It works the other way round, too.

Let's say you are playing a shell game.  There are three shells and one
pea under one shell.  You mix them and let someone pick a shell.  He has
1:3 chance of picking the shell with the pea under it and winning.

But you're a good guy and so when he picks a shell you lift another
shell with no pea.  That leaves two shells on the table, the shell he
picked and the other shell.  (There is no switching here.)

So you say to the guy, "See, you have 50:50 chance of winning.  It's
this one or that one.  Place your bet."

Does he really have a 50:50 chance of winning as you promised?

(If so, why does he lose 66% of the time? (-:  )



> -----Original Message-----
> From: piclist-bounces@mit.edu [mailto:piclist-bounces@mit.edu] On
Behalf
> Of wouter van ooijen
> Sent: 5. joulukuuta 2007 21:14
> To: 'Microcontroller discussion list - Public.'
> Subject: RE: [OT] The Monty Hall problem
> 
> > You know up front that the host is going to reveal one of the
> > "goat" doors, taking it out of play. Therefore, it's really a
> > two-door game from the outset, and you have a 50% chance of
> > picking the correct door.
> > There's /no/ advantage in switching.
> > -- Dave Tweed
> 
> You are the one behind the quiz in CCI? I hope you are kidding!
> 
> Alternatively: let's play this game a few times (I mean a very large
> few). You'll be the quiz host, I'll be the guest. For every car I
choose
> you pay me 45, for every goat I choose I pay you 55. If you are right,
I
> will go broke. If I am right, I will be rich. This was they way such
> issues were solved in the early days. The false ideas were weeded out
by
> financial darwinism-avant-la-lettre :)
> 
> Wouter van Ooijen
> 

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