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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.

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closed as unclear what you're asking by Fimpellizieri, Deusovi, 2012rcampion, dmg, CodeNewbie Feb 29 '16 at 8:25

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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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.

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  • $\begingroup$ The answer, but I would like to see an algorithm proving it, not just that. $\endgroup$ – user3836103 Feb 28 '16 at 23:23
  • $\begingroup$ Could you possibly think of something else? Yesterday when I tried to explain that to someone, they didn't want to listen... $\endgroup$ – user3836103 Feb 28 '16 at 23:39
  • $\begingroup$ Added a generalization, which allows for a more mathematical formulation of the result. $\endgroup$ – Zerris Feb 28 '16 at 23:49
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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 $1 - \frac13 = \frac23$

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