No you know the one the host has is not the correct one taking it out of the equation, thus it becomes 50/50, choosing to stay is just as risky as choosing to change.

Impossible, there are three lockets, so there can only be three outcomes, there is only one way to win, picking the correct locket, there are, in the first event, two ways to lose, picking either of the wrong lockets. </font>[/QUOTE]what are you yabbering on about?

Ignoring the host picking the right locket you can pick either of the wrong ones and the host picks the other wrong one, or you pick the right one and the host picks either of the wrong ones.

See, two possible outcomes, no?

NO

Four possible outcomes, as we're discarding any outcome in which the host picks the right one.


It's realy, realy easy!

Solution #1

There is a 1/3 chance that you'll hit the prize door, and a 2/3 chance that you'll miss the prize. If you do not switch, 1/3 is your probability to get the prize. However, if you missed (and this with the probability of 2/3) then the prize is behind one of the remaining two doors. Furthermore, of these two, the host will open the empty one, leaving the prize door closed. Therefore, if you miss and then switch, you are certain to get the prize. Summing up, if you do not switch your chance of winning is 1/3 whereas if you do switch your chance of winning is 2/3.
Solution #2

After the host opened one door, two remained closed with equal probabilities of having the prize behind them. Therefore, regardless of whether you switch or not you have a 50-50 chance(i.e, with probabilities 1/2) to hit or miss the prize door.


from your search (the #2 answer on google) Aragorn and Myself are arguing #2, you're arguing #1, both are valid.

[ 06-22-2005, 04:34 PM: Message edited by: Morgeruat ]

TERRY PASCAL

Alex,

I've been reading the Monty Hall problem and would like to post the answer to demonstrate where most people math is incorrect.

Given 2 doors, one with a prize and one without, which should you pick?

I think you'd agree that odds are 50-50 you'll be right. The fact that I may have had 3 doors initially and opened one, or 1000 doors and opened 998, does not change the current problem. There is still only a 50/50 chance of winning.

Because the probability of the problem can be expressed in three different ways.

P1 - 1/3 chance prize is door #1, 2/3 chance it is not
P2 - 1/3 chance prize is door #2, 2/3 chance it is not
P3 - 1/3 chance prize is door #3, 2/3 chance it is not

Most of the logic I've seen has the following flaw. As soon as the contestant picks door #1, they immediately assume that P2 and P3 are not valid. Well, they are. So they follow only P1. Regardless of which door I pick, all 3 are still valid. It is only when Monty opens a door that things change. By opening door #3 we now have

P3 - 0/0 chance prize is door #3, 3/3 chance it is not.

Now here everyone says, "Hey by P1, then the 2/3 chance all goes to door #2' But then why not, "By P2, then the 2/3 chance all goes to door #1"? Both are incorrect. When Monty picks #3, the whole problem is changed to:

P1 - 1/2 chance prize is door #1, 1/2 chance it is not
P2 - 1/2 chance prize is door #2, 1/2 chance it is not
P3 - 0 chance prize is door #3, 1 chance it is not

Look at it another way.

What's being said is that if I pick #1 and monty shows #3, then 2/3 of the time #2 will win. So If I had picked #2 to start with and monty opens #3, then the odds go 2/3 to #1. Why would they change? What if I don't have to tell monty? What if I could write it on a secret ballot?

Terry Pascal

I get what you guys are saying now!!

It depends on your interpretation of the events, as both use valid probability theory, and both can't be right.

[ 06-22-2005, 04:53 PM: Message edited by: Aragorn1 ]

Morgurat: The contestant doesn't know the host is limited to a choice of two. To him an _apparently_ random locket was just opened to reveal nothing. So if you mistakenly try to place yourself in the contestants place you erronically arrive to the conclusion that the choice is between the remaining lockets is indifferent.

However it is NOT indifferent. The host does NOT make a randomized choice like you do. The host makes a systematic choice based on the following
1) He never opens your locket.
2) He always opens an empty locket.

I made a simulation of this in Java last time someone refused to see the logic. If you fail to recognise the logic then know that it can be backed by an experiment. Switching is smartest.

My point is that you choose 1 of three, one will always be empty, this is always the hosts choice.

now stop and look at it as a fresh problem, you have 2 choices, one is wrong, and one is right. The third option has always been gone from this new problem thanks to the host, leaving only 2 choices, aka 50% of either choice being correct. Staying or switching doesn't matter as there are only 2 outcomes to this problem, right and wrong.

I realise that is where you and mr. Pascal has made an error. You think of it as a new problem with equal chance.
However it is NOT a new problem. You have information about the doors from the previous incidents.

Marilyn Vos Savant addressed this in a column a couple of years ago. And Bob Barker has done this on the Price Is Right for thirty-some years. Your best move is always to switch.

Would you rather have a 1/3 chance of being right, or a 2/3 chance?

If you stick with your original choice, you have a 1/3 chance of being right.

If you switch, you now have a 2/3 chance, even though one of the other two items is gone.

Why? Because originally, the other two items had a 2/3 chance of one of them holding the prize. All the host did was to eliminate one item that wasn't. So you're *still* picking the 2/3 option, even thought there's only one locket there.

Stick to the original problem. Don't put it into a new one.

<font color = lightgreen>Examine the problem as stated.

Suppose you choose an empty locket, which will happen 2/3 of the time. The host also picks an empty locket, which will happen 1/2 of the time (after your choice). These events are independent so you multiply the probabilities: (2/3)(1/2) = 1/3 = p(getting the prize by keeping your choice). P(getting the prize by switching) = 1 - p(getting the prize by not switching) = 1 - 1/3 = 2/3. Thus, it is smarter to switch once the host opens his empty locket.

Suppose you choose the locket with the prize, which will happen 1/3 of the time. The host chooses an empty locket, which happens all the time. Multiply these two events to get (1/3)(1) = 1/3. P(getting the prize by switching) = 1 - p(getting the prize by not switching) = 1 - 1/3 = 2/3. Thus, it is smarter to switch once the host opens his empty locket.

In either case, it is smarter to switch once the host opens his empty locket. Simple. [img]graemlins/petard.gif[/img] </font>