Monty Hall Problem is a subject which gave rise to a lot of arguments and had (still have as always) many oppositions . Many are there who disagree with it . The fact is that either they don’t understand it or they just don’t believe it . This is because, there are a lot of explanations , but few are clear . Here, is an explanation, which I think , is easier than the common one. The explanation is a bit elaborated, but only to make every point clear.
Coming to the problem , we have three doors or choices before us , two of them having goats inside, and only one with the prize , a car.
Lets mark each door 1 , 2 & 3 respectively. Here, each of the three items can be inside any of the doors with one and only one inside each of the doors.
Initially , we have to choose any one of the doors. Then, Monty Hall will open a door containing a goat and we are left with only two doors to choose from. The question comes next. To be the winner, should we stick to the one initially chosen or should we switch to the other one? Which has greater probability to make you a winner? Before considering these questions, we should know why the choices are not equally likely.
To explain why the choices are not equally likely, to begin with, we should consider a simple basic thing. That is, after our initial selection, Monty Hall will only be opening the door, containing a goat . And, we are the ones who therefore limit our choices just to two of the doors (Why ? What we need is the car….!). The third one is still there and if he had opened the door having the car, we would surely have chosen it (but not now as it’s a goat which is there). Since the prize is always inside the existing two doors, we avoid the third one, but, from what I said, it is clear that it can’t be ignored (i.e probability of selecting any of the rest is not 1/2 as a third one is there, clearly affecting their probabilities). Even if it’s not clear till now, I will make it clear.
Let’s come to the point. From what is said above, we have only two choices in the second stage, stick to what is chosen in the first stage(stay) or to switch to the other. Since the third one is avoided(as said earlier), these are the only possible choices. Therefore, let us represent these two cases as ‘stay’ and ‘switch’.
From all this, it is understood that for winning the price while ‘stay’ing in the second stage, we should, however, have chosen the car, in the first stage itself. We know that, probability of selecting a car in the first stage is 1/3, as there is only one car and choices, three.This clearly explains that the probability of winning while sticking to the initial choice is 1/3.
Similarly, to win the price while ‘switch’ing in the second stage, it is clear that we should have chosen a door which does not have the car inside ( only then, we are able to switch to the winning door). That is, we should have selected a door having a goat. Since we have two doors having a goat inside, we can ‘switch’ from these two and thus, the probability of winning by switching in the second stage become 2/3.
This shows that switching to the other door has a better chance of winning (2/3) than sticking to the initially chosen one(1/3). From all this, it is clearly seen that it is better to switch doors than to stick to the initial one , for winning the prize.
Alternately, what is said above can also be easily and clearly seen by listing out all the possible selections (total 9). Again, this also gives the same result. That is, chance for winning while ‘stay’ing is found to be 3/9 = 1/3 and the chance of winning while ‘switch’ing is 6/9 = 2/3. Again, this agrees with the above result.
Monty Hall problem deals not just with three doors, but we can extend it to any number of doors. Consider, instead of three, there are a hundred doors and we are to choose any. If 98 of all the doors(of course they contain a goat), are open, except one , we have to choose between two doors in the second stage . In this case, the probability of winning the prize while switching doors become 99/100 . This too can be explained as done above.
Rather than considering the selection in the second stage as present events, I just tried to explain the problem by considering the selections as pre-determined ones. I thought this could give an explanation to the problem from a different angle.
Rather than it’s mathematical application, Monty Hall problem applies to other dimensions too. It suggests that, while trying to find out a solution (of any kind) , whenever we get additional information, it is more likely to succeed if we update our decision based on it, than just sticking to the initial one.