*the* smart question

[mad=dog]

Seraph is correct.
The choice the host has is limited to two. If the prices is among those two he can only reveal the other - the empty one. Actually you can cut the possible outcomes down to
Initial pick. Possible outcomes [1] the price with 1/3 likelyhood or [2] empty with 2/3 likelyhood.
[1] In case you picked the price on the first choice you will always LOOSE if you change and always WIN if you stay. Remember this outcome has 1/3 likelyhood.
[2] In case you picked an empty closet the host will open the other empty closet leaving only the price. So you will always WIN if you change and LOOSE if you stay. Again we must recall that this outcome has 2/3 likelyhood.
In summary you have 2/3 chance of winning if you change and 1/3 chance if you stay.

The reason why the mind is tricked is that we have information that the contestant does not have. We KNOW that the host is limited to two choices, but the contestant doesn't know that.

EDIT: Sorry there were some critical typing mistakes.

[ 06-22-2005, 02:38 PM: Message edited by: mad=dog ]

[Q'alooaith]

Mad=dog, you do know your wrong don't you?

Once the first one has been removed you must recalculate, it's like working out how many days you get off a year and not counting days that overlap, just not the way to get the right awnser.

[Morgeruat]

Originally posted by mad=dog: The reason why the mind is tricked is that we have information that the contestant does not have. We KNOW that the host is limited to two choices, but the contestant doesn't know that. EDIT: Sorry there were some critical typing mistakes.

Are you saying that the contestant doesn't see which locket was opened, or what? this makes no sense...

[ZFR]

mad dog is right!

The host knows which box is empty. So when chosing between an empty and a full one he MUST pick the empty one.
i.e there is a 2/3 chance (when you pick an empty one) that he is "forced" to show you which is full by opening the one which is not.

[Q'alooaith]

Ahem

After he has chosen the host opens one of the other lockets and reveals it to be empty.

So by what logic do you take it that you have a better chance of choosing the right box by changing?

Can't you see the flaws in your logic, one of the lockets has been taken out of the equation, you already know it's a duff one, if the host does not revel the contents your back to the orginal one in three chance.

It's not that hard.

[Seraph]

There are only three possibilities, and they are all equally as likely.
The player picks empty locket one. The game host picks empty locket number two. Switching will win the car.
The player picks empty locket number two. The game host picks empty locket number one. Switching will win the car.
The player picks the prize. The game host picks either of the empty lockets. Switching will lose.
You win 2/3 of the time by switching, it is pretty straight forward.

Draw a probability tree, use the Bayes' Theorm, or (or better yet 'and') search "Monty Hall Problem" on google and Wikipedia if you don't belive me.

[Aragorn1]

No, there are two events to consider, not one as you are.

1st event: possible outcomes: 3
desired outcomes (x): 1

P(x)=1/3

Then a locket is removed:

2nd event: possible outcomes: 2
desired outcomes (x): 1

P(x)=1/2

Therefore it does not matter

My logic is firther confimerd by the fact that the probabilites of all possible outcomes add up to 1, thus i have covered all possible outcome.

[ 06-22-2005, 04:17 PM: Message edited by: Aragorn1 ]

[Q'alooaith]

ahh, no.

your counting two lose senerio's as one.

There are two ways to lose, and two ways to win. If you discount the host picking the right locket.

so again, a 1/2 chance of winning.

[Morgeruat]

A guest in a price show is shown three closed lockets. He is informed that only one of the lockets contain the price and is asked to make a pick. After he has chosen the host opens one of the other lockets and reveals it to be empty. He then asks the guest if he wants to change his choice. Which of the following is true 1) It is smartest to stick with his initial choice 2) It is smartest to switch his choice 3) It doesn't matter. The chance is equal

a couple of questions that are muddled by the question itself and haven't been sufficiently answered.

1: does the host reveal his pick to the guest, or just the audience
2: if it is shown to the guest is it revealed that the hosts choice is empty does it not remove it from the realm of possible choices bringing the number of possible locations for the locket down to 2 from 3 (the one he's chosen or the one he hasn't).
If he shows the emtpy one then there are 2 possibilities, it is in the one he chose, or it is not, and he then as 2 choices, stay or change. which still leaves a 50% chance of either being wrong.

If the host does not show the guest whether or not the one he (the host) chose then his chance is 1 in 3 of any particular choice being correct.

[Aragorn1]

Originally posted by Q'alooaith: ahh, no. you're counting two lose senerios as one. There are two ways to lose, and two ways to win.

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.