I just got done with a many page thread in an actuary forum debating the common explanations for the Monty Hall problem. In any case here is the conclusion I came to: Bayes' Theorem is dependent on there not being any correlation between the frequency with which given information is given, and the outcome. Whether or not you are aware of such a correlation. An example is a slight variation of the Monty Hall problem. Suppose you didn't know that Monty always shows a goat in response to your picking the car or either of the two goats. Here you would use Bayes' Theorem to calculate the probability that you chose a goat given that you did not choose the one he showed you. You would get a 1/2 chance to get the car by switching doors. But empyrically the chance to get the car by switching is 2/3. The formula: Prob(A|B) = (Prob(B|A)*Prob(A)*Prob(B was given|A))/(Prob(B)) / ((Prob(B|A)*Prob(A)*Prob(B was given|A))/(Prob(B)) + (Prob(B|A)*Prob(A)*Prob(B was given|A compliment))/(Prob(B)) Seems to work for adjusting when there is such a correlation.
Interesting formula ... but what is Prob(B was given...)? How is that different from Prob(B...)? The chance is only 2/3 empirically when Monty does always show a goat. If he only showed a goat when you chose the car, there'd be separate cases with very different probabilities.
Prob(B was given) means the probability that the information B was given to you. Like in the monty case, if you choose the car there is a 50% chance you are given Not goat 2, whereas if you choose the goat there is a 100% chance you are given not goat 2. Therefore there is a 66% chance you were given not goat 2 because you chose goat 1. With this formula I should be able to figure out any Monty Hall game variation. Do you know of any that have been tested already that I can try it out on?
Well, if A is the event of choosing the car and B the event of being shown the goat, the standard Bayes formula would give P(A|B) = P(B|A)*P(A) / P(B) = P(B|A)*P(A) / [P(B|A)*P(A) + P(B|A')*P(A')] Where A' is A-complement. P(A) = 1/3 and in the standard Monty Hall where a goat is always shown P(B) = P(B|A) = P(B|A') = 1, so P(A|B) = P(A) = 1/3 Since this is less than a half you would always swap when shown a goat. If you change the P(B|A) and P(B|A') to reflect how likely the host is to show the goat, depending on which door you've chosen, you'll get a different value. How is this different from your formula? I can't think of any situation where it would be wrong...
Do you know that the Monty Hall problem has been interpreted in the completely wrong way? There is a very simple mistake in a step. When I first came across the Monty Hall problem, I was astounded and also suspicious, and it took only a few days for me to see that it was flawed. I'm amazed that no one has noticed this so far. If you follow your common sense alone, you can see that a result of 2/3 is very stupid. But, you can analyze the problem carefully and see that the explanation is wrong. This topic appeared on sciforums a few months back, but I didn't get enough time to say this then. I still haven't got a lot of time on my hands, but if you post a detailed explanation of how the probability comes out to be 2/3, I'll point out where the problem is.
I just did that. Based on the assumption that a goat is always shown, event B is independent of event A, so the probability of your having chosen the car isn't changed by being shown the goat, i.e. 1/3. Since there's only one other option it must now have 2/3 probability. You'll only get a different answer if you change the initial assumptions; e.g. if a goat is only shown when you pick the car, being shown the goat gives a probability of 1 that you picked the car and you'd be crazy to switch. You just have to modify P(B|A) and P(B|A') - the probability of being shown the goat when the host knows that you picked the car, and when he knows you didn't, respectively.
Eh... huh? Your math indicates you mean (P(B|A)=1) the probability of being shown either goat given you chose the car, but you say "the" goat in writing. If your talking about one goat P(B|A)= .5 right? What exactly is event B? Anyways if you are just given that you did not choose a goat, then the math comes out to 1/2 chance of having chose the car. A: Chose Car B: Didn't choose goat 2 P(A|B) = 1*(1/3)/ (2/3) If you use that you are given that the goat was revealed according to the pre known rules of the game, you could probably work out the problem in a traditional method. Hopefully the way of doing that would end up looking the same as the formula I used. The point of the formula is more to demonstrate something. You see if you did not KNOW that way in which monty chose to reveal the goat he does, you would arrive at the above calculated 1/2 chance to have chosen the goat or car, and therefore the same chance to get the car by switching. Yet that would prove to be wrong through experiment. The explanation would be the conditional probability was shifted by the factor Prob(B was given|A) / (Prob(B was given | A) + Prob(B was given| A compliment)) If you want to call event B: (Host shows Goat 1 or Goat 2) thats fine, but thats not what I am talking about. When you play the game you are given (Host shows goat 1) or (Host shows Goat 2) which are not independent of the chance that you chose the car.
Yep, sorry, that's what I meant. I assumed the goats were indistinguishable and lumped them under the same event. If you choose a goat you get shown the other one; if you choose the car you get shown one of them randomly. Surely that depends on your assumptions? If you don't know, you still have to make some kind of guess at the basic rules. Unless you have data from previous quizes there's always the chance it might be wrong. I assume you're talking about P(B|A) and P(B|A') here? Why not? If you choose the car there's half a chance you'll be shown goat 1. If you didn't choose the car, you'll be shown the 'other goat', and again there's a 0.5 probability that this is goat1. Assuming you're equally likely to choose either goat, that is. Without that symmetry things get more difficult... Er, so I guess I should ask ... are you assuming that symmetry? Please Register or Log in to view the hidden image!
I'd be very interested in knowing what you think is the problem. If Monty always shows you a goat and allows you to switch, then I really don't see how the probability could be anything other than 2/3, thus favoring switching. Here's the argument (from Wikipedia): There are three possible scenarios, each with equal probability (1/3): * The player picks goat number 1. The game host picks the other goat. Switching will win the car. * The player picks goat number 2. The game host picks the other goat. Switching will win the car. * The player picks the car. The game host picks either of the two goats. Switching will lose. Two scenarios in three, each of those three having equal probability of occuring, wins on switching.
Well. First off, I typically look at the goats as seperate but if you don't you come to the same conclusion. It's just you arrive there at what looks like totally different ways. In one case you would have: A: Car not chosen, (Goat 1 OR Goat 2 chosen) B: Car chosen C: Monty Reveals (Goat 1 OR Goat 2) Prob(C) = 1 Prob(B) = 1/3 Prob(A) = 2/3 When Monty reveals (Goat 1 OR Goat 2), which you knew he would do, it doesn't change tell you anything new. The way I always look at it is: A: Goat 1 chosen B: Goat 2 chosen C: Car Chosen D: Monty reveals Goat 1 Prob(A) = Prob(B) = Prob(Car) = 1/3 Prob(C|D) = ((1/2)*(1/3))/(1/2) = 1/3 It seems to me like the first method isn't using all the information. Maybe the fact that all you care about is weather or not you chose the car makes it ok, but I don't really understand all the mechanics of what would make it ok in that case. In the math sense its independent either way, but in the second case its like the information goes both ways: Yeah the probability remains 1/3 but thats just a coincidence because at the same time you are told you didn't choose the goat you are given that it was less likely to choose the car. So if I change it such that there is a 2/3 chance of monty revealing goat 1 when you choose the car.... Prob(C|D) becomes ((2/3)*(1/3))/(5/9) = 2/5 And its not independent any more. The previous method of not caring about anything other than weather you chose the car gives you the incorrect result that you have a 1/3 chance to have chosen the car. So if it its ok to do it that way at all, its not just because you don't care about anything thats not the car its also because the goats are identical in every way including how often he shows them. So yeah its all dependent on the symmetry..
That is exactly where the problem is. I still can't belive you missed it! That <I>one</I> point is actually <I>two</I> separate points. The game host picks <I>either</I> of the two goats. There should actually be <I>two</I> options then, so that the three options are: * The player picks goat number 1. The game host picks the other goat. Switching will win the car. * The player picks goat number 2. The game host picks the other goat. Switching will win the car. * The player picks the car. The game host picks <I>goat number 1</I>. Switching will lose. * The player picks the car. The game host picks <I>goat number 2</I>. Switching will lose. So now the probablities are equal. Or alternately, you could have shown the choices as before, but assigned 50% of the probability to the third choice, since it could happen in two ways (Pick either goat = Pick goat 1 OR Pick goat 2). That too would lead to a 50-50 scenario. Why the heck didn't anyone notice this before???
What are the combined probability for the red cases? 1/2 or 1/3? Do you think that splitting the case into two can increase the likelyhood that the player picked the car? That presumes that the cases in red have the same probability as the ones in black, and they don't. What are the probabilities of the cases in black? Well, the cases are split according to the player's choice of door. And there's 2/3 chance that the player will choose a door with a goat. That leaves only 1/3 chance for the last case, regardless of how many cases you split it into. But that would be wrong: You can't simply assign 50% probability to the scenarios with the player choosing the door with the car. He'll only pick the car in 1/3 of the cases regardless of what Monty does next. Even if Monty opens all the doors. Many people did, I wager. It's still wrong. http://en.wikipedia.org/wiki/Image:Monty3.jpg
Why not? I'm not assuming any probabilities at the start. I'm listing all the possible events, and, since there is no preference for any of them, they <I>should</I> have equal probabilites. Forget the rest of my argument. It' the same thing presented in a different manner.
No, they shouldn't. Let's toss a coin. If it lands heads up, I might do a little jig, or I might not. If it lands tails up, I'll do nothing. You now have three scenarios: * Coin lands tail up. I do nothing. * Coin lands heads up. I do nothing. * Coin lands heads up. I do a jig. Do you think these are equally likely?
Because all of the events you have listed are more like 2 events occuring one right after another. In order for Monty to be able to choose between revealing goat 1 and goat 2, you have to choose the car first. The first thing that happens is the person chooses between the 3 doors and you have a 1/3 chance of each door. If you choose the car door, THEN you have a 1/2 chance that monty shows each goat. So thats a 1/3*1/2 = 1/6 chance of the last two events.
Funkstar, Kriminal99, I've posted a new thread to continue this discussion, and hopefully there will be more participants there (maybe even someone to support me!). You probably have seen that one already.
