On Sun, 29 Oct 2006 15:55:00 GMT, "rb608"
wrote:

wrote in message
Not as I see. As I see it, those supporting 50-50 odds aren't looking
at the situation properly.

For example, if a third person walks up at the point
after the first board is turned over and is offered a chance to get in
on things by picking one of the two remaining boards, their odds are
50-50, but they had different "rules" (this "each "hand's" _rules_ are
different" is why blackjack ain't a heads-up game).

As one who understands the mathematics, but still has difficulty
rationalizing the counter-intuitive nature of the answer, I think this
somewhat illuminates the crux. If the question is, "What is the probability
of selecting the correct answer from two remaining random choices?", the
answer is 1/2. That is the simplest and most understandable question.
Everybody gets it. But that's not the actual question posed by the problem,
nor are the choices random. The question posed is, "What is the probablity
of selecting the correct answer through this process?" The correct answer
to that is 2/3.

I don't see how it's (objectively) counter-intuitive, and I think
attempting to get too involved in "math" (beyond the basic) makes it
more, rather than less difficult - for example, if it had been 4 boards,
two were turned over revealing losers, and then the choice to change
were given, to me, common sense indicates the odds say change your pick
because of the same reasons I feel it does with 3. If you must have
"math," I'm fairly sure the formula would be that the odds in favor of
switching are pretty close to if not exactly x-1/x and the odds in favor
of sticking are always exactly 1/x, when x is greater than 2, but I'm
not a mathematician, so ??? Perhaps the odds in favor need to account
for the first pick when x is higher than 3 - such that it isn't quite
x-1/x - but it's always going to be better odds than 1/x. ****, that's
confusing...that's why, IMO, algebra isn't the way to figure this out.

About the only thing I can figure is that it is much like many threads
on ROFF in that most folks, myself included at times, don't always
_read_ what they are "reading," but rather, um, infer from what is
written by what they _think_ is being said. In this case, they are
simply ignoring that there are 3, not 2, boards and therefore, the
chances cannot be 1 in 2.

Heck, given the "game" as outlined by Wolfgang, there's nothing
presented in the "rules" preventing the person from choosing the
revealed losing board - they were simply offered a chance to change
their pick. It would be the chooser making the obvious choice not to
choose it because they can clearly see they won't win (they don't need
to know that the chance of winning is 0 in 3).

TC,
R

Joe F.