I found the easiest way to envision how this works is to extend this to a point where it is more intuitively obvious what is going on.
Imagine that instead of 3 door there are 100. Still only one prize though.
You pick one at random.
You have a 1 in 100 chance of having picked the right door.
Monty, knowing which door is the right one reveal 98 door that he knows did not contain the prize.
That means, If you picked a dud (99 out of 100 chance) he reveals all the door except the one that does contain the prize.
If you were lucky originally and picked the correct one (1 out of 100) he will reveal 98 of the 99 remaining duds leaving one dud.
So you are left with two doors one that you originally picked and has 1 in 100 chance of containing the prize and the one Monty left that has a 99 in 100 chance of containing a prize.
You would obviously want to switch at that point.
The basic thing to remember is that however many doors there are, Monty knows which door has the prize. If you didn’t pick the right right away he will show you which one does. If you did pick the correct one right away Monty shows you a dud.
So the single door Monty leaves has the prize if you guess wrong initially.
The chance that Monty’s door is the correct one is the same as the chance that you were incorrect in your initial guess.
With three door you have a 1 in 3 chance of being right with your first guess and the door Monty leaves has 2 in 3 chance that you were wrong initially and it now contains the door.
The more doors there are the more obvious it gets.
If you pick a lottery ticket you have a very low chance of guess right, but if someone were to offer you to switch tickets with you with one that represented all the numbers you didn’t pick and assured you that either the ticket they were holding or the one you were holding was a winner, that would be good deal.
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