# A game-show puzzle that's hard to beat

## The most famous brainteaser of the past 50 years is deceptively simple in appearance and yet diabolically counter-intuitive to solve, writes Jason England.

**The Monty Hall Problem is **without question the most famous and contentious brainteaser of the past 50 years. It’s been discussed and dissected in university classrooms, pubs and boardrooms, within the walls of the Central Intelligence Agency and even on the front page of the *New York Times*. It is deceptively simple in appearance and yet diabolically counter-intuitive to solve. If you’ve never heard of it before, you’re in for a cognitive treat – and new variations keep cropping up to extend the fun for old hands.

Here’s the way the standard problem is usually stated. You are a contestant on a game show. You are presented with three doors. Behind one of the doors is a new car. Behind the other two doors are goats. You are allowed to choose one of the doors but it isn’t opened yet. Instead, the host of the show (Monty Hall) opens one of the other doors, to reveal a goat. Two doors remain closed. Monty then gives you a choice: stick with your original door or switch to the other unopened door instead. Should you switch?

Before we go any further, there are a few things that must be clarified. The first is that the car is *always* placed randomly amongst the three doors. Second, the host *always* knows where the car is. The third is that the host will *always *open a door to reveal a goat. The last important detail is that the contestant (that’s you) knows all of this information ahead of time.

The popular reasoning goes like this: after opening a door to reveal a goat, Monty leaves you with two possibilities, so the odds must be 50/50 and therefore there is nothing to be gained by switching. You *could* switch, but there is no reason to do so. The popular reasoning is wrong.

As strange as it may seem, you should switch doors – it doubles your chances of winning! Without switching, you would only win a third of the time. By switching doors, you will win two thirds of the time.

How is that possible? Perhaps the easiest way to see that switching is the proper strategy is to work through an example. Let’s assume you choose Door 1. If the car is behind Door 2, then Monty has no choice but to reveal Door 3 (a goat). Switching from Door 1 to Door 2 wins for you. Likewise, if the car is behind Door 3 then Monty is forced to reveal Door 2. Switching from 1 to 3 wins for you. The only time you lose by switching is if you happened to choose the right door from the outset. But we know that only happens one third of the time. The other two thirds of the time switching is the correct strategy.

If you’re still not convinced, imagine there were 100 doors. Behind 99 of them are goats and behind one is the car. You choose a door. Monty then opens 98 of the doors with goats and leaves only your original door and one other. He then gives you the option of switching. What’s more likely, that you picked the right door at the very beginning or that Monty was avoiding the winning door when he was opening up all of the losers? Of course it’s the latter. In 99 times out of 100 Monty has to avoid the correct door. He’s practically pointing it out to you! In this extreme example, only one time out of 99 will a switching strategy lose for you.

The same logic applies to the standard three-door example. Statistically speaking, two thirds of the time you’ve chosen the wrong door initially and Monty is literally showing you the other wrong door. The one remaining is the correct one and you should switch to it as soon as you can.

Earlier I mentioned new variations of the Monty Hall Problem. There are dozens out there in maths and stats journals, but here we’ll consider two of the best.

**Problem 1**

The first involves a scenario where the car is *not* placed randomly at the outset. Let’s assume that Door 1 contains the car 75% of the time, Door 2 has it 15% of the time, and Door 3 has it the remaining 10% of the time. All other assumptions and procedures remain the same. What is the optimal strategy now?

**Problem 2**

The second variation uses four doors. You choose one. Monty then opens a losing door and gives you the option of switching or staying with your original door. After your decision, Monty opens another losing door and again gives you the option of switching or staying with your original door. What is the optimal strategy? Is it to never switch, to switch once as soon as you can but then to stay, to switch both times, or to switch only once at the very end?

**The answers**

In the first variation where the car is not placed randomly, the correct solution is initially to choose the door with the lowest probability of having the car, Door 3. Monty will then open one of the remaining doors – eliminating one of the car’s two more likely hiding places for you – and you would switch to the remaining door when given the chance. Your chances of winning are the sum of the chances of the two doors you did *not* choose at the beginning. In our example you would win 90% of the time using this strategy.

In the four-door example, any switching is better than none at all, but the best strategy is to wait until the final opportunity and then (and only then) switch – just like in the 100-door example, the odds are that Monty is signposting the winning door. In Monty Hall Problem literature this is known as the “Switch at the Last Minute” strategy and it works for any number of doors that follow a progressive elimination format. In our example, you would win 75% of the time using this strategy.

*Jason England is a magician based in Las Vegas and a renowned authority on casino gambling and card handling.*