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The Monty Hall problem.
#11
RE: The Monty Hall problem.
Adrian,

I remember this making sense to me once; I had a real aha moment where clarity came, but I was really high when that happened so this NOW makes not much sense to me.

Rhizo
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#12
RE: The Monty Hall problem.
(September 24, 2009 at 9:58 am)Eilonnwy Wrote: I'm not a big fan of statistics but visually in my head I get it.

The way I understand it is that the two doors you don't pick become a unit, so even if you know one of them has the goat, as a unit, there is still a 66% chance switching gives you the car because there was a 66% chance the goat was in one of the other two doors when you picked your door. And by revealing which of the two doors has the goat, there's still a 66% chance because there was always going to be a goat in one of those doors.

Am I understanding this correctly?

Yes, there's a 66.6..% chance that the car is in one of the two doors you didn't choose. Even when Monty gets rid of one of the doors the remaining door still has that 66.6..% chance.

It's a bit like being able to choose two of the doors instead of just the one.
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#13
RE: The Monty Hall problem.
Yay, I get it. Big Grin
"The way to see by faith is to shut the eye of reason." Benjamin Franklin

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#14
RE: The Monty Hall problem.
For those who don't get it, here is an explanation using the three options you can choose from:

1) Pick a door with a goat. Host reveals other goat. Switching wins.
2) Pick a door with a goat (the other one). Host reveals other goat. Switching wins.
3) Pick a door with a car. Host reveals one of the other doors (they both have goats). Switching loses.

The idea is that you have a 2/3 chance of picking a door with a goat at the start, so you are more to have picked a goat. When you have picked a goat, switching always wins. If you picked a car at the beginning, switching will always lose, but since you only have a 1/3 chance of picking a car at the start, the chances are against you. On the other hand, if you stay with your choice, 2/3 times you will lose, and 1/3 times you will win.

So it is more advantageous to switch Big Grin
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#15
RE: The Monty Hall problem.
Nice, I get that and I am not even high! Smile

Rhizo
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#16
RE: The Monty Hall problem.
Saying 'stick or choose' is not the same as saying 'pick one'? It's still a 50/50 chance.

I liken it to flipping a coin 100 times. If 99 times it comes out heads, what's the chance it will come out heads (or tails) on the 100th flip? It's 50/50. The previous flips (or door choices) are irrelevant.
Please explain how a previous choice has any bearing on the final choice of two doors.
I used to tell a lot of religious jokes. Not any more, I'm a registered sects offender.
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...the least christian thing a person can do is to become a christian. ~Chuck
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#17
RE: The Monty Hall problem.
But it's not because you've already made a choice of 1/3. That was the point. Most people discount their previous choice because it seems irrelevant, but the fact is your first choice plays a large factor in the result of your second, as I showed in my previous post.
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#18
RE: The Monty Hall problem.
I was posting as Adrian was.

However that seems only to make sense it after the 1st choice it was simply removed. But you said they revealed the goat behind that door, so now you know out of the remaining two doors one is a goat (tails) and one is a car(heads).

When presented with a new choice between the two, your odds are 50/50, which ever you choose, that the coin would come up 'heads'.
I used to tell a lot of religious jokes. Not any more, I'm a registered sects offender.
---------------
...the least christian thing a person can do is to become a christian. ~Chuck
---------------
NO MA'AM
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#19
RE: The Monty Hall problem.
Wikipedia has a good article on it. The point is that you are more likely to have already chosen a goat, so it isn't a 50:50 choice, because the other door is more likely (by another 1/3) to be the car.
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#20
RE: The Monty Hall problem.
More of my reasoning behind this:

You are basicly setting up two games. One is 1 out of 3. 33.33333% chance of 'winning'.

You make your choice, but instead of sticking to that choice the host removes one of the losing choices. Whether you picked a winning or losing door, one of the losing doors will be removed from the equation. So being more likely to have chosen a goat first time around is irrelevant. That would only be relevant if one of the two unpicked doors was removed without knowledge of it's contents (by the player).

Now you get a 2nd game. You are presented with two doors. One wins, one loses. Pick one. What are your chances or picking the winning one?

50/50
I used to tell a lot of religious jokes. Not any more, I'm a registered sects offender.
---------------
...the least christian thing a person can do is to become a christian. ~Chuck
---------------
NO MA'AM
[Image: attemptingtogiveadamnc.gif]
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