Probability

This repository is intended to prove a simple math probability problem.
Probably everyone heard about a problem of winning in the Monty Hall problem by switching doors.
But it seems wrong at first, so I decided to write a simple C code to prove it.

Problem

You are given 3 doors, 1 with a car and 2 with goats.
You randomly choose a door, then the host, who knows where the car is, opens one of the other doors with a goat.
And now you are given a choice to switch to the other unopened door or stick to your original choice.
Which strategy is better?
Intuition says to stick, but math says otherwise.

Proper Solution

Initially, the probability of choosing the correct door is 1/3, and incorrect is 2/3.
But after you are revealed a door with a goat, the chances inverse and become 2/3.
The following table will explain best.

Behind door #1	Behind door #2	Behind door #3	Result if staying at door #1	Result if switching to door offered
Goat	Goat	Car	Wins goat	Wins car
Goat	Car	Goat	Wins goat	Wins car
Car	Goat	Goat	Wins car	Wins goat

Also if this table doesnt tell you enough, you can consider the following:

• You are offered a 1,000,000 doors, behind one of them is a car and the rest are goats.
• You choose a door randomly.
• The host opens 999,998 doors with goats.
• Now you are offered to switch to the remaining door or stick to your original choice.
• Obvious answer is to switch, no way you guessed right with a chance of 1/1,000,000.

And even if 2 pieces of evidence from above dont convince you, just consider this:

• Initially, chances are of you choosing a car is 1/3, and in the rest 2/3, they are - well - 2/3.
• After host reveals a goat in that 2/3 of a right guess section, the section stays 2/3, but now with only 1 door.

Build yourself

• Via gcc:

gcc main.c -o prob
./prob

• Via clang:

clang main.c -o prob
./prob

• Via Visual Studio:

1. Open this file.
2. Somehow execute it via some green button.

Credits

• Wikipedia - Monty Hall problem