And if I remember the show correctly from when I was a child, sometimes Monty DID reveal the contestant's door FIRST.

So now let's apply that concept to the trick where the magician has an upward pointing knife or spike under 5 cups....It would be much 'fairer' if the magician did NOT know which cup actually held the spike. LOL, right.

No magician would perform that trick, as the probability of getting injured increases (I would like to say exponentially, but I haven't put a pencil to paper to determine whether this is really exponential) with each performance of the routine.

Do you seriously believe that you became a better guesser because TomasB opened a bunch of doors? You are aware, of course, that TomasB knows which door hides the car, and can carefully avoid opening it, yes?

Your original probability of getting the car was 0.01. After he opens 98 doors, your probability of having found the car(originally) - and, therefore, still having found the car - is . . . 0.01.

This situation arises in the game of contract bridge frequently; it is known as Restricted Choice.

It has been analyzed thoroughly, as thoroughly as the Monty Hall Problem.

You're better off switching. Always. In fact, the best Monty can do (for the show) is to make it 2/3 to 1/3 by showing no bias when you have picked the door with the car. If he shows any bias, (and you analyze it correctly), the probability is even more in your favor to switch.

Nope.

You failed to see that I changed the parameters, S2000. Monty doesn't know which door is correct, and neither do I.

Clearly, that's a different problem, as you say. In particular, it's not the Monty Hall Problem.

And I gave your comments more thought too, S2000. You are correct, the reveal of the other incorrect doors doesn't have any effect on the probability of my original choice being correct.

Yet when down to 2 choices where the prize hasn't been revealed by the opening of any number of doors, and I choose to stay with my original choice, that choice still has a 1 in two chance of being correct.

The choice to 'stay' with my original choice is now a 1 out of two choices, and if one door is correct, my chances of being correct, no matter whether choosing to switch or choosing the same door, are even.

so I really think the Monty Hall dilemma is just a matter of semantics, not a matter of math that everyone thinks it is.

I think I agree with you Kev... I will re-read all of this!!!

If the host knows which door hides the car (which is the case in the Monty Hall Problem), and if the host shows no bias in which door to open (when he has a choice), then you will win the car 1/3 of the time by staying, and 2/3 of the time by switching.

Seriously: it isn't a matter of semantics; it really is a matter of mathematical analysis. Under the conditions stated, you are correct that the final choice is one out of two, but you are mistaken in your implication that if it is one-out-of-two, each of the two has equal probability of hiding the car. They are not 1/2 and 1/2: they are 1/3 and 2/3

When you made your original choice, the door you chose had a 1/3 probability of hiding the car. That doesn't change: after the host shows you one door with a goat, you're no better a guesser than when you started, so there's still a 1/3 chance that your original guess was right; that leaves 2/3 for the unopened, unchosen door.

(But, you say, that door originally had only a 1/3 chance of hiding the car. Now you're telling me that it's changed to 2/3. How could that happen?)

Simply put: it's Monty's fault. He transferred the 1/3 chance from the door he opened to the remaining door.

(Absurd! How could he do that?)

He does that by conspiring to show you a door with a goat. When you choose a door with a goat, he doesn't have a choice: there's only one door he can show you, so he'll show you that door 100% of the time. When you choose the car, he has a choice: he'll show you one door sometimes, and the other door sometimes (and his best strategy is to do it randomly, 50/50).

Let's say that you originally choose door #1, and he shows you a goat behind door #3. He will do that 100% of the time when the car's behind door #2, but only 50% of the time when it's behind door #1. There's your ratio: 100% / (100% + 50%) = 2/3.

Viewed another way, when he opens door #3, that's prima facie evidence that he couldn't open door #2 (because if he could have, he might have; but, he didn't).

Any way you slice it, you're better off switching. Under the worst of circumstances, it's a 2:1 bet in your favor if you switch; it's better than 2:1 if the host displays bias.

Try this link, http://www.youtube.com/watch?v=o_djTy3G0pg&feature=related
'Testing out the Monty Hall Problem'
John

Those who believe the odds increase to 2/3 by switching are making a serious flaw in logic.

Let P(Door# = car) represent the probability that door number X contains the car.

In the beginning
P(1=car) = 1/3
P(2=car) = 1/3
and
P(3=car) 1/3

I choose door number 1.

Door number 3 is revealed to be a goat. Monty asks "do you want to change your choice?"

Here is where the logic gets flawed by so many. Because door 3 has been revealed, they believe that by changing their choice to Door 2, they get to add the probability from Door 3 to the probability of Door 2, but by keeping Door 1 choice, they are left only with the original probability of Door 1.

Hogwash. If you're going to add P(3=car), you'd have to add to both choices, D1 and D2.

Thus P(1 or 3 = car) = 2/3, and
P(2 or 3 = car) = 2/3

Both being exactly equal probability.

But Monty HAS revealed D3 does NOT = car, so this cancels out FOR BOTH CHOICES D1 and D2 (because it is added to both, not just to one).


Had Monty asked BEFORE revealing D3 = goat, whether you wanted to keep D1 or trade for BOTH D2 + D3, then I agree, switching gives you a 2/3 chance of winning. But this is because you get 2 doors instead of only 1. Once the reveal is made that D3 = goat, D3 is worthless, thrown out of the equation, and all that matters is D1 and D2.

Sorry, but the experts got it wrong. Anyways, one of the links above clearly stated Answer 2 was that both D1 and D2 had a 1 out of 2 chance of being correct. Which is what I concluded before reading everyone elses flawed logic.

Also, of the 2 doors not chosen, one ALWAYS has a goat. Since Monty does know where the car is hidden, he merely has to reveal one door with a goat.

If D2 = goat, and
D3 = goat

I agree, Monty can reveal either D2 or D3. He does this 1/3 of the time that D1 is chosen. Equal probability that he will reveal D2 or D3. 1/3 x 1/2 = 1/6

If D1 = goat,
and D2 = goat

Monty WILL NOT reveal D3 = car, he will reveal D2 = goat. He does this 1/3 of the time.

If D1 = goat and
D3 = goat
Monty reveals D3 = goat. He does this 1/3 of the time.

Thus, monty reveals D2 1/6 + 1/3 = 3/6 = half the time that D1 is chosen, and
Monty reveals D3 1/6 + 1/2 = 3/6 = half the time that D2 is chosen.

(Monty does each or rather one of the above 1/3 of the time because the participant chooses D1 only 1/3 of the time and the car is randomly put behind each of the 3 doors 1/3 of the time).

So your logic that by revealing D3 = goat was prima facae evidence that D2 was a car is incorrect. The probability that D2 = goat is exactly the same probability that D1=goat, as I just showed. "He might have, but he didn't" doesn't match the probabilities. Sorry.

Clarification: "Monty reveals D3 1/6 + 1/2 = 3/6 = half the time that D2 is chosen" should say "...half the time that D1 is chosen."

Kevin, are you talking about a situation where he knew that D3=goat, or a situation where he didn't know?

Your pick 2/3 Goat, His random pick 1/2 Goat
Your pick 1/3 Car, His random pick 1 Goat

This means that you both pick Goat 2/3 * 1/2 = 1/3 of the time, and you pick Car and he picks
Goat 1/3 * 1 of the time. So Kevin is correct; if Monty randomly selects and shows a Goat you have no reason to switch.

/Tomas

Which is to say, the information (D3 = goat) is worthless if his choice was biased (and thus the original probability (you were 2/3 to get it wrong the first time, and will win if you switch) is unchanged), but relevant if his choice was unbiased.