50-50 strategy and Let's Make A Deal
Abrams
Monty Hall Problem. On Let's Make A Deal, the big prize would be hidden
behind one of three doors. Dud prizes were behind the other two. Let's
assume the contestant picked Door #1. Monty might then say: "Good thing you
didn't pick Door #3...you would have won a donkey!" Then Monty would offer
the contestant the opportunity to stick with Door #1 or to switch and take
Door #2 instead. The question now is: Should he switch?? Where is the
winning prize?
The surprising answer is that the big prize is twice as likely to be behind
Door #2 than behind Door #1, so the contestant should definitely switch to
increase his odds of winning. The reason is that initially each door had a 1
in 3 chance of being the correct door. Monty had knowledge of this and
eliminated one of the incorrect doors. The probability that the winner is
behind Door #1 remains 1 in 3, while the remainder of the probability (i.e.
2 in 3) now gets transferred entirely to Door #2 (it had been previously
been split between #2 and #3, 1 in 3 for each). Thus, the odds favor
switching since Door #2 has twice the likelihood of being correct.
How does this relate to WWTBAM? Well, last night's question pertaining to
the Cracker Jack box dog was a perfect illustration of this. Neither Eric
Molnar nor his Dad (nor I) had the faintest clue as to the correct answer.
Therefore, the choices each had a 1 in 4 probability of being correct.
Eric's Dad said Chance, and this obviously became Eric's choice. He then
used the 50-50, and was left with Chance and Bingo. Since neither answer was
any better than the other, the final choice of Chance vs Bingo after the
other two possible answers were eliminated was equivalent to the choice
between Door #1 and Door #2 after the incorrect Door #3 was eliminated. Eric
went with his random pick of Chance, but he could have changed before
locking it in as his final ansa.
I immediately told my wife that the correct answer was probably Bingo, and
that he should switch: Bingo was 3 times more likely than Chance to be the
correct answer, and...bingo...it was!
This principle seems to apply to the 50-50 under the following
circumstances: (1) The contestant has no idea of the correct answer (i.e. it
is a total guess) and so the 50-50 does not remove choices that were
obviously wrong from the beginning; (2) He makes a random choice from the
beginning, and that choice is not eliminated by the 50-50. If so,
probability favors switching your choice at the end.
Any comments or analysis by mathematicians are welcome.
David Abrams
Zonked by the chronological order of Romanov czars on my qualifying call
tonight.
Abrams
James Allen wrote in message
<20000113221654...@ng-cp1.aol.com>...
>Not quite. The "Let's Make a Deal" problem assumes your initial choice will
not
>be eliminated.
>
>This is something the 50/50 doesn't guarantee.
>
Actually, I said at the end of my post :"...he makes a random choice from
the
beginning, and that choice is not eliminated by the 50-50." I think that
under these conditions, the Let's Make a Deal strategy works.
js
Darryl Tam wrote in message
<7%wf4.36149$g12.1...@news2.rdc1.on.home.com>...
>
>"Abrams" <jcab...@mail.utexas.edu> wrote in message
>news:85m4n8$nsv$1...@geraldo.cc.utexas.edu...
If he chooses A, then there are 12 possibilities:
Answers Left Correct Answer
------------ --------------
A B A
A B B
A C A
A C C
A D A
A D D
B C B
B C C
B D B
B D D
C D C
C D D
Elimianting the last six options leaves:
Answers Left Correct Answer
------------ --------------
A B A
A B B
A C A
A C C
A D A
A D D
As can be seen, half have A the origanal choice correct and half don't, so
switching has no gain.
James Allen
be eliminated.
This is something the 50/50 doesn't guarantee.
Let's look at the possiblities (assuming a totally blind random choice.)
1 out of 4 times you pick the right answer. You switch off after the 50/50 and
lose.
3 out of 4 times you pick the wrong answer. 2 of those three times your answer
is elimated by the 50/50. No advantage gained at all, you are back to a 50%
chance. The other time you pick a wrong answer that is not eliminated. You
switch off and win.
Thus, your odds are still 50/50.
Darryl Tam
"Abrams" <jcab...@mail.utexas.edu> wrote in message
news:85m4n8$nsv$1...@geraldo.cc.utexas.edu...
>
> James Allen wrote in message
> <20000113221654...@ng-cp1.aol.com>...
will
> not
> >be eliminated.
> >
> >This is something the 50/50 doesn't guarantee.
> >
>
>
> the
> beginning, and that choice is not eliminated by the 50-50." I think that
That only happens 50% of the time. There is still a 50% chance of winning.
The Monty Hall problem only occurs if
1) The contestant states his random choice, and
2) The computer eliminates 2 wrong answers, and only eliminates those 2
answers based on what the contestant says. If the computer doesn't
eliminate on what the guy thought, there is still a 50-50 chance.
James Allen
>not
>>be eliminated.
>>
>>This is something the 50/50 doesn't guarantee.
>>
>
>
>Actually, I said at the end of my post :"...he makes a random choice from
>the
>beginning, and that choice is not eliminated by the 50-50." I think that
>under these conditions, the Let's Make a Deal strategy works.
And I thought I explained that it didn't work that way.
Once again for the hard of hearing:
1/4 times you pick the right answer
3/4 times you pick the wrong answer. In 2 of those 3 cases, your wrong choice
will be eliminated by the 50/50.
So: Of the 4 possibilities, 2 of them leave you with your answer eliminated,
for the sake of your example we ignore these; that leaves 2 of the scenarios
where your random answer is not eliminated by the 50/50; in one of them you
picked the right answer, in one you have a wrong answer.
You can switch you answer if you like, but your chances are still 50/50.
BudMann9
..ready to chase a bottle of aspirin with a bottle of vodka....
Mike Caldarelli
of a chance to door 1's 1/3 chance...
Simple stats tells me that they both become equal 50-50. That's all there
is to it and I think that's what pretty much everyone who's responded to
this thread agrees too
--awesome
Tim Firman
>>> James Allen wrote in message
>>> <20000113221654...@ng-cp1.aol.com>...
>>> >will not be eliminated.
>>> >
>>> >This is something the 50/50 doesn't guarantee.
<snip>
>As can be seen, half have A the origanal choice correct and half don't, so
>switching has no gain.
This is analogous to the Monty Hall problem in a second way: This analysis
assumes that the people controlling the game act randomly.
In Monty Hall, the problem must be stated such that Monty must reveal a new
door which does not contain the prize for 1/3 to the answer (this condition
wasn't met on the original show)
In WWTBAM, the two chosen for 50/50 must not be correlated to your initial
guess for this to be valid. This always holds if you don't reveal your
suspicions, and I suspect that this holds regardless.
This is true if they _predetermine_ a false answer they expect is most
likely to be incorrectly guessed, and leave that one.
Regis(in the US) mentions that the 'computer' eliminates the two answers,
but that really doesn't indicate much.
If the choices aren't predetermined, unusual strategies become favorable.
What if the show deliberately kept whatever option the player claimed
to be favoring as one of the two final choices?
This would lead to lying strategies. Say that the player cannot choose
between A, B, and C, but can rule out D. The best strategy would be
to lie prior to using 50/50 by claiming to strongly suspect D is the
answer. If the producers use this alleged suspicion and ensure that D
remains after the 50/50, the player knows that the non-D answer is correct.
(It'll look wierd, but since each player may only use 50/50 once...)
Of course, the competitive equilibrium is for the show to ignore the
players' small talk prior to 50/50. The show may expect that most players
will not know and apply enough game theory to lie, however...
Also, I wonder if they tailor questions to circumstance. Do people get
asked hard questions in known fields of expertise (their claimed profession)?
Do telegenic persons get asked easier questions than less telegenic people?
Are the later questions easier if there hasn't been a big winner for some
time?
Tim
Larry Stone
In article <85o6cm$j0f$1...@knot.queensu.ca>,
Mike Caldarelli <7m...@qlink.queensu.ca> wrote:
>I'm still trying to understand why in Let's Make a Deal, Door 2 becomes 2/3
>of a chance to door 1's 1/3 chance...
>
>Simple stats tells me that they both become equal 50-50. That's all there
>is to it ...
No, that's not all there is to it. You randomly pick Door 1 of the three.
You have a 1/3 chance of being right and there's a 2/3 chance it's behind
Door 2 or Door 3. You also know that of Doors 2 and 3, the good prize
isn't behind at least one of them (this is a key point). Now, Monty, who
knows where the good prize is (another key point), has one of Doors 2 and
3 opened. If the good prize was behind 2 or 3, he opens the one one it
isn't behind (and if it's behind Door 1, he picks one at random). He of
course never picks the door you selected.
What have you learned at this point? Absolutely nothing. You already knew
the good prize wasn't behind at least one of Doors 2 and 3 and that's what
Monty just showed you. Since you haven't learned anything new, the
probabilties haven't changed. There's still a 1/3 chance it's behind Door
1 so there's a 2/3 chance it's behind the other remaining door. [Repeat
this paragraph until it sinks in!]
Now contrast that to WWTBAM. You select an answer (let's say A). Assuming
it's a random pick, there's a 1/4 chance of being right and a 3/4 chance
it's B, C, or D. Now you use 50-50. Two wrong answers are removed but it
could include the one you picked since they're pre-selected (which is
actually irrelevant to this issue). If this were the same as the Monty
problem, then two of B, C, and D would be removed. So in this case,
you've learned something new. Assuming that this is all random (which it
isn't since this whole game really isn't a random pick so that's another
reason this isn't comparable), if A was a wrong answer, there was a 2/3
chance it would have been removed (with Monty, there was no chance of it
being opened).
So no, they're not the same. Assuming all things random, after the 50-50,
it is 50-50 regardless of what you were thinking or said.
--
-- Larry Stone
la...@stonejongleux.com
http://www.stonejongleux.com/~lstone/
Abrams
Mike Caldarelli <7m...@qlink.queensu.ca> wrote in message
<85o6cm$j0f$1...@knot.queensu.ca>...
>I'm still trying to understand why in Let's Make a Deal, Door 2 becomes 2/3
>of a chance to door 1's 1/3 chance...
>
>Simple stats tells me that they both become equal 50-50. That's all there
>this thread agrees too
>
>--awesome
>
>
interested, they are
http://www.nadn.navy.mil/MathDept/courses/sm230/montyhal.htm
http://www.comedia.com/Hot/monty.html
http://129.138.4.51/~kmellis/monty.html
http://astro.uchicago.edu/rranch/vkashyap/Misc/mh.html
http://www.wiskit.com/marilyn.gameshow.html
BTW, I am still not convinced that if
(1)a random initial choice is made, and
(2)that choice remains after the two wrong answers are eliminated, that the
problem is not analagous to the Monty Hall problem. I am still trying to
"logic" it out.
Thanks for continuing this interesting thread.
David Abrams
Zonked today by the publication dates of Dr. Jekyll & Mr. Hyde (1886) and
Dracula (1897). Oh, well...try again tomorrow.
Dan Tilque
becomes 2/3
> of a chance to door 1's 1/3 chance...
>
> Simple stats tells me that they both become equal 50-50. That's all
there
> is to it and I think that's what pretty much everyone who's responded
to
> this thread agrees too
Sigh...
Think of this way. At the beginning of LMaD, you choose a random door
from three equal doors and everyone agrees the chance you are correct
is 1/3 and the chance that one of the other two doors are right is
2/3.
Let's assume that Monte is then required to show you a dud prize. He
is forbidden from showing the door you chose, so the odds on that door
have not changed. From your point of view, it's still 1/3 chance of
being right. What he has done is to collapse the chances of it being
one of the other two doors into one door. So odds for switching are
2/3.
--
Dan Tilque
Sent via Deja.com http://www.deja.com/
Before you buy.
James Allen
>>of a chance to door 1's 1/3 chance...
>>
>>Simple stats tells me that they both become equal 50-50. That's all there
>>is to it and I think that's what pretty much everyone who's responded to
>>this thread agrees too
>>
No it doesn't.
Let my try to put it the most clear way possible.
The problem assumes that Monty will _always_ reveal a wrong door after your
selection. You will then switch off to the remaining door.
If you originally picked the right door (1/3 of the time), you will switch off
to the wrong door and lose.
If you originally picked the wrong door (2/3 of the time), you will switch off
to the right door and win.
Thus you will win 2/3 of the time by switching.
You see, Monty eliminating one of the doors you didn't pick dramatically
changes the nature of the problem.
Example: The prize is behind Door A
Scenario 1: You pick door A. Monty eliminates either B or C. You switch and
lose.
Scenario 2: You pick door B. Monty eliminates door C. You switch and win.
Scenario 3: You pick door C. Monty eliminates door B. You switch and win.
You win 2 out of 3 times by switching.
Mike Caldarelli
provided by you and everyone else have helped me to understand... So thanks
However (and this question is for all).. does this entire strategy not
depend on WWTBAM knowing your initial suspiscion in order to give you a
chance to switch? Would this not imply that the producer/director is
sitting in the back room with the 50-50 computer and making sure it keeps
your suspected answer along with the right one (assuming your hunch was
incorrect)?
From what I know from reading this group in the past is that the 2 remaining
answers are predetermined before the show by the producers and are usually
the 2 most likely answers to stump a contestant... Therefore, there is no
"reaction" to the contestant's initial inkling...
So does the Monty Hall dilemma still apply if 50-50 answers are preset?
--awesome
Abrams
>So does the Monty Hall dilemma still apply if 50-50 answers are preset?
>
I would think so, hence my original post. I really believe that it applies.
David Abrams
Zonked by relative sizes of 4 characters in "Charlotte's Web"
Larry Stone
<jcab...@mail.utexas.edu> wrote:
> Mike Caldarelli <7m...@qlink.queensu.ca> wrote in message
> <85qikf$g1p$1...@knot.queensu.ca>...
>
> >So does the Monty Hall dilemma still apply if 50-50 answers are preset?
> >
> I would think so, hence my original post. I really believe that it applies.
No it doesn't. If the answers to be removed are pre-determined, then it
makes no difference what you are thinking or say. The "Monty Hall
deliemma", as it was put, only applies if you make a selection and if that
selection is wrong, the two removed are the two wrong answers from three
you didn't select. When you select one, you know at least two of the other
three are wrong. If two wrong answers are then removed, you haven't
learned anything new since you already knew at least two of the three were
wrong.
If the one you selected could be removed (as would be true in WWTBAM if
you first state or even think a choice before using the 50-50), then you
have learned something new. Specifically, if it were wrong, it had a 2/3
chance of being removed but it wasn't. That may not seem like you learned
something but you did.
Another way of looking at it is in the Monty Hall problem, the choice to
be removed is dependent on your choice. In the WWTBAM case, it isn't.
--
-- Larry Stone --- lst...@enteract.com
http://www.enteract.com/~lstone/
Roselle, IL, USA
Leszek Pawlowicz
what's behind either door, i.e. having no information on which choice
would be correct. Not true for this case ...
Dan Tilque
Leszek Pawlowicz <les...@nospam.pobox.com> wrote:
> More to the point, the "Monty Hall" dilemma relies on you not knowing
> what's behind either door, i.e. having no information on which choice
> would be correct. Not true for this case ...
But that's actually confusing the analysis. Lets say that on
Millionaire, you have absolutely no clue as to the answer or even if
one answer seems to be more likely than another. From your perspective
then choosing the answer is a matter of random guessing.
The important point then is whether you have to choose one before they
eliminate all but one of the others. If that's the case, as it is in
LMaD, then the times when you did not initially choose the correct
answer, you have exactly specified what selection(s) are to be
eliminated. That is, the host has to eliminate the wrong answers/dud
prize you did not choose. In LMaD, this happens 2 out of every three
times.
But if the ones to be eliminated are chosen beforehand, irrespective of
your choice, or even if you don't have to make a choice as I believe is
the case in Millionaire, then you've done nothing to restrict the
actions of the host and so neither selection left is more likely than
the other.
Of course, this is all dependent on you not having a clue as to what is
the answer. This situation probably seldom arises.
fred klein
> In WWTBAM, the two chosen for 50/50 must not be correlated to your initial
> guess for this to be valid. This always holds if you don't reveal your
> suspicions, and I suspect that this holds regardless.
> This is true if they _predetermine_ a false answer they expect is most
> likely to be incorrectly guessed, and leave that one.
>
> Regis(in the US) mentions that the 'computer' eliminates the two answers,
> but that really doesn't indicate much.
I seem to recall him mentioning "two wrong answers _at random_"
[emphasis mine]. I'll listen more carefully next time I watch
WWTBAM.
|snip other interesting questions|
Fred Klein
BudMann9
The 50-50 is not random.
There is no computer.
Writer(s) of questions pre-determined which 2 of the 4 choices are to be
eliminated should 50-50 be used.
Instructions are made to choose the "two likeliest answers."