What's a boy to do?

"riverman" wrote in message ...

Or just play by yourself:
http://math.ucsd.edu/~crypto/Monty/monty.html
--riverman

I got it right, 3 out of 5 times, by changing my selection each timed.

Op

"Opus McDopus" wrote in message
...

"riverman" wrote in message
...

Or just play by yourself:
http://math.ucsd.edu/~crypto/Monty/monty.html
--riverman

I got it right, 3 out of 5 times, by changing my selection each timed.

Op

Yep. A better way to convince yourself that changing doors is the best
strategy is to make a spinner out of a paper clip and a piece of paper. Draw
a circle divided in thirds, and unbend the paper clip so it works as a
pointer, and hold it in the center with the pencil when you spin it. Agree
beforehand that the prize is in a given section, and decide that you will
always switch. After about three spins, it becomes abundantly obvious how it
all works.

--riverman

On Sat, 28 Oct 2006 20:55:19 -0400, vincent p. norris
wrote:

When you chose the first, you had a 1/3 chance of being right, and nothing has changed that.

Scott, I assume you know what a Tontine is. Suppose you and two
friends form one. Overlooking health and age differences and the fact
that one smokes and drinks heavily, each of you has one chance in
three of winning.

Later, one of the others dies. Now, what is your chance of winning?

vince

Depends somewhat on your and their morality. There's a reason
tontines were outlawed...

Yeah, but I was presenting it as a question involving probabilities,
not morality. (Or did you accidently omit the "t" from "mortality"?

If one of the tontine has none of the former, the other is apt to
discover the quietness of the latter....

--
Antiquis temporibus, nati tibi similes in rupibus
ventosissimis exponebantur ad necem.

http://www.visi.com/~cyli

On Sun, 29 Oct 2006 11:02:22 +0800, "riverman"
wrote:



Or just play by yourself:
http://math.ucsd.edu/~crypto/Monty/monty.html

This puzzle right smack dab in the center of my realm, as its a regular
component of one of my classes. I can take you all to school on the solution
on several levels, but I'm not working today so you're off the hook.

Unfortunately, the first 4 or 5 times I tried it with not changing, I
was right every time. Then the odds started to work out, but I had
those early successes to work out of my mind.

--
Antiquis temporibus, nati tibi similes in rupibus
ventosissimis exponebantur ad necem.

http://www.visi.com/~cyli

"vincent p. norris" wrote in message
...
When you chose the first, you had a 1/3 chance of being right, and
nothing has changed that.

Scott, I assume you know what a Tontine is. Suppose you and two
friends form one. Overlooking health and age differences and the fact
that one smokes and drinks heavily, each of you has one chance in
three of winning.

Later, one of the others dies. Now, what is your chance of winning?

vince

If I'm one of the remaining members I'd say your chances were pretty damned
good....

john

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

On 28 Oct 2006 14:35:06 -0500, basil ratbone wrote:


wrote in message
news
On Sat, 28 Oct 2006 00:24:58 -0400, "Tim J."
wrote:


typed:
On Fri, 27 Oct 2006 23:39:09 -0400, "Tim J."
wrote:


typed:
On Fri, 27 Oct 2006 22:43:52 -0400, "Tim J."
wrote:


Wolfgang typed:
An interesting problem was recently brought to my attention.

I like that one. Here's another less thought provoking oldie:

We put you in a room and fill it with deaf people. Given the room
is now quite crowded, we remove dead people, but add bad people.
How many people are now in the room?

That could be BAFfling...

... but wrong.

How?

deaf-dead = 2, 2 + bad = baf, no?

May not be, but if not, ???

You never left the room.

Um, you didn't say "we you in a room and _add_ deaf people..." Phrased
the way you originally phrased it, wouldn't I be including in "deaf
people?" I'm not a math geek, and I've never heard of this "oldie," I
just thought it was more of logic thing with hex - f to d and back to f,
but ??? I could understand it if it spelled something, but including me
makes it bb0, no? Is that something hilarious to math prof-types or
something?

TC,
R
I would still knock you in the head and take your money along with
the others. I would be the baddest person in the room. there would be a
lot of dead people in the room . All those who woud remove them would also
be dead. so , the answer is no one alive. I would leave with yours and
theirs money. problem solved

BASIL?!...BASIL?!...BASIL?!....you sound like a flowery ****...

"asadi" wrote in message
...

If I'm one of the remaining members I'd say your chances were pretty
damned good....

john

Are we goin to remain on non-speaking terms forever?

Op

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.