Considering we all know about Monty Hall Problem of $2$ Goats and $1$ Car. Also, the solution of the probability $2/3$ concentrating on a single door.

Now, an alternate approach can be as follows:

Consider a person $A$ and doors $1,2,3.$ Now, also consider another person let say $B$.

$B$ comes into consideration only after Monty has shown the door behind which there is a goat.

Let's say $A$ always switches its door. $B$ selects the door which was previously selected by $A$.

Now in this scenario, as we say $A$ has probability $2/3$, whereas $B$ has probability $1/3$ of selecting door with the car.

But the problem is person $B$ comes into consideration where he has only $2$ choices one with the car and one with the goat, his probability of getting a door with the car behind should be $1/2$ and not $1/3$.

What am I missing here??

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    $\begingroup$ This is like asking "Why don't I win the lottery 50% of the time? There are only two outcomes, I win and I lose" $\endgroup$ – JMoravitz Dec 7 '17 at 6:09
  • $\begingroup$ In a lottery, you don't have only 2 tickets(you have more than 2 tickets in play and only one of them can win) so how can you win lottery 50% times. Whereas in above problem you have only 2 doors remaining. $\endgroup$ – Archies Dec 7 '17 at 6:39
  • $\begingroup$ You don't win 50% of the time, that's the point. If you don't like that example, how about this one: flip a fair coin repeatedly until it shows heads or you've flipped a total of four times. Count the number of tails flipped. There are five outcomes: 0 tails, 1 tail,...,4 tails. What is the chance of getting zero tails (i.e. what is chance of first flip heads)? It is not one in five. The point I am trying to make is that there are plenty of examples where the probability is not the ratio of the number of favorable outcomes to the total, as is the case here. $\endgroup$ – JMoravitz Dec 7 '17 at 6:47
  • $\begingroup$ I thought for sure you were introducing B entirely to make the Monty hall problem clearer to people who didn't get it and I thought you were very clever. Player B will have a 1 in 3 chance because there were three doors. Once he picked the door that was the end of it. Nothing else that happens makes any difference. I can't really understand why you or anyone would think his probability is 1/2. $\endgroup$ – fleablood Dec 7 '17 at 7:09
  • $\begingroup$ @JMoravitz Don't get my point wrong. In my problem, B has selected a door which is the remaining door of A's choice and Monty's display of door. Behind that door, there are only 2 possibilities A car or A goat (like head and tail). But there is an extra condition that if A gets Car, B gets Goat or A gets Goat, B gets Car. Only one of them is going to be true $\endgroup$ – Archies Dec 7 '17 at 7:10

Player $A$ and $B$ agree they they will pick the same door. As there are $3$ doors the probability of them picking the right door is $1$ in $3$. At that moment a pretty woman from the studio was hired to distract player $B$ and they step out to have a drink. Meanwhile Monty Hall and player $A$ do a bunch of stuff but Player $B$ doesn't pay any attention. He comes back and the game is over. He asks player $A$ "did we win?"

Nothing that happened while he was in the bar having a drink had anything to do with anything. So his odds of winning is still $1/3$. What difference would seeing a goat door do, if he was determined not to switch?

The entire point of the Monty Hall problem is that being shown a goat door does not change the probability of the door you picked. As he was going to show you a goat door anyway the probability was $1/3$ and stays one third. It just that the $2/3$ probability of being wrong is no longer represented by $2$ unknown doors with a $2/3$ probability of a car being behind one of them, but is represented by an unknown door with a $2/3$ probability of a car being behind it.

  • $\begingroup$ That was helpful. Thanks alot $\endgroup$ – Archies Dec 7 '17 at 7:27

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