Probability Challenge - Can You Solve It?
 

Snagged from an engineering site I use from time to time...

'On a game show, a contestant is offered the choice of three doors. A goat is behind two of them, a new car behind the third. The contestant picks one door, but the host opens a different door showing a goat. With two doors left, the host again asks, "Which door?"

To maximize the probability to win the car, should the contestant stay with the first guess, or switch to the other door? Explain the reason behind your answer.'

Always stick with the first.... at least, thats the way Price Is Right is done...

Switch doors! The original still has the 1/3 probability. So if you switch, you have a 2/3 probability.

I'd stick with the first choice. There's a 50/50 chance it has the car, but the offer of a second choice seems intended to give him the opportunity to pick the other door and not get the car.

The odds initially were 1/3.

With one choice revealed, whether or not he had the option to choose a new door, they have changed to 1/2. Changing his choice does not change the probability.

You sure? This is a old one, the answer is not obvious....but I think you are wrong. I first read it in Carolyn Vosavant (sp?) about 20 years ago.



Let me explain what happened on Let's Make a Deal. The contestant would choose one of three curtains. One would contain a very valuable prize and the other two smaller prizes. For the sake of argument let's say behind one curtain was a car and behind the other two a goat. Then Monty would always, I repeat ALWAYS, open up one of the two unchosen curtains to reveal a goat. After hundreds of shows this would imply that Monty Hall (the host) knew where the car was and deliberately opened a curtain that revealed a goat. Obviously when the player chose his curtain the probability it held the car was 1/3 and the probability one of the two unchosen curtains held the car was 2/3. Monty is then predestined to open an unchosen curtain containing a goat. Predestined is the key word here. Because Monty can not open the player's curtain at this stage the probability of the player's curtain reveals the car stays at 1/3. The probability an unchosen curtain reveales the car remains at 2/3, however it is now all on one curtain. So after a goat is revelead the probability the player's curtain has the car is 1/3 and the probability the other unopened curtain has the car is 2/3, making switching a wise choise.

well, if it is not a Porsche, i guess i would be satisfied with the other GTO.SmileWavy

That answer makes sense, but only because we are told that the host knows which curtain has the answer.

But it's so simple.

All I have to do is divine from what I know of you: are you the sort of man who would put the new car behind his own door or his contestants? Now, a clever man would put the new car into his own door, because he would know that only a great fool would reach for what he was given. I am not a great fool, so I can clearly not choose the door in front of you. But you must have known I was not a great fool, you would have counted on it, so I can clearly not choose the door in front of me.

:cool:

i hope this thread is never debleated.

lmao! great adaptation :D

I work with a guy that I suspect would rather win the goat, but that's another matter...

I don't think the probability changes through the whole scenario as it is described.
When the guy makes his first choice there is a 100% chance that one of the other doors will reveal a goat. This leaves him with the same probability when one goat is revealed as when the initial selection was made, hence changing or not changing his choice at this point will not change his odds.
As for the host knowing which door hides what it is irrelevant in this instance as we know the outcome. The only difference it would make in the game is if the host didn't know and opened a door to reveal a car, but we know that is not the case.

I disagree. The second choice is a second, independent trial with one less choice.

I'm with chris, I see a 50/50 on the second choice.

Then again, I have never won at cards, chess or checkers.

Because Monty can not open the player's curtain at this stage the probability of the player's curtain reveals the car stays at 1/3. The probability an unchosen curtain reveales the car remains at 2/3, however it is now all on one curtain. So after a goat is revelead the probability the player's curtain has the car is 1/3 and the probability the other unopened curtain has the car is 2/3, making switching a wise choise.

Don't agree with this logic. Since a curtain was revealed that does not have a car, the two remaining curtains have a 1/2 chance of having the car. My suggestion is being offered another chance makes sense if the original choice was good, assuming the host would rather not give away the car. That's a big assumption.

http://en.wikipedia.org/wiki/Monty_Hall_problem

I stand by my answer.

P.S. Now we know the answer, and we all qualify for 'The World's Biggest Loser' thread.

From the outset the probability of picking the car is 50%. The reason for this is the predetermined nature of the problem.

You see it is a FACT that one door will be opened to reveal a goat and you know that from the outset so situation only has two possible outcomes.

Outcome 1 - The contestants picks a goat
Outcome 2 - The contestant picks a car

Despite there being two goats and one car at the start there is still a 50% chance that the contestant will pick a car.

PREDETERMINED is the key.

One way of looking at it is when you initially pick one of the three doors you have a one in three chance of having the car. But the remaining two doors (lets call them Monty's doors) have a two in three chance that one of them contains the car. Obviously Monty will remove one option by revealing that goat but now you have one door from a group that had one in thee odds and Monty has one door from a group that had two in three odds. It's obvious that Monty's dorr is the better choice.

Maybe I'm a little slow today though but why does it matter if Monty knows which door has the car? If he forgets and gets lucky by still revealing a goat, are you not still being offered the only remaining door from the two in three group for one from the one in three group?

True, except that one option is already at a disadvantage of having the "prize" (the initial one in three chance it was the prize). Where aas the second option has the advantage of being the sole survivor from the group having two in three initial odds of having the prize.

The error is assuming the two doors are equal at the time the option to switch is given.

Just because the probability of the first choice being a winner has changed due to removing one of the losing doors doesn't matter in the least. It's merely an interesting observation.

If you flipped a coin nine times and it comes up heads every time, what are the odds that it will come up heads on the tenth? The answer is 50/50. It doesn't matter a lick what happened beforehand.