal events even though he cannot determine this...
1) [Hidden car D1][Guest guesses D1] ---> [Host flips a coin H = D2 T = D1] ---> [1/2 time goat 1 (H)]
2) ..................................................................................... ---> [1/2 time goat 2 (T)]
3) [Hidden car D1][Guest guesses D2] ---> [......Host forced to show.......] ---> [1/2 time Car]
4) ..............................................[...Host doesn't show this ever...] ---> [0/2 time Goat NOR Car]
5) [Hidden car D1][Guest guesses D3] ---> [......Host forced to show.......] ---> [1/2 time Car]
6) ..............................................[...Host doesn't show this ever...] ---> [0/2 time Goat NOR Car]

In both of the Guest perspectives above, they are 1/2 or 1/3 in the last. You and dionisos cancel out (4) and (6) in the third perspective without justice OR to assume that the weight of the 1/2 in the first case above is correct without recognizing that once the host reveals one of those goats, ONLY the other goat matters for a loss if you opt to stay instead and equals 1/3 leaving the car in the stay option as 2/3 also!

EDIT: In the last set, this is 1/2 if you count the values of the goats in the first case as 0 Cars. The 1/3 perspective is where you count the goats in the probability but exclude the ones that aren't ever shown for Goat NOR Car.

Humour me, mate. Do the experiment and then we'll talk some more. You don't seem to understand that mathematics can make no statement about the nature of reality because mathematics can only codify a particular narrative of reality as specified by the observer. If you're not willing to take my word for it then you can take it up with the great man himself.

"Mathematics can be used to prove ANYTHING".....Albert Einstein

Obvious Leo wrote:Humour me, mate. Do the experiment and then we'll talk some more. You don't seem to understand that mathematics can make no statement about the nature of reality because mathematics can only codify a particular narrative of reality as specified by the observer. If you're not willing to take my word for it then you can take it up with the great man himself.

"Mathematics can be used to prove ANYTHING".....Albert Einstein

Mate, if Einstein is correct, than even the math is incorrect no matter where it is used including any experiments that use it.

YOU CANNOT USE ANY REPEATED EXPERIMENTS and then average them out.
What is the odds of you being able to win a lottery if you could repeat a draw as many times you want?

Example, If I told you to toss a coin and seek a HEAD for a WIN, where upon I add that if you get a TAILS, you can toss again for each game without end?