3 ms·

But 1 in 3 times the host opens a door with a car. That changes the probabilities. Or are you assuming this never happens? Why?

• seriuslyguys 12y ago
  Let's suppose we are doing a computer simulation. 1. We simulate a million trials where the contestant chooses one of three door, one of which has a car and two a goat 2. We simulate the host randomly choosing one of the remaining doors to open. 3. We discard all trials that resulted in a car. We are left with the number of trials that resulted in a goat and store that number in a variable 'total' 4. In the remaining trails we switch and reveal what was behind the door. We store the number of times we saw a car in a variable called 'wins'. 5. The probability of winning after switching is 'wins' / 'total' It doesn't matter how many times the host shows a car because those trials are discarded.
  • dllthomas 12y ago
    Did you actually write that simulation? Because once up on a time (many years ago) I followed the same line of reasoning, wrote a simulation to show that I was right, and that simulation showed me I was wrong. If the host picks randomly and you discard rounds with a car, your odds switching are 50/50. If the host uses knowledge of where the car is to definitely reveal a goat, your odds switching are better.
  • seriuslyguys 12y ago
    @dllthomas I've written the simulation, and you're right. It is 50% when the host chooses randomly!