This sounds simple. What is the easiest solution to the Monty Hall problem you have which does not use listing out every probability and also requires you use algorithms? Also, no research or computers.
closed as unclear what you're asking by Fimpellizieri, Deusovi♦, 2012rcampion, dmg, CodeNewbie Feb 29 '16 at 8:25
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When you switch, if you had the car selected before, you now get a goat. If you had a goat selected, the other is revealed, so you now get the car. Thus, the probabilities after switching are exactly equal to the reverse of your original probabilities.
You originally had a $\frac13$ chance of getting the car, so switching gives you a $\frac23$ chance. Not switching leaves your original odds unchanged, at $\frac13$.
The generalization of this would be for a Monty Hall problem with $x$ doors, of which one is the "winning" door. After your initial guess, all but one other door is revealed. In this case, your initial guess has $\frac1x$ odds of being right, while switching has $1-\frac1x$ odds. As a result, switching improves your odds by $1-\frac2x$. Note that, for all $x > 2$, this is a positive change.
You have a:
$\frac13$ chance of picking the correct door, Monty'll show you one of the other doors, you switch and lose.
$\frac23$ chance of picking a wrong door, Monty'll show you the other wrong door, you switch and win.
EDIT: Now I have better understanding of what you've after.
You have a robot:
Preprogrammed to always swap.
First move: Get rid of a door at random. Say that as pick to Monty.
Second move: Get rid of door Monty shows.
Last move: Pick remaining door.
Now Monty will never show a winning door.
So the second move never gets rid of a win.
So the only way to lose is if the robot randomly picks the winning door in the first move with probably $\frac13$.
And therefore the probability of a win is $1 - \frac13 = \frac23$