What's a boy to do?
 

On 30 Oct 2006 16:17:56 -0800, "riverman" wrote:


First, list all the ways to throw three darts, A B and C.

ABC
ACB
BAC
BCA
CAB
CBA

Those aren't all the ways...think about it.

Remember, we are looking at a conditional probability; dart B has
already landed farther than dart A. So our list of outcomes is limited
to:

ABC
ACB
CAB

No, it isn't...think about it.

Our 'definition of success' is when dart C lands further than dart A,
which is clearly only the first two arrangements. So the probability of
throwing a third dart that lands farther than the first (given the
second dart has already landed farther than the first), is 2/3.

Its an unsettling conclusion, because people want to make the argument
that the distance from the bullseye affects the probability of each
outcome.

Well, perhaps it's because of that, or perhaps because it's
wrong...think about it.

However, every possible distance affects every outcome
equally, so they are all still equally likely, as counterintuitive as
it may be.

Maybe it would help you get on-target answer-wise if you tied a string
to your finger in exactly the same spot two days in a row...

HTH,
R

--riverman

What's a boy to do?
 

riverman wrote:
Its an unsettling conclusion, because people want to make the argument
that the distance from the bullseye affects the probability of each
outcome. However, every possible distance affects every outcome
equally, so they are all still equally likely, as counterintuitive as
it may be.

I agree with the basic premise and conclusions as you present them; but
I have to disagree that the six original possibilities are equally
possible as the original problem is stated. If all three darts are
thrown at the same time, and the probabilities are computed based on
the final position of the three darts, then I agree. But if the darts
are thrown sequentially, and with the implied effects of skill and
intent (and OBROFF, sobriety), then I'm unconvinced that the outcomes
are equally random.

I like the parallels of this question with the MH paradox. The
intuitive answer for an individual case is different than the
mathematical answer for a large sample size. I suppose that's what
makes them interesting.

Joe F.

What's a boy to do?
 

"Jonathan Cook" wrote in message
...
Wolfgang wrote:

An interesting problem was recently brought to my attention.

I think I've seen this before but never really thought about it.
Thanks.

I like you so I'm going to make this easier for you," and I remove board
number three to show you that it has a "you lose" tag under it.
Obviously,
the five dollars must be under one of the other two. "So," I say, "would
you like to stick with your original pick, or change your mind?"

It is a given that the game is not rigged in any way and you are not
being
fooled by anything ambiguous or otherwise misleading in the description.

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

I agree with everything that's been written here on what the "correct"
solution is, and how Marilyn made a bunch of PhDs and professors look
stupid (not that hard, really), but the problem has been irking me,
and I finally decided why.

You posed the problem _singularly_. One try. Probability is about
expected outcomes over lots of attempts. It breaks down in a singular
event.

Expermental probability is about expected outcomes observed over multiple
attempts, while theoretical probability is about expected outcomes
calculated by counting possible outcomes. But Probabilities in general ARE
about predicted outcomes for singular events.

As a _singular_ event, you either have the right board or not,
there is no "law of averages" to consider. And singularly, I'm not
convinced that it is worth switching boards (though I absolutely agree
that over lots of tries it is).


Put the numbers 1-100 in a hat, and draw one. It could be #1, or it might
not. It is correct to say that, the probability of it being #1 is 1/100
based on 100 draws, but as a singular event, its either #1 or its not. But
those two outcomes (that it is or its not) aren't equally likely. Its far
more likely to be something other than #1, so if I drew a number and said
"want to bet a horse that it's #1", you'd refuse the bet, because the
probabilities are well against you.

Think of it this way. What if you randomly turned over one of the
remaining boards, and when the board had "you lose" on it, you offered
to let me switch with the other one. Should I? Obviously in this case
it doesn't matter, because I still lose on the times the board with the
money is turned over, so switching or not I'll only win, on average, 50%
of the remaining ones, which is 1/3 of all of them. Randomness doesn't
add information to the problem like purposely turning over the losing
board does. But suppose I just play the game once, and the random
turnover displays a "you lose" board. Then it looks just like your
"play once" scenario.

If we RANDOMLY turned over a board, then it would be the winning board 1/3
of the time. In that case, you have a 0% chance of winning no matter what
you do. If the board we randomly turned over was a losing board, then you'd
have a 50% chance of winning by staying or switching. If the original
problem said 'you pick a board, then Wolfie randomly turns over another
board and it says 'you lose', then you might as well keep your original
choice. But that's not the problem being stated. Monty Hall (or Wolfgang, in
this case) is not RANDOMLY turning over a board: he knows the winning board
and he does NOT turn that over. Its not random, so not only does randomness
not add info to the problem, it doesn't even apply.


Probability is not the right analysis for a singular event. At least
that's what I think now, and now I'm going to bed...maybe I'll be wrong
in the morning ;-)

Yes, you will be. Sorry. :-)

--riverman

What's a boy to do?
 

wrote in message
...
On 30 Oct 2006 16:17:56 -0800, "riverman" wrote:


First, list all the ways to throw three darts, A B and C.

ABC
ACB
BAC
BCA
CAB
CBA

Those aren't all the ways...think about it.

Remember, we are looking at a conditional probability; dart B has
already landed farther than dart A. So our list of outcomes is limited
to:

ABC
ACB
CAB

No, it isn't...think about it.

Our 'definition of success' is when dart C lands further than dart A,
which is clearly only the first two arrangements. So the probability of
throwing a third dart that lands farther than the first (given the
second dart has already landed farther than the first), is 2/3.

Its an unsettling conclusion, because people want to make the argument
that the distance from the bullseye affects the probability of each
outcome.

Well, perhaps it's because of that, or perhaps because it's
wrong...think about it.

However, every possible distance affects every outcome
equally, so they are all still equally likely, as counterintuitive as
it may be.

Maybe it would help you get on-target answer-wise if you tied a string
to your finger in exactly the same spot two days in a row...


Thats not possible.

--riverman

What's a boy to do?
 

On Oct 27, 12:40 pm, "Wolfgang" wrote:
An interesting problem was recently brought to my attention.

Let us say that you and I are standing next to a table on which I have
placed three boards identical in every respect except that each has a
different number painted on it.....1, 2, and 3, respectively.

I say to you that if you turn your back I will place a five dollar bill
under one of the boards and a slip of paper that says "you lose" under each
of the others. You then turn back to face the table and point to or name
the board you think has the five dollar bill under it. If you're right, you
win the five bucks.

We proceed. You pick, say, board number one. I say, "O.k., tell you what,
I like you so I'm going to make this easier for you," and I remove board
number three to show you that it has a "you lose" tag under it. Obviously,
the five dollars must be under one of the other two. "So," I say, "would
you like to stick with your original pick, or change your mind?"

It is a given that the game is not rigged in any way and you are not being
fooled by anything ambiguous or otherwise misleading in the description.

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

Wolfgang

It don't matter. Play the game 300 times and you win 100 no matter how
you do it.
Initially your chances are 1/3.
With the give away the chances are expressed (1/2)*(2/3), which is
1/3 same as above. Again if you play 300 times you win 100.
If I remember correctly Poission figured all this out working for a
wealthy French nobelman who loved to gamble.

What's a boy to do?
 

Jonathan Cook wrote:
Wolfgang wrote:

An interesting problem was recently brought to my attention.

I think I've seen this before but never really thought about it.
Thanks.

You're welcome.

I like you so I'm going to make this easier for you," and I remove board
number three to show you that it has a "you lose" tag under it. Obviously,
the five dollars must be under one of the other two. "So," I say, "would
you like to stick with your original pick, or change your mind?"

It is a given that the game is not rigged in any way and you are not being
fooled by anything ambiguous or otherwise misleading in the description.

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

I agree with everything that's been written here on what the "correct"
solution is, and how Marilyn made a bunch of PhDs and professors look
stupid (not that hard, really),

I'll risk belaboring a point here because I believe it is an important
one. Ms. Savant did NOT make anyone look stupid. I think she was
certainly aware that many people would immediately jump to the wrong
conclusion.....else, why bother with what is in the final analysis a
very simple problem in mathematics? The whole point of the exercise is
that the answer IS counterintuitive. What made people look stupid
wasn't coming up with the wrong answer which was (and is) after all
something akin to falling into the trap of trying to figure out the
answer the question of where to bury the survivors of a plane crash
that occurs smack dab on an international border. What makes people
look stupid is insisting on the wrong answer after the correct (and
simple) one has been revealed and explained. Well, that and, in this
instance, Haddon's nasty little bias trick. I get to keep my shiny new
nickel because no one noticed.....or at least no one pointed
out.....that all of the outraged authors of the quotes he used were
associated with an institution of higher learning, 5 of 6 were
identified as Ph.D.s, and several references were made to mathematics
and mathematicians while none of the authors was identified as such. I
have a hard time believing that this is a representative sample of all
the letters sent to Ms. Savant in response to her exposition of the
Monty Hall problem.

but the problem has been irking me,
and I finally decided why.

You posed the problem _singularly_. One try. Probability is about
expected outcomes over lots of attempts. It breaks down in a singular
event. As a _singular_ event, you either have the right board or not,
there is no "law of averages" to consider. And singularly, I'm not
convinced that it is worth switching boards (though I absolutely agree
that over lots of tries it is).

The trouble here is at least partially one of semantics (I cannot for
the life of me understand why semantics is so widely accepted as a
pejorative term.....but that's another rant altogether). Playing the
game once constitutes a "single" event, not a "singular" one in any
meaningful sense. The difference is critical. A singular event is
something that happens only once......something like the evolution of
life on Earth, to pick a particularly controversial example. Playing
the game once may APPEAR to be singular if one stresses all the details
about who is involved, what they are wearing today, what they had for
breakfast etc. but, in all its essentials, it is identical to millions
of other events. It is NOT singular. The laws of probabilities
apply......MUST apply......not because this particular avatar is
repeated, but because it is repeatable and myriad others like it in
every essential detail have been repeated often enough for the
mathematically derived probabilities to be confirmed experimentally.

At any rate, you DO believe that probabilities apply to single events
and I can prove it easily. All we have to do is raise the
stakes......we don't even need to calculate the odds with any
precision. You are in a large airplane of a type famous for its
ability to glide like a brick, and the engines fail. You have a
parachute. You have never used a parachute before. I think we may
take it as a given that you are not likely to repeat the
experiment......and your internet connection is too slow for Google be
of much use in finding someone else who has. What do you
do......jump?.....or ride it out?


Think of it this way. What if you randomly turned over one of the
remaining boards, and when the board had "you lose" on it, you offered
to let me switch with the other one. Should I?

Yes, absolutely. Despite the fact that introducing a random element
has entirely changed the nature of the problem, as long as the random
pick turns up a loser, the outcome is identical to that of the
original. Changing your pick doubles the odds of winning.

Obviously in this case
it doesn't matter, because I still lose on the times the board with the
money is turned over, so switching or not I'll only win, on average, 50%
of the remaining ones, which is 1/3 of all of them. Randomness doesn't
add information to the problem like purposely turning over the losing
board does. But suppose I just play the game once, and the random
turnover displays a "you lose" board. Then it looks just like your
"play once" scenario.

Probability is not the right analysis for a singular event. At least
that's what I think now, and now I'm going to bed...maybe I'll be wrong
in the morning ;-)

Jon.

What's a boy to do?
 

Wolfgang wrote:
Jonathan Cook wrote:

....
Think of it this way. What if you randomly turned over one of the
remaining boards, and when the board had "you lose" on it, you offered
to let me switch with the other one. Should I?

Yes, absolutely. Despite the fact that introducing a random element
has entirely changed the nature of the problem, as long as the random
pick turns up a loser, the outcome is identical to that of the
original. Changing your pick doubles the odds of winning.

There was supposed to be more to this......yeah, I know, just what
everybody wanted to hear! :)

I'm still not used to posting on Google. Hit the "post message" button
too soon.

Will try again with the whole thing (I hope) shortly.

Wolfgang

What's a boy to do?
 

Jonathan Cook wrote:
Wolfgang wrote:

An interesting problem was recently brought to my attention.

I think I've seen this before but never really thought about it.
Thanks.

You're welcome.

I like you so I'm going to make this easier for you," and I remove board
number three to show you that it has a "you lose" tag under it. Obviously,
the five dollars must be under one of the other two. "So," I say, "would
you like to stick with your original pick, or change your mind?"

It is a given that the game is not rigged in any way and you are not being
fooled by anything ambiguous or otherwise misleading in the description.

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

I agree with everything that's been written here on what the "correct"
solution is, and how Marilyn made a bunch of PhDs and professors look
stupid (not that hard, really),

I'll risk belaboring a point here because I believe it is an important
one. Ms. Savant did NOT make anyone look stupid. I think she was
certainly aware that many people would immediately jump to the wrong
conclusion.....else, why bother with what is in the final analysis a
very simple problem in mathematics? The whole point of the exercise is
that the answer IS counterintuitive. What made people look stupid
wasn't coming up with the wrong answer which was (and is) after all
something akin to falling into the trap of trying to figure out the
answer the question of where to bury the survivors of a plane crash
that occurs smack dab on an international border. What makes people
look stupid is insisting on the wrong answer after the correct (and
simple) one has been revealed and explained. Well, that and, in this
instance, Haddon's nasty little bias trick. I get to keep my shiny new
nickel because no one noticed.....or at least no one pointed
out.....that all of the outraged authors of the quotes he used were
associated with an institution of higher learning, 5 of 6 were
identified as Ph.D.s, and several references were made to mathematics
and mathematicians while none of the authors was identified as such. I
have a hard time believing that this is a representative sample of all
the letters sent to Ms. Savant in response to her exposition of the
Monty Hall problem.

but the problem has been irking me,
and I finally decided why.

You posed the problem _singularly_. One try. Probability is about
expected outcomes over lots of attempts. It breaks down in a singular
event. As a _singular_ event, you either have the right board or not,
there is no "law of averages" to consider. And singularly, I'm not
convinced that it is worth switching boards (though I absolutely agree
that over lots of tries it is).

The trouble here is at least partially one of semantics (I cannot for
the life of me understand why semantics is so widely accepted as a
pejorative term.....but that's another rant altogether). Playing the
game once constitutes a "single" event, not a "singular" one in any
meaningful sense. The difference is critical. A singular event is
something that happens only once......something like the evolution of
life on Earth, to pick a particularly controversial example. Playing
the game once may APPEAR to be singular if one stresses all the details
about who is involved, what they are wearing today, what they had for
breakfast etc. but, in all its essentials, it is identical to millions
of other events. It is NOT singular. The laws of probabilities
apply......MUST apply......not because this particular avatar is
repeated, but because it is repeatable and myriad others like it in
every essential detail have been repeated often enough for the
mathematically derived probabilities to be confirmed experimentally.

At any rate, you DO believe that probabilities apply to single events
and I can prove it easily. All we have to do is raise the
stakes......we don't even need to calculate the odds with any
precision. You are in a large airplane of a type famous for its
ability to glide like a brick, and the engines fail. You have a
parachute. You have never used a parachute before. I think we may
take it as a given that you are not likely to repeat the
experiment......and your internet connection is too slow for Google be
of much use in finding someone else who has. What do you
do......jump?.....or ride it out?


Think of it this way. What if you randomly turned over one of the
remaining boards, and when the board had "you lose" on it, you offered
to let me switch with the other one. Should I?

Yes, absolutely. Despite the fact that introducing a random element
has entirely changed the nature of the problem, as long as the random
pick turns up a loser, the outcome is identical to that of the
original. Changing your pick doubles the odds of winning.

Obviously in this case
it doesn't matter, because I still lose on the times the board with the
money is turned over,

Don't look now, but you've just claimed that it doesn't matter
what you do in one situation because something else might happen in a
different situation. If logic counts for anything, you have just torn
the fabric of the universe asunder.

so switching or not I'll only win, on average, 50%
of the remaining ones, which is 1/3 of all of them.

O.k., I'll stipulate that your analysis of the numbers is correct.
So what? All you've done devise a scenario in which the odds of
winning are 50:50. This has no bearing on the original problem in
which the point of the whole thing is that the odds are NOT 50:50.

Randomness doesn't
add information to the problem like purposely turning over the losing
board does.

Correct.....and it isn't adding anything to understanding or
elucidating it either.

But suppose I just play the game once, and the random
turnover displays a "you lose" board. Then it looks just like your
"play once" scenario.

Sure it does. But the important thing is that it plays out like it
too......it doubles the odds of winning.

Probability is not the right analysis for a singular event. At least
that's what I think now, and now I'm going to bed...maybe I'll be wrong
in the morning ;-)

Well, you're right about the utility of probabilities in predicting
singular events. But you are wrong in thinking that this is what we
are dealing with.

Toivo and Aino are back. This time they're in Vegas and they're
playing a somewhat different game. There are ten thousand boards and
only one of them has a five dollar bill under it. As before, Toivo
gets one pick and Aino gets the rest. They are going to play the game
just one time. You don't get to play at all. You just get to bet on
who is going to win. You get to bet just one time.

Who are you going to bet on?

Wolfgang
and how much would you be willing to risk? :)

o.k......that's all of it.

What's a boy to do?
 

As soon as you lifted #3 and exposed it as "you lose", the problem was
over.

Now we have a new one:
Two boards -- one with a five and one without.
By asking me if I wish to change my mind, the new problem is simply one
of choosing #1 or #2. I do this by saying yes or no.
My probablity of getting the $5 is simply 0.5

SO, in answer to the question: "what do I do", I flip a coin.

The dart problem is indeterminate -- not enough information about
unstated variables.

cheers

oz -- there's these two trains, heading towards each other with a bee
flying............

What's a boy to do?
 

MajorOz wrote:
As soon as you lifted #3 and exposed it as "you lose", the problem was
over.

Well, not quite.....there was still the matter of making a
choice.....AFTER figuring out what the best choice is.

Now we have a new one:
Two boards -- one with a five and one without.
By asking me if I wish to change my mind,

Huh? Who is asking you to change your mind about what? The scenario,
as stated, gives no hint that you have done, said, or otherwise decided
anything about which to change your mind.

the new problem is simply one of choosing #1 or #2.

Huh? What was the old problem? (um......is anybody else seeing a whole
bunch of words here that aren't showing up on my screen?)

I do this by saying yes or no.

What are you saying "yes" or "no" to? Is it perhaps #1?.......or maybe
#2?.....something invisible to mere mortals?

My probablity of getting the $5 is simply 0.5

O.k........if you say so.

SO, in answer to the question: "what do I do", I flip a coin.

Toward what end?

The dart problem is indeterminate -- not enough information about
unstated variables.

We await the detailed analysis with bated breath......or
palpitations......or something.

cheers

Prosit!

oz -- there's these two trains, heading towards each other with a bee
flying............

Huh?

Wolfgang
who is beginning to think that perhaps brother skwalid has a point
after all.......this universe is starting to get a disturbingly skewed
look to it. :(