Your method of listing the possible outcomes is flawed which is why you are arriving at the wrong answer.
In each group of 4 lines you are considering two scenarios where the player picked the door with the Car. Those are actually the same scenario. Whether the host chooses to show you Door 2 or Door 3 has no impact on the probability you are trying to measure.
The host knows what is behind all the doors and when the initial choice is correct, the host can open any of the remaining two doors randomly. So I think both scenarios have to be accounted for in the probability.Is it the correct approach?
The host knows what is behind all the doors and when the initial choice is correct, the host can open any of the remaining two doors randomly. So I think both scenarios have to be accounted for in the probability.Is it the correct approach?
I know the rules of the problem, and you are incorrect. The player does not know whether the host had more than one choice when opening the door. If the player chooses door 1, and the car was there, that is one scenario. Whether or not the host chooses to show door 2 or 3 has no impact on the likeliness of "Stay" vs. "Switch" succeeding.
Fine. I got the point.
Whenever I think of this problem, i get more confused. The player does not know whether he has got it right on the first attempt and hence the opening of one door does not alter the probability of staying or switching the answer. That means in the scenarios that I have posted, half of those scenarios can be eliminated but still the probability remains 50% isn't it? What am I missing here?
WOW!,this is very beautiful sanath1987!
If there are 3 doors, no matter which door you pick it has 1/3 chance of winning. This gives, the rest of the doors has *combined* 2/3 chance of winning. So: You can stay with your original 33.33% chances, or switch to 66.66% chanses of winning. The flaw in your thinking is 50%, this number does not exist in this problem at all, but our brain wants it to, because it seems logical.
G G C
G C G
C G G
G=goat, C=car. This is the only lines you need, because the car can be behind only one door, and goats in the other two. Let's say you pick door one in each possible configuration.
1. You picked goat. Stay, get goat. Switch, 50% car. (Nope. Goat removed. You get car.)
2. You picked goat. Stay, get goat. Switch, 50% car. (Nope. Goat removed. You get car.)
3. You picked car. Stay, get car. Switch, 100% goat.
If you switch, 1 and 2 gives car, 3 gives goat. Thus, twice chance when you switch, because one goat is removed each time...
If you don't switch, you get goat at 1 and 2, and car only at 3...
We all have heard about the classic Monty Hall problem. If you have not heard about it, please check out this Wikipedia link.
https://en.wikipedia.org/wiki/Monty_Hall_problem
There has been a lot of debate around this problem and based on mathematical proofs and computer simulations, it has been accepted that switching the choice gives two-thirds chance of winning the car while staying with the original choice gives only one-third chance of winning the car.
After going through all the explanations, I was convinced that switching the choice gives the best chance. But today morning I came across this problem again and when I thought about it for a moment, I felt that there is 50% probability in winning the car if we switch the choice or stay with the original choice. Switching and staying gives same probability and I tried to list of all possibilities to figure it out. Below is the list of all the possibilities.
From the above list, it is clear that the probability of winning the car is 50% regardless of whether we switch or not. So the question is, have we got it all wrong before?
There are a lot of mathematicians in this website who are very good at statistics and probability theory. What is your opinion on this Monty Hall problem?