### Help! I Can’t Handle GMAT Probability and Combinatorics (Part 1)

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There’s a classic brain teaser called the Monty Hall problem. It’s named after the host of an old-timey TV game show, who used it to confound contestants. He’d present each contestant with three closed doors. Behind one door was a new car, and behind the other two doors were goats.

Monty invited the player to pick one of the three doors. Whichever door the player chose, Monty would then open a different one, revealing a goat, not the car. Then, he would offer the player a choice. If the player wanted, he could switch doors, picking the other unopened door. Or, he could stick with the door he picked in the first place. Whichever decision he made, he would win the prize behind the door he chose.

The question is, does switching doors give the contestant a higher probability of picking the car? It should seem obvious that it doesn’t. The car is equally likely to be behind any of the doors. So, it seems like once one of the three doors is open, you have 50/50 odds of picking the car, regardless of whether you switch or stay.

However, that isn’t true. Believe it or not, switching actually increases the probability of picking the car to 2/3, while staying on the same door means that you only have a 1 in 3 chance of picking it. If you didn’t figure that out on your own, you’re in the company of many very smart people, including a number of famous mathematicians.

From a GMAT perspective, here’s what to take away from this little anecdote. GMAT probability and combinatorics are counterintuitive for everyone. I majored in math in college – I even took a couple of courses on probability – and I still can’t solve these problems by intuiting which formulas to use. It seems crazy to me that switching doors after the fact would make you more likely to pick the car. So, what do I do – and what should you do? I think about almost all GMAT probability and combinatorics problems in terms of counting out possibilities.

Let’s give it a try with the Monty Hall problem. Imagine that you play three games with Monty. Additionally, imagine we’re in probability land, where everything happens exactly according to its probability. Since there are three possible scenarios, and they’re all equally likely, here’s what your three games will look like.

Now, Monty opens one of the two doors you didn’t pick. Remember that Monty always opens a door with a goat behind it. So, in two out of the three games, Monty is stuck: there’s only one door he can choose.

In both of these games, you started with a goat. If you switch, you’ll win!

In game 3, Monty could pick either door, since they both have goats. Let’s say he flips a coin to decide which one to pick. If he gets heads, he’ll open the first mystery door. If he gets tails, he’ll open the second one. This scenario only comes up when you picked the car in the first place – so, Monty only needs to worry about it 1/3 of the time.

Interestingly, it doesn’t matter what Monty flips. No matter what, if Monty flips a coin at all, and you switch, you’ll lose. If you always switch, you’ll win 2 out of your 3 games, and lose 1 of them. That means that switching gives you a 2/3 probability of winning.