Behind Door Number 1, Door Number 2, or Door Number 3
redleathers
Is the prize (vacation, car, boat, etc.) behind door number 1, 2, or
door number 3.
Understandably the constestant would moan and groan, wince and flinch,
under the stress of choosing. But choosing he or she must. Let's say
in this example that the contestant chose door number one. So Monte
would make some kind of deal with him or her for some money or
something and he would then reveal the contents behind one of the
doors other than the one the contestant chose. Let's say in this
example it was door number 3. Well low and behold behind door 3 would
be a barnyard of chickens, a goat, some ducks, or something similar.
Next Monte would ask the question: Now aren't you so glad that you
didn't pick door number 3? And the contestant would say, whew that
was a close call. Then Monte would ask another question: Do you want
to keep the choice you made, door number 1 or would you like to change
your choice and instead take door number 2?
So now this brings us to the point where you get to get involved.
Would you stay with door number one or switch and take door number
two?
Note: This has *everything* to do with the how and the why of my
winning craps methodology.
Agent777
"redleathers" <REVS...@aol.com> wrote in message
news:452144d9.03071...@posting.google.com...
Switch to two. Your original pick was a one-out-of-three chance of picking
the right door. You now have a 50-50 chance.
Mason
Now, it applies to craps? Utter Bilge.
Monty KNOWS which door has be booby prize and can ALWAYS reveal it.
So the fact that he shows it provides no new information. This is the
most important fact in the problem.
Suppose there were 100 doors. Behind 1 is the grand prize and behind
99 are the booby prizes. Contestant is offered his choice and makes
it.
Contestant territory is 1 door. Unclaimed territory is 99 doors.
(Monty opens 98 doors ... each has a booby prize.) This provides no
new information since Monty KNOWS the prize locations and can always
do so.
The contestant is offered the opportunity to switch to the unclaimed
territory.
Choices ....
keep the claimed territory ... with 1 chance in 100 of having the
grand prize
or
switch to the unclaimed territory ... with 99 chances out of 100 of
having the grand prize.
It is the same with three doors in the choice set. Keep the claimed
set with 1 chance in three of having the grand prize or switch to the
unclaimed set with 2 chances of 3 of having the grand prize.
Relevance to CRAPS is ZERO because no one knows what is behind the
doors at the crap table.
But the problem is fun and instructive. That is why it is a classic.
"redleathers" <REVS...@aol.com> wrote in message
news:452144d9.03071...@posting.google.com...
John Kerr
Bother@forgedpostsanonymous.unorg Cymbal Man Freq.
$5,000 in today's dollars), I'd take the money despite knowing that door number
1 or 2 may contain a prize worth 3 times that much. Admission was free except
for the costume and the long wait in line with all the other loonies.
If I won a prize worth $15,000 in today's money, like a vacation cruise, I'd
be liable for the tax on all that and I would only be happy with that if I had
cash included in the prize that covered the tax bite and living expenses while
gone on the trip, and cash for the weeklong hangover after that.
This strategy may cut me out of the final "Let's Make A Deal", but it will
keep me from having yet another chance to win those barnyard animals....they
don't always leave the set after the first appearance.
Remember, winning a car is just buying it for the price of the tax on it,
which could amount to more than 1/3rd of the retail price of the prize. So if he
gave me $5,000 in today's dollars to keep me from choosing a door, but that was
enough to get me into the final round, I could win a car and use that $5,000 in
cash I received earlier to pay PART of the tax due on a new car (plus the
taxable portion of the $5,000).
redleathers
Tom Williams
I saw an old nature film about how male snakes travel through the
jungle looking for nookie. Nature/instinct taught male snakes to
travel in a straight line through a huge forest to maximize chances of
encountering a female snake. Zig-zagging reduced the odds! And
Mason, snakes don't know statistics.
So stay put & don't fall for Monte's zig zag.
boneyard
know about this paradox
"Tom Williams" <run...@outback.com> wrote in message
news:3f124c09...@news4.sucknews.com...
Bruce
REVS...@aol.com (redleathers) wrote:
>> Switch to two. Your original pick was a one-out-of-three chance of picking
>> the right door. You now have a 50-50 chance.
>
>Incorrect.
The only thing incorrect is you and your logic. He is absolutely correct
and is trivially proved correct via experimentation.
billb
> Switch to two. Your original pick was a one-out-of-three chance of
picking
> the right door. You now have a 50-50 chance.
>
>
switch or not it's still 50/50.
--
billb
billb
--
billb
Bother@forgedpostsanonymous.unorg Cymbal Man Freq.
redleathers
> ask Marilyn.
It is the same with three doors in the choice set. Keep the claimed
set with 1 chance in three of having the grand prize or switch to the
unclaimed set with 2 chances of 3 of having the grand prize.--Mason
Yep. Switch to door number 2 for the reason Mason just stated.
stacy_friedman
> Monte Hall ...
[snip]
Ahh, yes, the venerable Monty Hall problem.
First things first. The man's name is Monty, not Monte. And second,
under the common interpretation of the problem (that Monty never
reveals the prize), the correct strategy is to *always* switch doors.
But I'm not going to rehash what's already been written. This link,
http://barryispuzzled.com/zmonty.htm, has a thorough explanation of
the possible interpretations of the puzzle, including the usual one
where Monty always shows you a goat in one of the doors you didn't
pick. That's the interpretation used by Marilyn vos Savant as well
when she answered this question 13 years ago. Check out this link for
her story, as well as good examples of why this is a tough problem to
wrap one's head around:
http://www.willamette.edu/cla/math/articles/marilyn.htm
> Note: This has *everything* to do with the how and the why of my
> winning craps methodology.
No, it most certainly does not. The beauty of the Monty Hall problem
is that it defies intuition. Consider your intuition defied. When
you understand why the correct answer to the Monty Hall problem is
correct, you'll go a long way toward understanding craps math as well.
This includes why your methodology isn't a winning one by any
standard definition of "winning".
tom home runs are boring p
> This includes why your methodology isn't a winning one by any
> standard definition of "winning".
????????
what other "standard" definition(s) of "winning" are you refering to??
winning (as it relates to craps) means coming out ahead (in dollars),
doesnt it? having more money gambling at craps than you started with?
red claims he is winning--that he has more money after using his
"method" than he had before. now, of course, whether you BELIEVE him
is a totally separate question than what i am asking.
do you have a different "definition" of winning as it relates to
craps?
tom "home runs are often boring" p
Bruce
REVS...@aol.com (redleathers) wrote:
>Yep. Switch to door number 2 for the reason Mason just stated.
Is it me or have you just arbitrarily switched sides?
QuiGon
news:71b15e3e.03071...@posting.google.com...
> REVS...@aol.com (redleathers) wrote:
> > Monte Hall ...
> [snip]
>
> Ahh, yes, the venerable Monty Hall problem.
>
> First things first. The man's name is Monty, not Monte. And second,
> under the common interpretation of the problem (that Monty never
> reveals the prize), the correct strategy is to *always* switch doors.
> But I'm not going to rehash what's already been written.
And 3rd, by an amazing coincidence, Monty Hall is going to be at Foxwoods in
September to play this game with some people..!! So that's the relevancy to
craps..!!
QuiGon
news:3f1237a8...@news.houston.sbcglobal.net...
>
> The only thing incorrect is you and your logic. He is absolutely correct
> and is trivially proved correct via experimentation.
I forget who is on which side of the argument, but here's the facts and
logic... Look at it this way: Assume there are 100 doors, 1 has a winner
and 99 have booby prizes... Furthermore, we must assume that Monty knows
which one is the winner (this is a HUGE and NECESSARY component).
Say the player chooses door number 56. He has a 1% chance of being right.
Therefore, the chances he's wrong are 99%. So Monty eliminates 98 of the 99
doors he DIDN'T choose. But the important thing to remember is that Monty's
elimination process is NOT a random element - he KNOWS where the prize is.
Now... following those facts, does anyone on this NG mean to tell me that
door 56 has a 50/50 chance of being the correct one...? Because that would
be wrong. Sorry, but door number 56 is still 1%. The smart thing to do is
switch.
If we go back to the original problem, the first door chosen has a 33%
chance of being right and a 67% chance of being wrong. Therefore, after one
door is eliminated, the smart thing to do is switch.
billb
hullabaloo.
--
billb
"QuiGon" <Qui...@nospam.com> wrote in message
news:_0SQa.69252$H17.20608@sccrnsc02...
billb
> elimination process is NOT a random element - he KNOWS where the
prize is.
>
but the player choice IS a random selection.
switching will cost you the prize as often as it earns you the prize.
--
billb
"QuiGon" <Qui...@nospam.com> wrote in message
news:_0SQa.69252$H17.20608@sccrnsc02...
> "Bruce" <br...@nospam.com> wrote in message
> news:3f1237a8...@news.houston.sbcglobal.net...
> >
> > The only thing incorrect is you and your logic. He is absolutely
correct
> > and is trivially proved correct via experimentation.
>
> I forget who is on which side of the argument, but here's the facts
and
> logic... Look at it this way: Assume there are 100 doors, 1 has a
winner
> and 99 have booby prizes... Furthermore, we must assume that Monty
knows
> which one is the winner (this is a HUGE and NECESSARY component).
>
> Say the player chooses door number 56. He has a 1% chance of being
right.
> Therefore, the chances he's wrong are 99%. So Monty eliminates 98
of the 99
> doors he DIDN'T choose. > Now... following those facts, does
stacy_friedman
> stacy_f...@hotmail.com (stacy_friedman) wrote in message.
>
> > This includes why your methodology isn't a winning one by any
> > standard definition of "winning".
>
> ????????
> what other "standard" definition(s) of "winning" are you refering to??
>
> winning (as it relates to craps) means coming out ahead (in dollars),
> doesnt it? having more money gambling at craps than you started with?
There's a big difference between "winning," the verb, and "winning,"
the adjective as it applies to "system" or "methodology".
Suppose I play the age-old stock-picking scam at the craps table with
unsuspecting people. I tell the first guy to bet $10 pass and the
second guy to bet $10 don't, $1 boxcars. One of them will win. Is
that a winning system for the guy who won?
> red claims he is winning--that he has more money after using his
> "method" than he had before. now, of course, whether you BELIEVE him
> is a totally separate question than what i am asking.
Whether I believe his results is irrelevant. I'm up lifetime in craps
mostly due to a single monster roll I had in Tahoe. I was playing
straight passline, pressing my bets, no odds. Did I win? Is that a
winning system?
Over 99.8% of people who bring $5500 to a $5 table and play the
Martingale will make $5. Did they win? Is that a winning system?
Mason
> but the player choice IS a random selection.
> switching will cost you the prize as often as it earns you the
prize.
No. You are confused. Switching earns the prize twice as often.
Since the previous explanations posted elude your understanding, do
this simple analysis.
Here are ALL of the possibilities for the prize locations
(Three doors Y=prize n=not prize)
P1:Y-n-n
P2:n-Y-n
P3:n-n-Y
Total possiblities of location/choice = 9
Strategy :Player >Keeps< Initial Selection
Player Choice
P1:Door 1 Win-Lose-Lose
P2:Door 2 Lose-Win-Lose
P3:Door 3 Lose-Lose-Win
Total Player Wins :3 (of 9)
Strategy :Player >Changes< Initial Selection
Player Choice
P1:Door 1 Lose-Win-Win
P2:Door 2 Win-Lose-Win
P3:Door 3 Win-Win-Lose
Total Player Wins :6 (of 9)
A careful examination of the above or, better still, doing it for
yourself should go a long way toward resolving your confusion.
billb
> P2:n-Y-n
> P3:n-n-Y
>
> Total possiblities of location/choice = 9
>
irrelevant. there are 3 possibilities. 2 are losers. that's all you
need to know.
you have the same chance of winning as you do buying the 4 or 10.
case closed.
'Elves and Dragons! I says to him. cabbages and potatoes are better
for me and you. Don't go getting mixed up in the business of your
betters, or you'll land in trouble too big for you,'
--
billb
John Kerr
swich....I would swich <g>!
JB
Stuart Woods
> mostly due to a single monster roll I had in Tahoe.
I am up lifetime mostly due to 3 monster rolls - $4000 in Vegas, $6000
in AC and $4000 in Sandia. But right now I am on such a losing streak
I'm seriously
considering going over to the dark side for the first time in my life.
Mason
the
prize."
I posted a simple analysis of the problem and then I said "A careful
examination of the above or, better still, doing it for yourself
should go a long way toward resolving your confusion."
Obviously, I was wrong. Stick with the elves and dragons.
As my old daddy used to say, "Son (he always called me Son), you can
lead a horse to water but, if you can get him to float on his back,
you got something!"
"billb" <gl...@spackle.org> wrote in message
news:Jo%Qa.10870$u51.7756@fed1read05...
John Kerr
Re: Behind Door Number 1, Door Number 2, or Door Number 3
Group: rec.gambling.craps Date: Tue, Jul 15, 2003, 3:39pm (CDT-2) From:
gl...@spackle.org (billb)
P2:n-Y-n
P3:n-n-Y
Total possiblities of location/choice = 9
irrelevant. there are 3 possibilities. 2 are losers. that's all you need
to know.
you have the same chance of winning as you do buying the 4 or 10.
case closed.
'Elves and Dragons! I says to him. cabbages and potatoes are better for
me and you. Don't go getting mixed up in the business of your betters,
or you'll land in trouble too big for you,'
--
billb
billb, you are about to get the infamaous "Inumerate" title! And come to
think of it...you might just deserve it in this case <g>! Just foolin
with ya...but you are wrong!
JB
SEE REAL ADDRESS BELOW
<mrzer0_...@sbcglocal.net> writes:
>It is the same with three doors in the choice set. Keep the claimed
>set with 1 chance in three of having the grand prize or switch to the
>unclaimed set with 2 chances of 3 of having the grand prize.
>
>Relevance to CRAPS is ZERO because no one knows what is behind the
>doors at the crap table.
>
>But the problem is fun and instructive. That is why it is a classic.
It is probably also irrelevant, but in craps after the comeout roll, it is like
a second game... in the first one (the first roll) the advantage is very much
in the player's favor. He has twice as much chance of winning than losing...
once he gets past that, then there is the unfavorable part where the house has
a large advantage. I can't think of any way to take advantage of this, but
maybe somebody else has done so. It is very similar to Parrondo's Paradox,
which is not supposed to apply to games of chance (such as craps).
Jim (The System Man)
No email may be sent to Want...@aol.com
Use this: JimFerr1[TAKE THIS OUT]@aol.com
FREE Gaming Strategies & Methods
No catch, not selling anything
Look here: http://hometown.aol.com/jimferr/profile.html
SEE REAL ADDRESS BELOW
(redleathers) writes:
>It is the same with three doors in the choice set. Keep the claimed
>set with 1 chance in three of having the grand prize or switch to the
>unclaimed set with 2 chances of 3 of having the grand prize.--Mason
>
>Yep. Switch to door number 2 for the reason Mason just stated.
Isn't it one chance out of two? The original had a 1/3 chance of being
correct. Then you were told that one of the others is not correct. This
leaves a 50-50 chance? I have heard this one many times with the answer being
to switch. I still don't understand it. You had a 1 in 3 chance originally.
He shows a losing one which you did not select. So, after that event of him
showing which is a loser it seems that it should be a 50-50 chance whether you
take the other one or keep the one that you have... that is when the real
choice is made. You took one risk when you decided to look at one of the
doors... at that time it was 1 in 3. After that selection it is 50-50 either
way... I know that the "correct" answer is to switch, but I don't know why?
I remember this being on the puzzle newsgroup faq, but I just looked for it and
could not find it. This is a classic puzzle.
Mason
>
> >It is the same with three doors in the choice set. Keep the
claimed
> >set with 1 chance in three of having the grand prize or switch to
the
> >unclaimed set with 2 chances of 3 of having the grand prize.--Mason
> >
> >Yep. Switch to door number 2 for the reason Mason just stated.
>
> Isn't it one chance out of two? The original had a 1/3 chance of
being
> correct. Then you were told that one of the others is not correct.
This
> leaves a 50-50 chance?
No. This has NO effect on the original chance. Since Monty can
ALWAYS reveal a booby prize among the unchosen doors because he KNOWS
were the prize is and there will always be at least one booby prize in
the unchosen set, his action gives no new information.
If your original chance was 1/3, the remaining chances represented by
the unchosen doors MUST BE 2/3 (sum of all chances is 1). It is your
advantage to switch sides. Monty's actions are irrelevant.
The essential question being asked when you are given the choice to
switch or not is .... Do you want what is behind one door or what is
behind two doors. Forget Monty.
SEE REAL ADDRESS BELOW
<mrzer0_...@sbcglocal.net> writes:
>No. This has NO effect on the original chance. Since Monty can
>ALWAYS reveal a booby prize among the unchosen doors because he KNOWS
>were the prize is and there will always be at least one booby prize in
>the unchosen set, his action gives no new information.
>
>If your original chance was 1/3, the remaining chances represented by
>the unchosen doors MUST BE 2/3 (sum of all chances is 1). It is your
>advantage to switch sides. Monty's actions are irrelevant.
>
>The essential question being asked when you are given the choice to
>switch or not is .... Do you want what is behind one door or what is
>behind two doors. Forget Monty.
Mason, that's the part that I don't understand. Since the game is played that
way... pick one out of 3 and then Monty will eliminate one more, isn't it
always a 1 out of 2 chance? The first choice is irrelevant... the important
one is when there are two of them and you get to pick one of them... you pick x
by keeping the first choice, and y by changing... that is 50-50, or a 50%
chance of being correct in what you do... what am I missing?
SEE REAL ADDRESS BELOW
merely chose A every time and did not switch I was correct one time out of
three. On the other hand, if I chose A every time and then switched every
time, I was then correct two times out of every three. I switched off of the
correct one on one of the three but switched onto the correct one two times out
of three! Absolutely amazing! Now, how do we apply this to craps? In one of
the messages I believe that Red said this is the key to his winning method.
In article <20030715212430...@mb-m15.aol.com>,
Richard
news:20030715213538...@mb-m15.aol.com:
> I followed instructions and tried it on paper. This is what
> happened... if I merely chose A every time and did not switch I was
> correct one time out of three. On the other hand, if I chose A every
> time and then switched every time, I was then correct two times out of
> every three. I switched off of the correct one on one of the three
> but switched onto the correct one two times out of three! Absolutely
> amazing!
What's absolutely amazing is how much you can learn in 11 minutes if you
actually try the experiment. Congratulations.
Richard
Michael P. Casey
> irrelevant. there are 3 possibilities. 2 are losers. that's all you
> need to know.
>
> you have the same chance of winning as you do buying the 4 or 10.
>
> case closed.
Want to put money on it?
MPC
tom home runs are boring p
> > stacy_f...@hotmail.com (stacy_friedman) wrote in message.
> >
> > > This includes why your methodology isn't a winning one by any
> > > standard definition of "winning".
> >
> > ????????
> > what other "standard" definition(s) of "winning" are you refering to??
> >
> > winning (as it relates to craps) means coming out ahead (in dollars),
> > doesnt it? having more money gambling at craps than you started with?
>
> There's a big difference between "winning," the verb, and "winning,"
> the adjective as it applies to "system" or "methodology".
thanks,
tom p
QuiGon
news:20030715213538...@mb-m15.aol.com...
> I followed instructions and tried it on paper. This is what happened...
if I
> merely chose A every time and did not switch I was correct one time out of
> three. On the other hand, if I chose A every time and then switched every
> time, I was then correct two times out of every three. I switched off of
the
> correct one on one of the three but switched onto the correct one two
times out
> of three! Absolutely amazing! Now, how do we apply this to craps? In
one of
> the messages I believe that Red said this is the key to his winning
method.
While I enjoyed this mathematical problem, I don't see how it applies to
craps other than the fact that Monty Hall will indeed be playing Let's Make
A Deal at Foxwoods Casino this September.
SEE REAL ADDRESS BELOW
<mpc...@yahoo.com> writes:
Now that I know that switching every time is the thing to do, I wish that I
knew a way to apply it to craps.
SEE REAL ADDRESS BELOW
<aesc...@takeout.comcast.net> writes:
>> I followed instructions and tried it on paper. This is what
>> happened... if I merely chose A every time and did not switch I was
>> correct one time out of three. On the other hand, if I chose A every
>> time and then switched every time, I was then correct two times out of
>> every three. I switched off of the correct one on one of the three
>> but switched onto the correct one two times out of three! Absolutely
>> amazing!
>
>What's absolutely amazing is how much you can learn in 11 minutes if you
>actually try the experiment. Congratulations.
>
>Richard
Correct. I have been trying to figure this out for a long time, but now
somebody suggested an article at:
http://www.willamette.edu/cla/math/articles/marilyn.htm and it explained that
you must try it and it will come to you.
billb
there are 2 choices. each of the remaining choices has a 50% chance
of being right.
So why switch?
--
billb
"John Kerr" <jbk...@webtv.net> wrote in message
news:12117-3F1...@storefull-2151.public.lawson.webtv.net...
billb
> ALWAYS reveal a booby prize among the unchosen doors because he
KNOWS
> were the prize is and there will always be at least one booby prize
in
> the unchosen set, his action gives no new information.
>
Exactly. And in the absence of new information, switching changes
nothing.
But if a large segment of the population believes it does, maybe IGT
should come up with a machine that will exploit this belief.
It could put 3 boxes on the screen one with a prize and then let
customers change their mind after revealing on loser. Better yet,
give players the option of paying to have a cartoon Monty come out
and remove a loser before they execute that rich valuable option of
switching. Call it the Ask Marilyn machine. All the self styled
geniuses like yourself will feed in $100 bills like they're going out
of style. Bill Bennett (a TRUE genius) will be there of course, to
play the machines denominated in $1000 increments.
This could make enough money to build another mega resort.
> The essential question being asked when you are given the choice to
> switch or not is .... Do you want what is behind one door or what
is
> behind two doors.
Yes, but you could just as easily have argued that the first pick
INCLUDED the Monty door and hence the FIRST pick had a 2/3 chance of
winning.
Or, break it out another way. Say the first pick has a 50% chance of
winning because the remaining 2 picks include an irrelevant Monty
door that will be discarded after the first pick is made. Thus the
2/3 pick is 50% rotten and the first pick is 100% good. Again,
equality of pick whether you switch or not.
Explain why not.
Yours and Marilyn's argument hinges on the ridiculous notion that by
switching we are somehow betting on 2 doors instead of one.
I don't buy it.
--
billb
billb
50/50 prop.
if so, why switch?
--
billb
"SEE REAL ADDRESS BELOW" <want...@aol.comJimFerr> wrote in message
news:20030716002039...@mb-m19.aol.com...
SEE REAL ADDRESS BELOW
writes:
>before you said that since Monty eliminates one choice it is always a
>50/50 prop.
>
>if so, why switch?
I said that before I tried it on paper... now I know that switching every time
is a good bet... unless somebody can show me that I did it incorrectly.
SEE REAL ADDRESS BELOW
writes:
>Explain why not.
>
>Yours and Marilyn's argument hinges on the ridiculous notion that by
>switching we are somehow betting on 2 doors instead of one.
>
>I don't buy it.
Just do it on paper and you will see.
SEE REAL ADDRESS BELOW
writes:
>here's how I see it. there are 3 choices. if we take away one loser
>there are 2 choices. each of the remaining choices has a 50% chance
>of being right.
>
>So why switch?
Because out of 3 times you will switch from a loser to a winner two of those
times, and from a winner to a loser one time. I still don't know why, but by
doing it on paper as an experiment it just works out that way.
billb
you pick and hold. If you make any other pick and then switch you
have a 2/3 chance of winning.
The weird thing is you get a 66% chance of winning starting with 3
choices, when a person faced with 2 equal choices only gets a 50%
chance.
I think that's the part that strikes people as not making any sense.
--
billb
"SEE REAL ADDRESS BELOW" <want...@aol.comJimFerr> wrote in message
news:20030716002039...@mb-m19.aol.com...
billb
if you don't switch you don't take advantage of the revealed loser.
once the loser is eliminated the other choice has a 2/3 chance of
winning simply because guys like you always pick losers at first.
--
billb
"SEE REAL ADDRESS BELOW" <want...@aol.comJimFerr> wrote in message
news:20030716012128...@mb-m22.aol.com...
billb
> Just do it on paper and you will see.
>
>
I don't have to. I accept Marilyns argument. My complaint was only
that the assertion was being made that after the reveal, the choice
was something other than 50/50. That point was really never in
dispute.
It's the way the problem's worded that causes the hassle.
If the problem was stated simply, there's one prize and you get 2
picks or one pick, it would be easier to understand.
--
billb
Mason
If the problem was stated simply, there's one prize and you get 2
picks or one pick, it would be easier to understand.
--
billb
-----------------------------------------------------------------
Yes, it would be easier to understand since thinking would not be
required. Of course, then, it wouldn't be a problem. The name
"The Monty Hall Problem" should give you a broad hint that some
thinking is required to solve it.
You have clearly demonstrated that the hint escaped your notice
or the requirements exceeded your capabilities.
Rander the Roller
>
Unfortunately, a fair amount of probability theory is non intuitive.
A great book on probability that avoids getting bogged down in the technical
swamps is Warren Weaver's _Lady Luck_.
Rander
"Las Vegas is merely America carried to it's logical conclusion"
--Rander the Roller
SEE REAL ADDRESS BELOW
writes:
>Marilyins argument is, I gather, you have a 1/3 chance of winning if
>you pick and hold. If you make any other pick and then switch you
>have a 2/3 chance of winning.
>
>The weird thing is you get a 66% chance of winning starting with 3
>choices, when a person faced with 2 equal choices only gets a 50%
>chance.
>
>I think that's the part that strikes people as not making any sense.
That is absolutely amazing and contrary to what your instinct would tell you.
I wonder if there is anything on the dice table that works that way. I really
doubt it.
SEE REAL ADDRESS BELOW
writes:
>because picking one and sticking always has 1/3 chance of winning.
>if you don't switch you don't take advantage of the revealed loser.
>once the loser is eliminated the other choice has a 2/3 chance of
>winning simply because guys like you always pick losers at first.
Actually, you would be giving up the winner and going to a loser one time out
of three, and giving up a loser and going to a winner two times out of three...
which makes it a good bet. You are giving up a 1 in 3 chance of winning in
exchange for a 2 out of 3 chance of winning... the 50/50 sounds so logical, but
it is wrong.
billb
the original problem. it's just the way this problem is formulated
and worded that causes the problem.
--
billb
"SEE REAL ADDRESS BELOW" <want...@aol.comJimFerr> wrote in message
news:20030716115213...@mb-m03.aol.com...
> In article <fH5Ra.10944$u51.8712@fed1read05>, "billb"
<gl...@spackle.org>
> writes:
>
> >because picking one and sticking always has 1/3 chance of winning.
> >if you don't switch you don't take advantage of the revealed
loser.
> >once the loser is eliminated the other choice has a 2/3 chance of
> >winning simply because guys like you always pick losers at first.
>
> Actually, you would be giving up the winner and going to a loser
one time out
> of three, and giving up a loser and going to a winner two times out
of three...
> which makes it a good bet. >
>
billb
--
billb
stacy_friedman
> Or, break it out another way. Say the first pick has a 50% chance of
> winning because the remaining 2 picks include an irrelevant Monty
> door that will be discarded after the first pick is made. Thus the
> 2/3 pick is 50% rotten and the first pick is 100% good. Again,
> equality of pick whether you switch or not.
>
> Explain why not.
That's not the way things work. If there are three doors with a prize
behind exactly one of them, you have exactly 1/3 chance of being right
and 2/3 chance of being wrong. So if you could pick the two doors at
once, and keep both items inside, you'd have a 2/3 chance of winning
the car, right?
That's exactly what Monty does for you, if you think about it. He
opens one of the other doors and says "Look, here's a goat. Do you
want to open up the remaining door and take that prize, or stay with
the one you have?" Your chances of winning if you stay put are still
1/3. Since there's only one other option, if you switch you will now
have a 1/3 chance of *losing*. But a 1/3 chance of losing = 2/3
chance of winning, so switching doubles your chances to win the car if
you know which door *not* to pick.
If you had never picked a door at first, the odds after Monty opens
one would now be 50-50 of you winning with one of the unopened doors.
But the fact that you had already picked a door means that Monty would
not have opened it. His choices were limited by your actions, and
that changes the problem dramatically.
SEE REAL ADDRESS BELOW
writes:
>the 50/50 IS right once the loser is revealed, but that isn't part of
>the original problem. it's just the way this problem is formulated
>and worded that causes the problem.
The one loser is *always* revealed. That is what makes it work out at 2/3 in
favor of the player when the player switches sides. The only time that the
player loses is the one time in three that the player was on the winning one to
start with. Since there is only one other choice, then on the other two times
the player is switching from a loser to a winner. Just writing that clarified
it for me, I hope that it also helps you.
SEE REAL ADDRESS BELOW
stacy_f...@hotmail.com (stacy_friedman) writes:
>If you had never picked a door at first, the odds after Monty opens
>one would now be 50-50 of you winning with one of the unopened doors.
>But the fact that you had already picked a door means that Monty would
>not have opened it. His choices were limited by your actions, and
>that changes the problem dramatically.
Very good, Stacy. That clarifies it.
Pond Scum
>"billb" <gl...@spackle.org> wrote:
>> Or, break it out another way. Say the first pick has a 50% chance of
>> winning because the remaining 2 picks include an irrelevant Monty
>> door that will be discarded after the first pick is made. Thus the
>> 2/3 pick is 50% rotten and the first pick is 100% good. Again,
>> equality of pick whether you switch or not.
>> Explain why not.
>1/3. Since there's only one other option, if you switch you will now
>have a 1/3 chance of *losing*. But a 1/3 chance of losing = 2/3
>chance of winning, so switching doubles your chances to win the car if
>you know which door *not* to pick.
I've got billb killfiled, so I missed his original posting, but is he having
trouble understanding the Monty Hall problem???
The followup above is correct. You always switch. By switching doors,
you increase your chances of winning the car from 1 in 3 to 2 in 3.
There are many ways of explaining this problem. If you always switch:
1/3rd of the time you will pick the car on your first pick,
and by switching you win nothing, 100% of the time:
.33 * (100% * 0) = 0
2/3rds of the time you will not pick the car on your first pick.
By switching you will guarantee that the car lies behind one of
the two remaining doors. Since Monty shows you which one of the
two remaining doors does not have the car behind it, you are
guaranteed that it is behind the remaining door. So for the
2/3rds of the time that you initially pick wrong, you will
win the car 100% of the time if you switch:
.67 * (100% * 1) = .67
If you sum the product of the probabilities and the payoffs
over all possible outcomes, you find the expected value is .67
if you switch. If you don't switch the expected value is .33.
-Jonathan jh...@sonic.net
Peregrine
So many people seem to be missing the key element of the "Monty Hall"
problem.
Once one of the losing choices is revealed, the odds on your original
choice being the winner does not stay at 1 in 3, but now becomes 1 in
2. So there is no difference in expectation between switching and
staying with the original choice.
To all those people who still think that switching is 2 to 1
favourite, what odds will you offer me that sticking with the original
choice is correct. I'll gladly take 7 to 4 for whatever stakes you
like.
Peregrine
SEE REAL ADDRESS BELOW
<Pere...@nospam.blueyonder.co.uk> writes:
>To all those people who still think that switching is 2 to 1
>favourite, what odds will you offer me that sticking with the original
>choice is correct. I'll gladly take 7 to 4 for whatever stakes you
>like.
I don't bet on sure things, so you may keep your money. Let's say that we have
3 doors labeled A, B & C. We will do this three times. The first time we will
select door A, the second time door B and the third time door C. The car is
behind door A each time. So, the first time we have the correct door (A), and
Monty shows us door B to be a goat, and we switch to door C, and lose the car.
The second time we have door B and the car is behind A... Monty shows us the
goat behind door C, and we switch to B, which has the car and we win. The
third time we have door C and the car is behind A... Month shows us the goat
behind door B, and we switch to C and win... so we won two times and lost one
time.
Basically, we will lose when we get the car the first time, but we will win
when we do not... and we have 2 chances out of 3 of not getting it right the
first time... which then gives us 2 out of 3 chances of getting it right the
second time... so we will win 2/3 of the times and lose the car 1/3 of the
times... see how much money you saved!
PFels401
door 3. Monte opens door 2 and it is empty.
Do bothe players gain by switching to each others door?
SEE REAL ADDRESS BELOW
(PFels401) writes:
That is a different puzzle. We are talking about one player and three doors,
with Monty telling that player which of the other two doors is not the car.
This has nothing to do with craps... so, may we get back on topic?
Pond Scum
>So many people seem to be missing the key element of the "Monty Hall"
>problem.
>Once one of the losing choices is revealed, the odds on your original
>choice being the winner does not stay at 1 in 3, but now becomes 1 in
>2. So there is no difference in expectation between switching and
>staying with the original choice.
That was a troll, right. You couldn't possibly be that stupid, could you?
Revealing a losing choice does not change the probabilty of having
selected a winner initially. It's still 1 in 3. Carry that to the extreme,
1 million doors, with 999,998 losers exposed after your initial selection.
Do you think that you selected the single winning door, out of 1 million
doors, 50% of the time? The probability is 1 in 1 million that you chose
correctly, and 999999 in 1000000 that the car is behind the other door.
-Jonathan jh...@sonic.net
salmoneous
There is an assumption here that isn't always stated in the question.
Namely, that Monte knows where the car is and shows you a door that he
knows has a goat behind it. If Monte doesn't know where the car is -
he just picked a random door and it happened to have a goat - then
there's a 50% chance of the car behind either of the remaining door
and it doesn't matter if you switch.
redleathers
down in it wallowing and swallowing.
Suddenly all the mathophiles are saying it does matter what happened
before to determine where you are now.
Suddenly, they are all talking player behavior, not just the hard
choice.
If there was no door three. If there were only two doors and one had
the prize and the other did not then the answer would be 50/50.
But if there were three doors to begin with in ancient time, and the
monty hall of their time showed the goat behind door three and the
volcano errupted and preserved those ancient citizens for tens of
thousands of years and now they return to the problem with but two
doors left--one prize and one not--the mathematics *still prove* that
he should switch to door number two to gain a distinct mathematical
advantage.
Don't you see the larger point is that just because something works
out mathematically--improves the odds by doing a certain behavior--has
nothing at all to do with the real live choice now 10,000 years in
front of you. Door number one or door number two. hahahaha It isn't
that the math is wrong or another math is right. We know the 1/3rd v.
2/3rds works out mathematically. It is the real life choice now
between two doors. Just imagine what the odds would be if the player
and everyone else had forgotten that there even was a 3rd door--a
third choice. hahahaha reality, yes even mathematically provable
reality is a mental construct not the reality as our white western
hemisphere minds imagine.
billb
> that changes the problem dramatically.
right.
--
billb
Mason
Why am I not surprised?
"billb" <gl...@spackle.org> wrote in message
news:dymRa.11057$u51.5833@fed1read05...
John Kerr
fingernails till the thread is stretched to the limit!
A lesson in life...but more suited to craps "If you are wrong, admit
it!"...another..."Never try and defend an undefendable position!"
redleathers
Oh for sure the correct answer is: switch from door one to door two
and double your chances of winning now that Monty has revealed door
three.
It's just a certain view of mathematics, of existence, of reality,
that makes it correct. Again, I ask, what if the participants and
Monty all just mentally forgot--nothing physically changes--just
forgot that there was ever a 3rd door to begin with, to ever choose?
Mason
>
> Oh for sure the correct answer is: switch from door one to door two
> and double your chances of winning now that Monty has revealed door
> three.
>
> It's just a certain view of mathematics, of existence, of reality,
> that makes it correct. Again, I ask, what if the participants and
> Monty all just mentally forgot--nothing physically changes--just
> forgot that there was ever a 3rd door to begin with, to ever choose?
The answer is as obvious as Readleathers is sophomoric.
--------------------------------------------------------------------------------
-
Here are ALL of the possibilities for the prize locations
(Three doors Y=prize a/b =not prize)
P1:Y-a-b
P2:a-Y-b
P3:a-b-Y
Total possibilities of location/choice = 9
Strategy :Player >Keeps< Initial Selection
Player Choice
P1:Door 1 Win-Lose-Lose
P2:Door 2 Lose-Win-Lose
P3:Door 3 Lose-Lose-Win
Total Player Wins :3 (of 9)
Strategy :Player >Changes< Initial Selection
Player Choice
P1:Door 1 Lose-Win-Win
P2:Door 2 Win-Lose-Win
P3:Door 3 Win-Win-Lose
Total Player Wins :6 (of 9)
--------------------------------------------------------------------------------
-
Memory of the exposing of the door is not a factor, is it?
Unfortunately, no one knows what is behind the doors at the crap table
so this discussion has no relevance to this NG.
stacy_friedman
> It's just a certain view of mathematics, of existence, of reality,
> that makes it correct. Again, I ask, what if the participants and
> Monty all just mentally forgot--nothing physically changes--just
> forgot that there was ever a 3rd door to begin with, to ever choose?
So it's a choice between keeping the unopened door you have, or
switching to the goat that Monty's showing you? That's a silly
question.
Or, do you mean that Monty showed you the goat behind door #3, but
then aliens came and abducted the goat and the door, and wiped away
your memory? In that case, you're still better off switching (since
the math hasn't changed), but you won't know it. This happens a lot
in poker, where one player is drawing dead to another. In that case,
you're better off folding but you don't know it.
stacy_friedman
In order to do this, you must first recognize that you are, in fact, wrong.
Mason
news:71b15e3e.03071...@posting.google.com...
"This happens a lot
in poker, where one player is drawing dead to another. In that case,
you're better off folding but you don't know it."
Bite your tongue.
Tom Williams
Contestant relinquishes door #1 and then chooses door #2 because of
better odds. But if he mentally relinquishes door #1 then repicks
door #1 his odds are not improved? Isn't it a new bet every time
Monte shows a door? I thought this was the "independent events"
crowd? The odds for the new bet is 1 in 2 (not 2 in 3.)
Talk about pretzel logic.
Have you all been smoking Monte's goat's fur??
Tom Williams
SEE REAL ADDRESS BELOW
Williams) writes:
>It's semantics.
No it is not... read my postings and you will understand it. Since this
discussion is off topic, I will not repeat it. You really do have 3 from which
to select the first time, and then only ONE the second time... this changes it
from 1 in 3 and it becomes 2 out of 3 chances of getting the car. You now know
one door which you will not pick (the one which Monty showed). In order to
have the better chance you *must* change to the other door... as illogical as
it may sound. You will get the car 2/3 of the times (you lose it the 1 in 3
times that you get it right the first time), and you will get it the other two
times since there are now only two doors so you *must* be going from a goat to
a car those two times.
Mark Rafn
>It's semantics.
No, it's really not. If the rules are that monty will always pick a junk door
and show it to you before offering the choice, then you should switch.
He reveals information about the remaining doors by opening one. He does NOT
reveal anything about the door you originally picked. To use his revelation,
you must switch doors.
>Contestant relinquishes door #1 and then chooses door #2 because of
>better odds. But if he mentally relinquishes door #1 then repicks
>door #1 his odds are not improved?
No. Door 2 + door 3 have a 66% chance of having the prize. Picking door 1
again brings you back to the 33% chance to win.
>Isn't it a new bet every time Monte shows a door?
Nope, he's reduced the frequency of losers in a subset of doors (the ones you
haven't picked).
>I thought this was the "independent events" crowd?
True for dice, not true for many other wagers. In this case, there is a
removal taking place.
>The odds for the new bet is 1 in 2 (not 2 in 3.)
If he randomly picked a losing door to show, and that included YOUR door 1/2
the time, you'd be right. You'd have a 50% chance. However, since he shows
you a door from a limited subset of doors, only that subset is affected by his
choice.
--
Mark Rafn da...@dagon.net <http://www.dagon.net/>
Mason
"Tom Williams" <run...@outback.com> wrote in message
news:3f172d6c...@news4.sucknews.com...
> It's semantics.
>
No. There are only nine possible distinct situations after the initial door is
chosen(three different door choice arrays and three different contestant
selected doors in each door choice array). Picking an door and keeping it is a
successful strategy in three of the starting situations. Switching is
successful in six of them.
A contestant switching from his original choice is twice as likely to succeed.
That is all there is to the Monty Hall Problem even with sophomoric addenda
about memory loss.
There is no application of the Monty Hall Problem to craps because since no one
knows what is behind the doors in a game of random and independent trials so
there is no reason to prefer one door over any other. They all hide equally
likely possibilities.
billb
> times that you get it right the first time), and you will get it
the other two
> times since there are now only two doors so you *must* be going
from a goat to
> a car those two times.
Monty will know that everybody switches cause they all read ask
Marilyn, and will move the car to the original choice when they are
distracted, so you will never win.
--
billb
stacy_friedman
> It's semantics.
>
> Contestant relinquishes door #1 and then chooses door #2 because of
> better odds. But if he mentally relinquishes door #1 then repicks
> door #1 his odds are not improved? Isn't it a new bet every time
> Monte shows a door? I thought this was the "independent events"
> crowd? The odds for the new bet is 1 in 2 (not 2 in 3.)
No they're not. Here's hopefully a clearer explanation:
If you pick a door randomly, you have a 1 in 3 chance of winning.
If you stand back without picking, Monty eliminates one non-prize door
randomly, and then you pick, you now have a 1 in 2 chance of winning.
But if you pick before Monty eliminates one door, his elimination
isn't random. Monty has rules that he must play by, and they are:
1) Don't show the contestant a prize.
2) Don't show the contestant what's behind the door they picked.
If you've already picked a door and it's the car, Monty is free to
show you either of the other two doors with goats. He won't show you
that you've won the car because that would violate the rules. That
happens 1/3 of the time. But if you've picked a goat, Monty has *only
one option* -- to show you the other goat. If he opened anything
else, he'd violate one of the rules.
Picking a goat first happens 2/3 of the time. So 2/3 of the time, if
you switch to the door Monty didn't open, you'll win the car.
If you didn't pick a door right away, Monty always has two goat doors
to choose from. When you do pick first, 2/3 of the time he has only
one door to choose from. It's a different problem.
Stacy
QuiGon
>
> It's the way the problem's worded that causes the hassle.
>
> If the problem was stated simply, there's one prize and you get 2
> picks or one pick, it would be easier to understand.
No offense, but it sure sounds to me like you were wrong, realized you were
wrong, but instead of *admitting* you were wrong, you're now blaming it on
the wording of the question.
The KEY point here is that Monty knows where the car is, and which doors
have the booby prize. That is what makes the problem work. But even after
I stated that point most emphatically in a previous post of mine, you still
disagreed.
Now, here's a statement I know could re-ignite this controversy: IF Monty
doesn't know where the car is, THEN you are in a situation where switching
or staying is just a 50/50 guess. The fact that Monty's elimination is NOT
random is what makes the problem work.
Joe Schmoe
> stacy_f...@hotmail.com (stacy_friedman) wrote in message news:<71b15e3e.03071...@posting.google.com>...>
>
>>Whether I believe his results is irrelevant. I'm up lifetime in craps
>>mostly due to a single monster roll I had in Tahoe.
>
>
> I am up lifetime mostly due to 3 monster rolls - $4000 in Vegas, $6000
> in AC and $4000 in Sandia. But right now I am on such a losing streak
> I'm seriously
> considering going over to the dark side for the first time in my life.
"Give yourself to the dark side; it is the only way"
-Darth Vader
Joe Schmoe
> between two doors. Just imagine what the odds would be if the player
> and everyone else had forgotten that there even was a 3rd door--a
> third choice. hahahaha reality, yes even mathematically provable
> reality is a mental construct not the reality as our white western
> hemisphere minds imagine.
So what are the three choices at craps; do, don't or ?? layoff?
I don't see Monty's routine applies...