The Monty Hall problem
Swami Gulagulaananda said:
"You showed me a goat and made me a bakra"
There was an interesting discussion at office today. I'd heard about this problem before, but didn't know that the name of the problem is "Monty Hall" problem. The problem goes like this:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? [Source: Wikipedia]
If you have not heard of this problem, take some time and think about it. See what the solution is, according to you. Then continue reading... You will encounter views and counter-views.
Spoiler Alert: If you continue reading beyond this part, you will encounter answers
Apparently the correct answer to the problem is that you should switch doors. Why? It's apparently because there is a 2/3 chance that you will win if you switch. Don't believe it? I didn't either, and frankly, I am having trouble even now, though I read the solutions and even simulated the problem by writing a program to simulate it. The answer is indeed 2/3 - You can read about all the solutions on this Wikipedia page.
My Confusion
But despite staring at the answers in the face, I am having a great deal of difficulty appreciating the solution. And this is a problem that plenty of others also seem to have faced. So it's really strange. You guessed it right, I am one of those who feels that the answer is - It doesn't matter if you switch because after eliminating the wrong door, the probability becomes 1/2 as against remaining 1/3.
My approach was simple. Any door I pick in the beginning is right with a probability of 1/3. Now, when you eliminate one wrong door, you are left with two doors. The probability of being right is 1/2, because there are only two doors left and only one of them has the car.
But according to the solutions, the door you picked has a probability of 1/3 in the beginning, while the two others collectively has 2/3. On eliminating the wrong door from the remaining two, that 2/3 is retained to the remaining door and is not shared - Now this is where I am having trouble.
My argument is that - Assume that you had a 100 doors. As you keep eliminating wrong doors, shouldn't the probability keep getting rearranged and distribute equally over the remaining unopened doors including yours? The more number of incorrect doors that you eliminate, the greater the chance of mine being correct as well as the ones that remain.
To simplify this, assume person A is asked to choose among the three doors. After the wrong door has been eliminated, if I ask person B, who has no idea about 3 doors having been present, is asked to pick a door, then, as far as he is concerned, isn't the probability 1/2 ? How does introducing a wrong door and then discarding it increase the probability of getting it?
The counter-argument that was made is that - the information of the wrong door being removed is not being considered at all when the incorrect door was discarded. I don't know, I guess I've to read a lot more of probability to get some things cleared up. Well, since my program shows 1:2 chances of winning for without switching to switching, I guess I'll admit that the answer is indeed what's given in Wikipedia.
Confusion Cleared
While I was able to understand the mathematics behind it, I wasn't feeling it right intuitively. Anyhow, a really good example given by my friend Niyaz is as follows. Assume that there is a pile of 1000 rocks among which one is a diamond. Now you are given an opportunity to pick one rock randomly in the darkness. I keep the remaining 999. Now, 999/1000 times I am going to get the diamond. Now, if I throw out 998 rocks among the 999, it is still with the same probability that I have the diamond.
Interestingly, now it seems so obvious and clear.
Confusion Cleared
While I was able to understand the mathematics behind it, I wasn't feeling it right intuitively. Anyhow, a really good example given by my friend Niyaz is as follows. Assume that there is a pile of 1000 rocks among which one is a diamond. Now you are given an opportunity to pick one rock randomly in the darkness. I keep the remaining 999. Now, 999/1000 times I am going to get the diamond. Now, if I throw out 998 rocks among the 999, it is still with the same probability that I have the diamond.
Interestingly, now it seems so obvious and clear.
Here's the program I wrote:
With Switch: 664
Without Switch: 336
[Finished in 0.1s]
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