Problem :
You are given a choice of three doors by an Angel. You can choose only one of the doors among the three. Out of these three doors two contains nothing and one has a jackpot.
After you choose one of the doors angel reveals one of the other two doors behind which there is a nothing. Angel gives you an opportunity to change the door or you can stick with your chosen door.
You don’t know behind which door we have nothing. Should you switch or it doesn’t matter?
Soltuion:
You choose one of the door. So probability of getting the jackpot – 1/3.
Let’s say that the jackpot is in Door no 1 and you choose Door no 1. So the angel will either open door no 2 or door no 3. Let’s look at the sample space of this Puzzle.
Case -> Door1 Door2 Door3
Case 1 : Jackpot Nothing Nothing
Case 2 : Nothing Jackpot Nothing
Case 3 : Nothing Nothing Jackpot
Want to keep your guess:
Let’s suppose that you guessed correctly. Then it makes no difference what the game show host does, the other door is always the wrong door. So in that case, by keeping your choice, the probability that you win is 1/3 x 1 = 1/3.
But let’s suppose you guessed incorrectly. In that case, the remaining door is guaranteed to be the correct door. Thus, by keeping your choice, the probability of winning is 2/3 x 0 = 0.
Your total chances of winning by keeping your guess is: 1/3 + 0 = 1/3.
Want to change your guess:
Again, let’s suppose that you guessed correctly. By changing your guess the probability that you win is 1/3 x 0 = 0.
But let’s suppose you guessed incorrectly. Again, this means that the remaining door must be the correct one. Therefore by changing your choice, the probability of winning is 2/3 x 1 = 2/3.
Your total chances of winning by changing your guess is: 2/3 + 0 = 2/3.
Hence it is advisable to switch.
This in an old question whose answer has been tricked on through generations, even the most educated ones take this explanation to be true. But this includes a flaw. Let me try to show you another view point before I cite the flaw:
lets say the prize is actually behind door 2. If we choose door 1, they open 3. If we choose 3 they open 1. Had we chosen door 2 they still had 2 options to open- door 1 and door 3.
Our guess in the first two scenarios is incorrect whereas in the later two it is correct. Whether we change it or not, the probability remains 2/4, i.e. 1/2; it is neither 1/3 nor 2/3.
The problem is that they don’t consider the fact that hosts open one wrong door always. So there remains only 2 doors. One has the prize and the other doesn’t, as simple as that.
Shown one door containing nothing, there are 2 doors with equal probability of having the prize. It’s just trick to keep you thinking.
Remember the dummy coins we added while solving Probability exercises for IIT-JEE. The extra door here is the dummy coin.
You are ignoring the additional information you get when one of the wrong doors is eliminated. Imagine the same problem scaled up to 10 doors. Your initial pick has a 1/10 chance of being correct. Then 8 of the remaining wrong doors are eliminated, leaving you with your initial choice or the option to switch to the final door. The chances of you having picked the right door on the first guess do not change, so 1/10, and with only 1 choice remaining, the probability of it having the prize must now be 9/10. Same logic applies in the 3-door case.
Your choice in the first attempt has no relevance in the second pick irrespective of how many you scale it to. Once the angel opens one door, your odds are 1 in what ever is remaining whether you stay or change. In your example it is always 1/9.
nice comment
Exactly there is no such answer
I think there is a misconception here,
on just going by a simple analysis like taking these 3 cases
Case -> Door1 Door2 Door3
Case 1 : Jackpot Nothing Nothing
Case 2 : Nothing Jackpot Nothing
Case 3 : Nothing Nothing Jackpot
there are 3 possible doors I can pick up in my first attempt making it a total of 9 possible situations, and me trying to change the door will increase another 2*3 = 6 choices in each case.
So, if I don’t opt to change the door and stick to the same door my probability of winning would be 3/9 which is 1/3.
But if I opt to change the door I increasing my likely hood of getting the correct door but also increasing the number of attempts to get it right proportionally. What I meant while saying this is, if I opt door 2 in case 1 I will be getting a chance to change which might increase my winning chances but also increase the number of attempts to choicing door 3 or 1.
This makes my chances of winning a jackpot if I change the door will increase to 3/9 for each case making it 9/27 possible cases of winning a jackpot which is again 1/3 so its the same probability of winning a jackpot in either of the cases.
So there is no point in opting for a change.