<Language> [Fwd: [Fwd: MONTY HALL GAME ODDS....]]

[H. Mark Hubey]

<><><><><><><><><><><><>--This is the Language List--<><><><><><><><><><><><><>

Just in case, someone brings up this Monty Hall problem
again, here is the solution. We might even discuss
Bayes Theorem (formula). Some philosophers are Bayesians
and reason using probability theory instead of logic. So
the usage of this formula is very important.

Disclaimer: I hope the solution is correct :-)

> From: Ben Bullock <ben at hayamasa.demon.co.uk>
> Organization: http://www.hayamasa.demon.co.uk/
> Newsgroups: sci.stat.math

> I've corrected this solution to the Monty Hall problem.
>
> I do believe that this solution is correct.
>
> This solution is partly based, by the way, on one found in a book
> called `Understanding Statistics' by Upton and Cook, which is a book
> for teaching A-level statistics published by Oxford University Press.
>
> The `Monty Hall' problem
> ========================
>
> In a certain game show, behind one of three doors 1, 2 and 3 is a
> valuable prize.  A contestant must pick one door to open.  Only if he
> picks the right door will he win the prize.  The contestant picks one
> door.  The host then opens one of the two doors which the contestant
> didn't pick, and shows the contestant that there is no prize behind
> it.  He then asks the contestant if he would like to switch to the
> remaining, unopened, door, instead of the one he first picked.  What
> should the contestant do?
>
> The paradox
> ===========
>
> Argument 1 (`no point switching') runs as follows:
>
> ``If the prize was equally likely to be behind any of the doors, then
> opening one of the wrong doors only shows the contestant that the
> prize is not behind that door.  Therefore it is equally likely to be
> behind either of the other doors.''
>
> Argument 2 (`always switch') runs as follows:
>
> ``Two out of three times the contestant picks the wrong door.  In
> those cases, the host will show him the other wrong door.  If the
> contestant always switches to the other unopened door, he will be
> switching away from an empty door to the prize door 2 out of 3 times.
> If he doesn't switch, he will loose 2 out of 3 times.  So he will have
> better odds of winning if he always switches''.
>
> Which of these arguments is right?
>
> The solution
> ============
>
> Neither of the above two arguments is exactly wrong or exactly right.
> Both of them contain a hidden assumption about what the host will do.
> (This is a good counterexample which shows the danger of `hand waving'
> arguments such as the above two).  To correctly understand this
> problem it is necessary to use Bayes' theorem.
>
> Suppose that the contestant chooses door 3 initially, and that the
> host opens door 2 (in any other case the result is the same, just by
> relabelling doors).  The chance of the prize being behind door 3,
> given that the host opened door 2, is, by Bayes theorem,
>
>                 P(O2|P3) P(P3)
>      P(P3|O2) = --------------
>                     P(O2)
>
>                               P(O2|P3) P(P3)
>               = ------------------------------------------------         (1)
>                 P(O2|P1) P(P1) + P(O2|P2) P(P2) + P(O2|P3) P(P3)
>
> where the events are
>
>         O2: host opens door two,
>         P1: prize is behind door 1,
>         P2: prize is behind door 2,
>         P3: prize is behind door 3.
>
> If the prize is behind door 2 the host won't open door 2 of course:
>
>      P(O2|P2) = 0
>
> If the prize is behind door 1 the host must open door 2:
>
>      P(O2|P1) = 1
>
> The probabilities of the prize being behind the different doors are
>
>      P(P1) = P(P2) = P(P3) = 1/3.
>
> If we abbreviate P(O2|P3), the chance of the host opening door 2 when
> the prize is behind door 3, as
>
>      P(O2|P3) = p,
>
> then
>
>                    1/3 p
>      P(P3|O2) = -----------
>                 1/3 p + 1/3
>
>                   p
>               = -----
>                 1 + p
>
> from equation (1) above.  If the host opened door 2, the prize must be
> behind either door 1 or door 3, so
>
>      P(P1|O2) + P(P3|O2) = 1
>
> and
>
>      P(P1|O2) = 1 - P(P3|O2)
>
>                       p
>               = 1 - -----
>                     1 + p
>
>                   1
>               = -----.
>                 1 + p
>
> Now, the contestant doesn't know what p is: it can be anything between
> 0 and 1.  Consider the possibilities:
>
> If p = 0,
> ---------
>
> in other words, if the host never opens door 2 when the
> prize is behind door 3, then the probability that the prize is behind
> door 1, given that the host opened door 2,
>
>                  1
>     P(P1|O2) = -----
>                1 + 0
>
>              = 1,
>
> so the prize *must* be behind door 1.
>
> If 0 < p < 1,
> -------------
>
> then
>
>      1/(1+0) > P(P1|O2) > 1/(1+1),
>
> or
>
>            1 > P(P1|O2) > 1/2,
>
> in other words, the prize is more likely to be behind door 1 than door
> 3, and so although we are not sure where the prize is, there is a
> benefit to changing doors.
>
> If p = 1/2,
> -----------
>
> in other words, if the host is equally likely to pick either door 1 or
> door 2 to open when the prize is behind door 3, then
>
>                    1
>      P(P1|O2) = -------
>                 1 + 1/2
>
>               = 2/3.
>
> This is the hidden assumption in argument 2 above.
>
> If p = 1,
> ---------
>
> in other words, if the host will certainly open door 2, not door 1,
> when the prize is behind door 3, then P(P1|O2) = 1/2, and there is no
> benefit in switching.  This is the hidden assumption in argument 1
> above.
>
> The contestant's `best guess'
> =============================
>
> The `best guess', given that the contestant does not know the value of
> p, is to switch, since even in the case p = 1 this does not decrease
> his chance of winning, and in the case p < 1 it increases it.
>
> --
> Ben Bullock (home page http://www.hayamasa.demon.co.uk/)
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