On Sat, 21 May 2011 19:55:55 +1000, "Trevor" wrote:


"John Williamson" wrote in message
...
Suppose for the moment that there are two contestants. One picks door
two, and the other picks door one. Then the moderator opens door three
and shows everyone that there is a donkey behind that door. Now, will it
make any difference if the other two switch their initial picks or not?
And, if they do swap doors, with they both enjoy a 2/3 chance of
winning?

In this scenario, once the third door is opened, they each have a
50/50 chance of winning, and there is no advantage in swapping
choices. Before the revelation their chances of winning were 1/3. If
it seems illogical that the odds are changed by this revelation,
remember that one time in three the host will reveal, not a goat, but
the car.

Only if the host opens a door at random, which isn't the case in the
classic Monty Hall problem. The host knows where the car is, and always
opens one of the other doors.

Er, the host can't open one of the other doors now, since they have both
already been selected by the two contestants. If he does however, and
reveals a goat, then that contestant now has a ZERO chance of winning
obviously, with the remaining contestant on 66%.
Try actually re-reading the new scenario presented.

Trevor.

I think maybe you need to re-read. Each contestant picks a door, then
the host opens the remaining door. If he exposes a goat, then at least
one of the contestants gets a car. In fact either of the contestants
will get the car, with a 50/50 chance.

d

"Don Pearce" wrote in message
...
Suppose for the moment that there are two contestants. One picks door
two, and the other picks door one. Then the moderator opens door three
and shows everyone that there is a donkey behind that door. Now, will
it
make any difference if the other two switch their initial picks or
not?
And, if they do swap doors, with they both enjoy a 2/3 chance of
winning?

In this scenario, once the third door is opened, they each have a
50/50 chance of winning, and there is no advantage in swapping
choices. Before the revelation their chances of winning were 1/3. If
it seems illogical that the odds are changed by this revelation,
remember that one time in three the host will reveal, not a goat, but
the car.

Only if the host opens a door at random, which isn't the case in the
classic Monty Hall problem. The host knows where the car is, and always
opens one of the other doors.

Er, the host can't open one of the other doors now, since they have both
already been selected by the two contestants. If he does however, and
reveals a goat, then that contestant now has a ZERO chance of winning
obviously, with the remaining contestant on 66%.
Try actually re-reading the new scenario presented.


I think maybe you need to re-read. Each contestant picks a door, then
the host opens the remaining door. If he exposes a goat, then at least
one of the contestants gets a car. In fact either of the contestants
will get the car, with a 50/50 chance.


NOPE, IF the host picks the remaining door he now has a 1/3 chance of the
big prize, and the other two also have only a 1/3 chance each of the big
prize, NOT 50:50.

trevor.

"Trevor" wrote in message
u...
"Don Pearce" wrote in message
...
Suppose for the moment that there are two contestants. One picks door
two, and the other picks door one. Then the moderator opens door
three
and shows everyone that there is a donkey behind that door. Now, will
it
make any difference if the other two switch their initial picks or
not?
And, if they do swap doors, with they both enjoy a 2/3 chance of
winning?

In this scenario, once the third door is opened, they each have a
50/50 chance of winning, and there is no advantage in swapping
choices. Before the revelation their chances of winning were 1/3. If
it seems illogical that the odds are changed by this revelation,
remember that one time in three the host will reveal, not a goat, but
the car.

Only if the host opens a door at random, which isn't the case in the
classic Monty Hall problem. The host knows where the car is, and always
opens one of the other doors.

Er, the host can't open one of the other doors now, since they have both
already been selected by the two contestants. If he does however, and
reveals a goat, then that contestant now has a ZERO chance of winning
obviously, with the remaining contestant on 66%.
Try actually re-reading the new scenario presented.


I think maybe you need to re-read. Each contestant picks a door, then
the host opens the remaining door. If he exposes a goat, then at least
one of the contestants gets a car. In fact either of the contestants
will get the car, with a 50/50 chance.


NOPE, IF the host picks the remaining door he now has a 1/3 chance of the
big prize, and the other two also have only a 1/3 chance each of the big
prize, NOT 50:50.

I should have added that yes IF (and only IF) the host reveals a goat, the
other two will now have a 50:50 chance, but surely that is obvious, and
remains so whether they both switch or both stay, so is NO Longer like the
original MH problem at all.

Trevor.

On Tue, 24 May 2011 17:06:22 +1000, "Trevor" wrote:


"Trevor" wrote in message
. au...
"Don Pearce" wrote in message
...
Suppose for the moment that there are two contestants. One picks door
two, and the other picks door one. Then the moderator opens door
three
and shows everyone that there is a donkey behind that door. Now, will
it
make any difference if the other two switch their initial picks or
not?
And, if they do swap doors, with they both enjoy a 2/3 chance of
winning?

In this scenario, once the third door is opened, they each have a
50/50 chance of winning, and there is no advantage in swapping
choices. Before the revelation their chances of winning were 1/3. If
it seems illogical that the odds are changed by this revelation,
remember that one time in three the host will reveal, not a goat, but
the car.

Only if the host opens a door at random, which isn't the case in the
classic Monty Hall problem. The host knows where the car is, and always
opens one of the other doors.

Er, the host can't open one of the other doors now, since they have both
already been selected by the two contestants. If he does however, and
reveals a goat, then that contestant now has a ZERO chance of winning
obviously, with the remaining contestant on 66%.
Try actually re-reading the new scenario presented.


I think maybe you need to re-read. Each contestant picks a door, then
the host opens the remaining door. If he exposes a goat, then at least
one of the contestants gets a car. In fact either of the contestants
will get the car, with a 50/50 chance.


NOPE, IF the host picks the remaining door he now has a 1/3 chance of the
big prize, and the other two also have only a 1/3 chance each of the big
prize, NOT 50:50.

I should have added that yes IF (and only IF) the host reveals a goat, the
other two will now have a 50:50 chance, but surely that is obvious, and
remains so whether they both switch or both stay, so is NO Longer like the
original MH problem at all.

Trevor.


Of course - and that is precisely the new scenario presented. As I
said - re-read. And no, it is nothing like the original MH problem.

d

"Don Pearce" wrote in message
...
NOPE, IF the host picks the remaining door he now has a 1/3 chance of
the
big prize, and the other two also have only a 1/3 chance each of the big
prize, NOT 50:50.

I should have added that yes IF (and only IF) the host reveals a goat,
the
other two will now have a 50:50 chance, but surely that is obvious, and
remains so whether they both switch or both stay, so is NO Longer like the
original MH problem at all.

Of course - and that is precisely the new scenario presented. As I
said - re-read. And no, it is nothing like the original MH problem.


Why do I need to re-read, that is exactly what I said!!!!!!!!!!!!!!!!!
What the hell are you objecting to??????????????????

Trevor.

On Tue, 24 May 2011 17:00:57 +1000, "Trevor" wrote:


"Don Pearce" wrote in message
...
Suppose for the moment that there are two contestants. One picks door
two, and the other picks door one. Then the moderator opens door three
and shows everyone that there is a donkey behind that door. Now, will
it
make any difference if the other two switch their initial picks or
not?
And, if they do swap doors, with they both enjoy a 2/3 chance of
winning?

In this scenario, once the third door is opened, they each have a
50/50 chance of winning, and there is no advantage in swapping
choices. Before the revelation their chances of winning were 1/3. If
it seems illogical that the odds are changed by this revelation,
remember that one time in three the host will reveal, not a goat, but
the car.

Only if the host opens a door at random, which isn't the case in the
classic Monty Hall problem. The host knows where the car is, and always
opens one of the other doors.

Er, the host can't open one of the other doors now, since they have both
already been selected by the two contestants. If he does however, and
reveals a goat, then that contestant now has a ZERO chance of winning
obviously, with the remaining contestant on 66%.
Try actually re-reading the new scenario presented.


I think maybe you need to re-read. Each contestant picks a door, then
the host opens the remaining door. If he exposes a goat, then at least
one of the contestants gets a car. In fact either of the contestants
will get the car, with a 50/50 chance.


NOPE, IF the host picks the remaining door he now has a 1/3 chance of the
big prize, and the other two also have only a 1/3 chance each of the big
prize, NOT 50:50.

trevor.




The host doesn't pick the remaining door, he opens it. He reveals a
goat/donkey whatever. That means the two contestants have door each,
one of which has a car behind it. They now each have 50/50 odds. There
is nothing in this scenario that puts one contestant's odds higher
than the other since they both picked a door each at the start.

d

"Don Pearce" wrote in message
...
The host doesn't pick the remaining door, he opens it. He reveals a
goat/donkey whatever. That means the two contestants have door each,
one of which has a car behind it.

ONLY if it is not behind the one the host already opened. Since he no longer
has a choice of doors, he must have a 1/3 chance of showing the car.

They now each have 50/50 odds. There
is nothing in this scenario that puts one contestant's odds higher
than the other since they both picked a door each at the start.

Right, where did I say otherwise, IF the host has not already shown the car?
The whole point is that the game is now no longer like the Monty Hall
scenario in any way.

Trevor.