What's a boy to do?

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

"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 Fri, 27 Oct 2006 22:45:54 -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...

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

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

"Wolfgang" wrote in message
...
An interesting problem was recently brought to my attention.
The question.......what should you do?

Wolfgang




When you first picked, you had a one in three chance of being right.

With one board removed, your pick 'still' has a one in three (33 1/3 %)
chance of being right.

However, with only two boards left, you can chance your chances. If you pick
a different board (the remaining) you will now have a one in two (50%)
chance of being right. it behooves you to re-pick.

john

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

"Wolfgang" wrote in message
...

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

The easiest way to convince yourself of the correct answer (since it's
non-intuitive) is to play the game with someone. After a short while,
you'll realize that the only way you can get it right if you don't switch is
if you picked it right from the beginning - in other words, 1 chance in 3.

jeffc wrote:
"Wolfgang" wrote in message
...

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

The easiest way to convince yourself of the correct answer (since it's
non-intuitive) is to play the game with someone. After a short while,
you'll realize that the only way you can get it right if you don't switch is
if you picked it right from the beginning - in other words, 1 chance in 3.

Sure, first they take all yer shiny new nickels.......

Hey, I've only got just so many five dollar bills, ya know!

Anyway, that's right.....so long as we stress the "convince" part, as
opposed to learn. Unless and until you become familiar with the
correct solution to the Monty Hall problem (whether it's explained to
you or you work the logic out for yourself) intuition can lead you down
a long and, if it's presented as a betting game, ruinous road.

Wolfgang
who wouldn't bet any of his few remaining shiny new nickels on the
prospect of selling these revolutionary analyses to the folks who run
vegas.

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"?

vince

When you first picked, you had a one in three chance of being right.

Right.

With one board removed, your pick 'still' has a one in three (33 1/3 %)
chance of being right.

John, care to respond to my question, posted above?

vince

"jeffc" wrote in message
.. .

"Wolfgang" wrote in message
...

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

The easiest way to convince yourself of the correct answer (since it's
non-intuitive) is to play the game with someone. After a short while,
you'll realize that the only way you can get it right if you don't switch
is if you picked it right from the beginning - in other words, 1 chance in
3.


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.

MEANWHILE: how about this cherry;

You toss three darts at a target. Dart A misses the target, then Dart B
misses by even more. What is the probability that Dart C will miss by more
than Dart A?

--riverman