# The Gameshow (Monty Hall problem)

Created | Updated May 22, 2003

### The Gameshow

If you're uncomfortable with Schroedinger's cat that's dead and alive at the same time, why not consider this exercise in classical probability, the nicest and simpest probability there is. (This is the one that says a die has a 1 in 6 chance of landing on any particular side)

This next thought experiment is described more fully in the Monty Hall problem. Imagine you're the contestant on a gameshow. You've made it to the final prize round, a simple game of chance. You are presented with three doors, marked A, B, and C. Behind one door is the star prize, behind the others, nothing.

All you can go on is guesswork, there is no "skill" to the guess. The game producers will not move the prize about. The gameshow host does not know where the prize is.

For argument's sake, assume you choose door B. You know the odds of choosing the correct door are 1 in 3.

Now the host shows you that there is nothing behind door A (Assume he gets a message through his ear piece). He asks you; "Do you want to change your mind?" Now you know that door A is wrong, so only doors B or C can have the prize behind them. Is it worth changing you mind? Will it make any difference. Most people will think like this:

There were three doors, each with a 1 in 3 chance of being correct. Now there are only two doors, so they each have a 1 in 2 chance of being correct. Therefore there is no incentive to change my mind. There is also no incentive to keep the choice I have made. Both remaining doors have an equal chance of being correct.

**Wrong!** Changing your mind and selecting door C **doubles** your chance of being correct. Here's why:

At the start, each door has a 1 in 3 chance. The host shows you door A is wrong. **But** it is still a valid choice. Yes, it's a dumb choice, but there's nothing to stop you choosing it. So door A has a **0** in 3 chance, door B (your choice) still has a 1 in 3 choice. Now classical probability must be balanced, all choices adding up to 1. So the remaining door, C, must have a 2 in 3 chance of being correct. Read that bit again. I'll wait.

If you think that's stupid, talk to Maryln vos Savant, the smartest person alive with an IQ in excess of 200. She's the one who came up with this. Experiments in labs and mathematical models all show that this is correct.