# I Monty Hall - Multiple solutions via direct calculation?

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1. Mar 8, 2016

https://en.wikipedia.org/wiki/Monty_Hall_problem#Direct_calculation

I understand the problem and why it is better to always switch. Now, I want to prove it by myself via a direct calculation. Before I start I wonder if the direct calculation on Wikipedia is the only solution or are there multiple ways of getting there. This question goes for other probability problems too.

I'm thinking there are more calculations possible depending on the question you ask?

2. Mar 8, 2016

### .Scott

There are always more than one way to work the arithmetic. You can also change the question: Is there a way that Monty can choose which door to open that would make it harder to win? ... or easier to win?

3. Mar 10, 2016

### MrAnchovy

There is always another way, and the one on Wikipedia is frankly ridiculous. This is all you need:

• P(choose right first time) = $\frac 13$
• P(prize is behind your door) + P(prize is behind the other unopened door) = 1
• P(prize is behind your door) = P(choose right first time)
• P(prize is behind the other unopened door) = $1 - \frac 13 = \frac 23$

4. Mar 10, 2016

Argh, spoiler. Thanks though, but it isn't a formal proof incorporating switching doors or not.

5. Mar 10, 2016

### MrAnchovy

Oh sorry if that is along the lines you were thinking, I was assuming that you had been misled by the Wikipedia article into performing unnecessary calculations with decision trees and conditional probabilities and God knows what.

6. Mar 10, 2016

### FactChecker

The Wikipedia article is applying Bayes' Rule. That is the standard fundamental method to adjust probabilities when new information is obtained. It breaks down the probabilities into individual parts that can be determined (relatively) routinely and then added up. Other approaches are either disguising Bayes' Rule or are using special logic without explaining. The use of special logic allows alternative answers that cause people to argue endlessly.

7. Mar 10, 2016

### MrAnchovy

There is no "special logic" in the approach in my post #3, although I accept it could be stated more rigorously.

8. Mar 10, 2016

### FactChecker

In your second line, all the probabilities are conditional on there being no prize behind the open door. So their values have changed from the original probabilities. In this case, the probabilities are easy to adjust, but they are not the same as those same probabilities were before the door was opened. So you are mixing probabilities before and after the door was opened without a notation change. That is just using Bayes' Rule without the clear conditional notation.