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Re: [Phys-l] Never Trust a Theoretician! - Monty Hall Revisited.



While I've never been on Wheel of Fortune, I would imagine that contestants
try to "aim" their spins rather than spinning the wheel randomly. I know
that's what I'd do if I had a chance. That being said, is it really
feasible to test the fairness of the wheel based on contestant-generated
data?

On Thu, Nov 24, 2011 at 8:26 PM, chuck britton <cvbritton@mac.com> wrote:

While the memories of TV quiz show rigging of the '50's may be fading
fast, I suspect that the sort of
rigging suggested here would be hard to implement.

How hard would it be to verify that the 'Wheel' of Fortune is 'fair'
is spite of many people thinking otherwise?

Many years worth of data must available.
.
At 7:59 AM -0600 11/24/11, brian whatcott wrote:
Memo from Uber-BeanCounter:
To: Monty Hall.

With immediate effect, arrange all doors to be empty as the default
scenario.
Set one prize behind the chosen door once in ten trials for the
optimal return on media investment.

Sincerely
UBC

On 11/24/2011 6:26 AM, Philip Keller wrote:
Memo
> From network bean-counters
To Monty Hall

We see that you have been giving away cars about 1/3 of the time.
In an effort to reduce costs please start revealing a wrong door
and offering the player a chance to switch. Do this almost every
time they have guessed correctly on their initial guess and only
occasionally when their guess was wrong. This will save us money
because there are many websites that claim that switching is always
better. Hopefully our contestants will have read those.

Paul Nord<Paul.Nord@valpo.edu> wrote:


The answer is 2/3 probability if you switch.

Imagine the game this way:
You: Monty, I would like to choose both doors #2 AND #3.
I want you to give me the best prize that's behind either of those
doors.
Monty: I'm sorry but you have to pick just one door.
You: I could play by your rules. But let me tell you
ahead of time that I'm going to choose door #1. And then you will
show me that either door 2 or 3 is a non-winner. At that time I
will switch my choice to the the other door -- the best of doors #2
and #3.

Or extend the game to 100 doors with only one prize. You choose a
door and Monty can always reveal 98 non-winning doors. (Better
yet, picture playing Minesweeper on a 10x10 grid with one mine.
You pick a square. I show you 98 squares with no mine.) The odds
of winning by not switching are 1/100. But when you switch you get
the best choice of all of the other 99 doors - 99/100.

Paul


On Nov 23, 2011, at 10:12 PM, brian whatcott wrote:

I seem to recall quite a protracted discussion here on the merits of
the
Monty Hall Paradox.
This concluded that in the three door variant - with the empty door
revealed, I seem to
recall that if you stuck, you rated a one in three chance of winning:
but if you swapped
your door choice after the third empty door was opened, your chances
rose to one in two.

But tonight I saw a Mythbusters experiment, nicely done *, that made
the
relative chances
one in three for sticking, and TWO in three for swapping doors.

Bah humbug!

Brian W
* Discovery Channel (HD) 9-10 pm CST. Wed Nov 23rd 2011


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