What's a boy to do?

Scott Seidman wrote:
"Wolfgang" wrote in :


The question.......what should you do?

Wolfgang


If I recall correctly, you should change your mind. When you chose the
first, you had a 1/3 chance of being right, and nothing has changed that.
If you change your mind now, that gives you a 2/3 chance of being correct.
I'm pretty sure I'm dead wrong on that 2/3 number, though, but the chance
is more than 0.5.

Actually, I believe your exactly right about the 2/3.

Um.....well, it's been a couple hours since I last looked at the
solution, so I could be wrong.

The key is that the removal process is not random.

Yep.

Wolfgang

"Wolfgang" wrote in message
ups.com...
Actually, I believe your exactly right about the 2/3.

Um.....well, it's been a couple hours since I last looked at the
solution, so I could be wrong.

The key is that the removal process is not random.

Yep.

Wolfgang

Yes, the key from a pure mathematical probability standpoint is that the
removal process is not random. However, from a human nature standpoint the
fact that the removal is not random could also dictate in some circumstances
that I should not switch.
If the rules are that you must always reveal one of the losers, then the MH
problem solution dictates that it is in my interest to switch. However such
a rule was not stated in the question you posed. If I suspect that you are
aware that I most likely am familiar with the MH problem solution, and if I
also think that you think that I am unaware that you have that knowledge
then, if you reveal one of the losers, it is probably not in my interest to
switch (geez what a tortured sentence). Conversely, with those respective
mindsets, if you choose not to reveal one of the losers I probably should
switch. Of course after playing a few times in the absence of a rule to
always reveal one of the losers, the activity would quickly go to each of us
trying to second guess the other.

Bob Weinberger

On Sat, 28 Oct 2006 06:35:31 GMT, "Bob Weinberger"
wrote:


"Wolfgang" wrote in message
oups.com...
Actually, I believe your exactly right about the 2/3.

Um.....well, it's been a couple hours since I last looked at the
solution, so I could be wrong.

The key is that the removal process is not random.

Yep.

Wolfgang

Yes, the key from a pure mathematical probability standpoint is that the
removal process is not random.

Not as I see. As I see it, those supporting 50-50 odds aren't looking
at the situation properly. I vaguely remembered the puzzle with Marilyn
vos Savant, and of the explanations I saw, none really phrased the
explanation both simply _and_ accurately (not that they aren't out
there, I just didn't see them). Here's my explanation:

Look at in reverse. Given the way Wolfgang phrased it, by switching,
you essentially get to pick two boards. Let's change the phrasing such
that if once you had picked board #1, he had given you the choice of
sticking to that choice (1 chance in 3) or switching to pick _both_ of
the two remaining boards (2 chances in 3), most people can easily see
the odds advantage of switching and as such, would switch to picking two
boards. You know one of the two will not and cannot have a five under
it.

At that point, Wolfgang turns one board of your two over and it's the
one that isn't the five. He now offers you the choice of switching back
to board #1. If you switch, you have traded your two-board pick back to
Wolfgang for your original one board (1 chance in 3) pick. The fact that
one board of your two-board (2 chances in 3) pick is now revealed is not
material to the odds. You knew and expected that one of the two boards
of your two-board choice couldn't and wouldn't have the five under it,
so why would the fact that things are as you expected and as they have
to be influence your new choice?

This explanation is for the puzzle as Wolfgang explained, not all
possible variants. For example, if a third person walks up at the point
after the first board is turned over and is offered a chance to get in
on things by picking one of the two remaining boards, their odds are
50-50, but they had different "rules" (this "each "hand's" _rules_ are
different" is why blackjack ain't a heads-up game). And secondly, if
you continue to play and Wolfgang is free to "change the rules" in every
"game," the proper choice could change depending on what he does or
doesn't do.

TC,
R

wrote in message
Not as I see. As I see it, those supporting 50-50 odds aren't looking
at the situation properly.

For example, if a third person walks up at the point
after the first board is turned over and is offered a chance to get in
on things by picking one of the two remaining boards, their odds are
50-50, but they had different "rules" (this "each "hand's" _rules_ are
different" is why blackjack ain't a heads-up game).

As one who understands the mathematics, but still has difficulty
rationalizing the counter-intuitive nature of the answer, I think this
somewhat illuminates the crux. If the question is, "What is the probability
of selecting the correct answer from two remaining random choices?", the
answer is 1/2. That is the simplest and most understandable question.
Everybody gets it. But that's not the actual question posed by the problem,
nor are the choices random. The question posed is, "What is the probablity
of selecting the correct answer through this process?" The correct answer
to that is 2/3.

Joe F.

On Sun, 29 Oct 2006 15:55:00 GMT, "rb608"
wrote:

wrote in message
Not as I see. As I see it, those supporting 50-50 odds aren't looking
at the situation properly.

For example, if a third person walks up at the point
after the first board is turned over and is offered a chance to get in
on things by picking one of the two remaining boards, their odds are
50-50, but they had different "rules" (this "each "hand's" _rules_ are
different" is why blackjack ain't a heads-up game).

As one who understands the mathematics, but still has difficulty
rationalizing the counter-intuitive nature of the answer, I think this
somewhat illuminates the crux. If the question is, "What is the probability
of selecting the correct answer from two remaining random choices?", the
answer is 1/2. That is the simplest and most understandable question.
Everybody gets it. But that's not the actual question posed by the problem,
nor are the choices random. The question posed is, "What is the probablity
of selecting the correct answer through this process?" The correct answer
to that is 2/3.

I don't see how it's (objectively) counter-intuitive, and I think
attempting to get too involved in "math" (beyond the basic) makes it
more, rather than less difficult - for example, if it had been 4 boards,
two were turned over revealing losers, and then the choice to change
were given, to me, common sense indicates the odds say change your pick
because of the same reasons I feel it does with 3. If you must have
"math," I'm fairly sure the formula would be that the odds in favor of
switching are pretty close to if not exactly x-1/x and the odds in favor
of sticking are always exactly 1/x, when x is greater than 2, but I'm
not a mathematician, so ??? Perhaps the odds in favor need to account
for the first pick when x is higher than 3 - such that it isn't quite
x-1/x - but it's always going to be better odds than 1/x. ****, that's
confusing...that's why, IMO, algebra isn't the way to figure this out.

About the only thing I can figure is that it is much like many threads
on ROFF in that most folks, myself included at times, don't always
_read_ what they are "reading," but rather, um, infer from what is
written by what they _think_ is being said. In this case, they are
simply ignoring that there are 3, not 2, boards and therefore, the
chances cannot be 1 in 2.

Heck, given the "game" as outlined by Wolfgang, there's nothing
presented in the "rules" preventing the person from choosing the
revealed losing board - they were simply offered a chance to change
their pick. It would be the chooser making the obvious choice not to
choose it because they can clearly see they won't win (they don't need
to know that the chance of winning is 0 in 3).

TC,
R

Joe F.

wrote in message
I don't see how it's (objectively) counter-intuitive,

There are two remaining choices, switch or don't switch. I think 50:50 is
an easy conclusion to draw from that.

About the only thing I can figure is that it is much like many threads
on ROFF in that most folks, myself included at times, don't always
_read_ what they are "reading," but rather, um, infer from what is
written by what they _think_ is being said.

No doubt many ROFFians are guilty of that; but that is also the intentional
nature of "brain teasers".

In this case, they are
simply ignoring that there are 3, not 2, boards and therefore, the
chances cannot be 1 in 2.

Not exactly, but similar IMO. There are, in fact, 2 choices remaining. The
failure is that of seeing the two choices as random for the original player.
Most folks, I expect, see the problem from the perspective of your "third
person" for whom they *are* random.

Joe F.

rb608 wrote:
wrote in message
I don't see how it's (objectively) counter-intuitive,

There are two remaining choices, switch or don't switch. I think 50:50 is
an easy conclusion to draw from that.

Exactly......and that is precisely what makes the Monty Hall problem
interesting......well, that's a part of it, anyway (more about that in
just a moment). It isn't the math. Hell the math is simple enough
that even I (no math wiz......by ANY stretch of the imagination) have
no trouble at all in understanding and accepting various permutations
of the explanation. Anyone adept at mathematics and who takes a moment
to think it through will invariably come up with the right answer and,
doubtless, find the whole thing rather silly.

Somewhat ironically, it takes a basic knowledge of the fundamental laws
of probability to figure out the wrong answer.....you have to know that
tossing a coin will, in the long run, result in something very close to
half heads, half tails. Anyone who doesn't know this can only
guess.....and is as likely to guess right as wrong.....50:50 chance!
Sweet! There is no doubt in my mind that both Craig Whitaker and
Marylin vos Savant were well aware of this when the former posed the
question and the latter decided to answer it.

At least a couple of people have made references to the rules as I
stated them in posing the problem. In fact, there were NO rules.
There was simply a question about how one should proceed in a precisely
and unambiguously stated situation. Suggestions and speculations about
how to work through more or less similar situations (changing the
"rules") may or may not be interesting in their own right, but they
have nothing whatsoever to do with the original problem. I suspect
that most of them have something or other to do with a certain level of
discomfort engendered by the decidedly counterintuitive correct
solution to the original.

When all is said and done, the whole thing is a trick question. What
makes it exquisitely delicious is that, as stated at the outset, I, the
expositor, was not playing any kind of trick on the player......well
not directly, anyway. No, what tricks the player is his or her own
knowledge of probabilities and a lightning quick recognition of an
absurdly easy problem.

Right, Ken?

O.k., that last bit was just a little unfair. Um......or was it?
After all, Haddon said the pretty much same thing. Did anybody else
see it? A shiny new nickel to the first to point out Haddon's own
sorta nasty little trick. (hint: it's in the quotes......more or less)


Bottom line? The Monty Hall problem really isn't much of a
mathematical puzzle at all. What it IS......in spades......is a
beautifully elegant probe into human psychology!

As for illustrating the logic behind the correct solution, here's my
own humble contribution:

Let us change the scenario a bit. Instead of a single player who gets
to decide whether to change his or her pick after one of the losers is
exposed, let's have TWO players......Toivo and Aino. Toivo gets to
pick one of the three possibilities.....Aino automatically gets the
other two. All three positions are exposed. Any one may be the
winner, but no one should have any difficulty in seeing that the smart
money would bet on Aino. Whether in a single round or in repeated
play, the odds are clearly in his favor to the tune of two to
one.......67% to 33%.....not too roughly. Now, let us suppose that
rather than exposing all the possibilites at once, the expositor turns
over one of the boards at random. How does this change the odds?
Clearly, it has absolutely no effect on the odds. O.k., so, at least
one of Aino's two possibilities HAS TO be a loser.....right? After
all, there are three positions and only one of them is the winner.
Alright, so, if the expositor first turns over one of Aino's
possibilities, which is one of the losers, how does this affect the
odds? Again, it cannot possibly affect the odds......the winner and
both losers are where they are.....NOTHING can affect the fact that
Aino wins two times out of three.....more or less......in the long run.

Now, let's go back to the problem as originally stated. Toivo is the
only player. As long as he sticks with his original choice when given
the option, nothing, in essence, is any different than it was in the
two player game......he loses two times out of three......Aino has
simply become invisible. Obvious......right? Right. O.k., so, what
if Toivo chooses to jump one way one time and another the next? Beats
the **** out of me (and you too, if there's an honest bone in your
body). Ah, but what if Toivo changes his choice EVERY time? Well,
then he quite simply BECOMES Aino!!

Wolfgang
dunkenfeld knew.

wrote in message
...

I don't see how it's (objectively) counter-intuitive, and I think
attempting to get too involved in "math" (beyond the basic) makes it
more, rather than less difficult - for example, if it had been 4 boards,
two were turned over revealing losers, and then the choice to change
were given, to me, common sense indicates the odds say change your pick
because of the same reasons I feel it does with 3. If you must have
"math," I'm fairly sure the formula would be that the odds in favor of
switching are pretty close to if not exactly x-1/x and the odds in favor
of sticking are always exactly 1/x, when x is greater than 2, but I'm
not a mathematician, so ??? Perhaps the odds in favor need to account
for the first pick when x is higher than 3 - such that it isn't quite
x-1/x - but it's always going to be better odds than 1/x. ****, that's
confusing...that's why, IMO, algebra isn't the way to figure this out.

About the only thing I can figure is that it is much like many threads
on ROFF in that most folks, myself included at times, don't always
_read_ what they are "reading," but rather, um, infer from what is
written by what they _think_ is being said. In this case, they are
simply ignoring that there are 3, not 2, boards and therefore, the
chances cannot be 1 in 2.

No comment on any of that. I just wanted to repost it because it may be the
most beautiful thing I've ever seen!

Heck, given the "game" as outlined by Wolfgang, there's nothing
presented in the "rules" preventing the person from choosing the
revealed losing board - they were simply offered a chance to change
their pick. It would be the chooser making the obvious choice not to
choose it because they can clearly see they won't win (they don't need
to know that the chance of winning is 0 in 3).

Um.....well, o.k., this may be even beautifuler.

Wolfgang
hoo boy!