Like Chris said, "Look from the host's perspective."  The whole time the host knows that in the end he will have a 1/2 chance of giving the prize away.

No, the host has a 2/3 chance of giving the prize away if the contestant is smart and always switches.

If you knew that the host was going to open one door, the whole time you would have a 1/2 chance of winning the cash because you would know that eventually you will have a choice of two.

No, you have a 2/3 chance of winning if you play right. Iniitially you have a 1/3 chance that the prize is behind the door you picked. You have a 2/3 chance that the prize is not behind the door that you picked. The host shows you one door without the prize. But there is still a 2/3 chance that the prize is not behind your original door. That has not changed.

Do the experiment 1,000 times or 10,000 you get close to the same result. Read back through the thread. The probability tree I drew IS mathematically correct. Do a Google search for "Monty Hall problem". Read what is said.

I believe that both opinions are right; the argument is derived from a failure to recognize that we are all trying to find one answer to two different questions.  Does anybody agree? And can they guess what I mean?

I came across this idea when I finally asked myself this:  Does revealing one of the empty doors change the odds that the first door chosen would contain the prize?  No, it does not.  Because when you made that choice, you had three choices, obviously giving you a 1/3 chance of winning.  Now if you stayed with that choice, you would still be considered as having a 1/3 chance, because your odds of winning mathematically don't change just because a door was opened.  Many of you agree with this.  But, if you switch doors after the empty door was revealed, your odds are changed to 1/2, because you are exercising your right to change doors, and when you were given that option you had two doors, thus giving you a 1/2 chance of winning.  It's not a question of how many times you win out of three games if you stick with the same door, it's a question of how many options did you have when you made your choice.  Still I must admit that there is no one correct answer for this problem.  Because everybody wants a universal answer for a problem, and there cannot be because it all depends on your perspective.  I just believe my perspective to be good.  I wish I had more time to explain more because there is much more to say, but I am very short on time.

I agree with Chris. There is only one answer to this problem.

samed wrote:

Does revealing one of the empty doors change the odds that the first door chosen would contain the prize?  No, it does not.  Because when you made that choice, you had three choices, obviously giving you a 1/3 chance of winning.  Now if you stayed with that choice, you would still be considered as having a 1/3 chance, because your odds of winning mathematically don't change just because a door was opened.  Many of you agree with this.  But, if you switch doors after the empty door was revealed, your odds are changed to 1/2, because you are exercising your right to change doors, and when you were given that option you had two doors, thus giving you a 1/2 chance of winning.

ok, so you say:
don't switch - prob of 1/3 to win.
swtich - prob of 1/2 to win.

Don't probabilities have to add up to 1? You only have 2 options.

samed, you are NOT right. The only way you could be right is if you began changing the definitions and principles associated with probability theory. If you are so certain we are both right, then show me the math. Show me the math that says the probability on the switch comes out as either 1/2 or 2/3.

Did you do the experiment? What happened.

In a previous post you stated that the answer is 2/3 and 1/2.

Typo. 2/3 and 1/3.

The problem has a solution. The solution has been presented. You have said nothing that has convinced me you are right. I ask once again: Show me some math. This IS a math problem. ALL probability IS math. It's not theory. You can't answer a math question with hand-waiving and expect to convice me.

samed wrote:

... I don't think any aspect of the problem has an association with any 2/3 probability.

Samed, perhaps the discussion (which includes an explanation using conditional probability) at Who’s the Goat . . . Marilyn or the Mathematicians? (<—click on this link) will convince you.

2/3 staying put, and 1/3 switching. It's the same problem, just replace prize with nothing.

I'm not sure about the logic involved in this thread (and I can't help but question the validity of the experiment used above). I think it's safe to say that there is a 100% probability that at least one of the curtains has no money behind it. Regardless of which curtain you choose and which curtain actually contains the money, the host always has the option of selecting a curtain that offers no prize. This option is available to him/her with a 100% probability regardless of which curtain you choose. Therefore the host's selection of this curtain does not affect the probability that one of the other curtains DOES contains the money.
That sounds confusing, so I think we should look at this logic another way. Imagine that your state lottery decides to change it's rules. Rather than announce a winning number to all, the ticket holders must present their tickets to a lottery representative to see if they have won. After every elimination, the remaining ticket holders have the choice of holding onto their ticket or switching with someone else. It may be true that the probability of having the winning ticket increases as other contestants are eliminated, assuming you are not one of them. And if you are lucky enough to be the prize winner, it would certainly seem to you that your chances of winning increased every time you changed your ticket, but I think your fellow contestants would disagree with your conclusion. After all, switching tickets certainly did not turn out favorably for them. In the end there is only one lottery winner, so the odds of the game have not changed.
   To cover all the bases, let's say the lottery office always knows who has the winning ticket, and so randomly selects one person who is NOT the winner, and allows all remaining contestants to switch tickets if they so choose. This still doesn't change which ticket is the correct one, nor can the contestants deduce which is correct. After each round of elimination, each remaining contestant may have a higher PROBABILITY of having the right ticket at that time, but the overall probability of correctly selecting the right ticket at the end is unaffected by including the variability that you can change your ticket after every elimination; either you have the correct ticket when your time comes, or you do not.

   In short, I believe that the addition of ongoing variability changes nothing with regards to chance; the probability of being correct is determined by the number of contestants remaining in the lottery (or the number of curtains available to choose from), not which ticket you actually have (or which curtain you currently favor) and which are remaining. It's a process of elimination, and your odds of being a winner increases as the number of other possible winners decreases.

Last edited by Gungnir (2007-03-08 02:45:31)