If you google "Monty Hall Problem" you'll find a lot of sites that
simulate this (you pick a door, are shown a goat (or other booby
prize), then decide to switch or stay -- do each a bunch of times,
and you can see your success rate over many trials.
At 5:18 PM +0000 12/17/03, Michael Barrow wrote:
>Joe does not get the "Let's make a deal" analysis quite right.
>There were three boxes, one contained a prize. The contestant chose
>a box, the host would then open one of the other boxes and reveal an
>empty box. The contestant was now invited to switch his or her
>choice to the other unopened box.
>
>"It does not make any difference if you switch" is the wrong, though
>attractive, answer. The logic is there are two boxes and one prize,
>hence a 50:50 chance. Hence switching doesn’t matter. The correct
>answer is as follows: one third of the time the contestant initially
>chooses the correct box, so switching would be lead to the booby
>prize. But two thirds of the time the wrong box is chosen, and
>switching leads to the valuable prize. Hence you should switch.
>Psychologically however, most people would probably not switch ?
>you’d look a fool (on TV!) if you switched away from the right box.
>
>Mike Barrow.
>
>--On 17 December 2003 12:09 -0500 Joe Nowakowski <[log in to unmask]> wrote:
>
>>The first problem sounds like the "Let's Make A Deal" contest. This was a
>>popular television show in the US several years ago. At the end of each
>>show, contestants are shown three curtains. Behind two are 'fabulous'
>>prizes, and behind the third is a used umbrella or something. The
>>contestant picks a curtain from the three, then the host opens one of the
>>other two curtains behind which is a nice prize. The contestant is then
>>given the chance to choose between the remaining two doors.
>>
>>A couple of years ago there was a running debate about whether the
>>contestant should switch or not. One camp concluded that the chances of
>>winning are always improved by switching after the prize is shown behind
>>one of the curtains. To me, it seemed that at the end of the game there
>>were two curtains, each of which had an equal chance of hiding the real
>>prize. Switching or not therefore resulted in a 50% probability of
>>winning.
>>
>>In the prisoner's example, it seems to reduce to one of two going free
>>after the jailor reveals that the third is certain to go free.
>>
>>In the second problem, I think it helps to reverse the question and ask
>>the probability of a CD passing the inspection: .99 x .97 x. 98 x .99 =~
>>.93. So the probability of a CD failing the inspection is appr. .07. In a
>>sample of 100, 7 would be rejected. For the last part, however, I think
>>we need to know the true defective rate. You can complete the 'tested'
>>side of a probability tree to get the joint probability of 'tested and
>>defective,' but to compute the conditional probability you need the
>>overall probability of a CD being defective. If you assume the testing
>>rate is the same as the true rate of defects, then the conditional
>>probability that a CD was tested, given it was defective, if the
>>probability of tested and defective, .02, divided by the probability of
>>being defective, .07, or about 28.5%.
>>
>>I hope this helps.
>>
>>Take care,
>>
>>Joe Nowakowski
>>
>>.At 09:00 AM 12/15/03 -0800, you wrote:
>>>Dear friends, I have some doubts in solving the following problems on
>>>probability. Anyone knows the solution, please email me. Stirzaker
>>>Three persons A, B, and C are held in solitary confinement. The jailor
>>>tells each of them that two are to be freed, the third is to be flogged.
>>>Prisoner A, say, then knows his chance being released is 2/3. At this
>>>point the jailor reveals to A that one of those to be released is B,
>>>this jailor is known to be truthful. Does this alter A’s chance of
>>>release? After all, he already that one of B or C was to be released.
>>>Can it that knowing the name changes the probability? 2) This problem is
>>>in the book 'Probability and Statistics' by Walpole, Myers Before
>>>the distribution of certain statistical software every fourth CD is
>>>tested for accuracy. The testing process consists of running four
>>>independent programs and checking the results. The failure rate for the
>>>4 testing programs is, respectively, 0.01, 0.03, 0.02, and 0.01. i)
>>>What is the probability that a CD was tested and failed any test? ii)
>>>2 or 3? iii) In a sample of 100, how many CDs would you expect to be
>>>rejected? iv) Given a CD was defective, what is the probability that it
>>>was tested? Thank you. A.Muthusamy & Computer Applications PSG
>>>College of Technology Coimbatore-641004 INDIA www.psgtech.edu Do you
>>>Yahoo!?
>>>New Yahoo! Photos - easier uploading and sharing
>>Joseph M. Nowakowski
>>Associate Professor of Economics
>>Muskingum College
>>New Concord, OH 43762
>>(740) 826-8206
>
>
>
>Michael Barrow Tel: +44 (0)1273 606755
>School of Social Sciences Fax: +44 (0)1273 673563
>University of Sussex
>Falmer, Brighton, UK
>BN1 9QN
--
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