Monty Hall Problem

The Monty Hall problem is a great brain teaser stemming from Hall’s game show “Let’s Make a Deal”. Here’s the premise:

  • There are 3 doors; behind each door is either one of two goats or a car.
  • You pick a door (call it door A).
  • The host then opens one of doors B or C which reveals a goat. Let’s say he opens door B.

Hall then gives you a choice of changing your pick of doors. So now you have to choose between door A (your original choice) or door C. The immediate response is probably one of trying to minimize regret. Naturally, we would feel worse if we change our answer and the original choice turns out to be the correct choice (assuming you want the car). Probably around the same time, your mind calculates that the odds are 50/50 so since it’s a ‘coin toss’ you decide to stick to your original choice, door A.

So as it turns out, good old Monty reveals that door A indeed reveals a brand new car! Winner, winner, chicken dinner!

Congrats on beating the odds! Huh? But did you make the correct choice based on probability? Forget about the outcome of you winning and focus on the process here. Or in this case your thought process.

Mathematically your odds of winning the car would increase to 66.7% (assume randomization of the cars and goats each time) if you decided to switch your original answer.

Why?

This seems totally counter-intuitive and there have been many attempts at explaining why it is so. I recommend you look at a few video explanations such as the following:

The easiest way for me to think about it is to grow the game. Increase the number of doors to 100 doors, 99 goats and 1 car. If you pick door number 1 and Hall reveals 98 doors of goats, do you make the switch? Visualize in your mind the one door you have chosen and colour it red; colour the other 99 doors as green and note the sheer difference in size between the area covered in red doors vs green doors. You are essentially picking one door (door A) vs 99 other doors (it just so happens Hall revealed the other 98). It becomes clearer that switching would be advantageous.

The critical thing here is to recognize that the original three door problem does not become a 50/50 proposition because we know something about the choices. If you are originally given only two doors to choose from, then it is indeed a 50/50 prop.

What’s the takeaway here?

  • Reevaluate your probabilities when you are given new information and if required change your probability calculations.
  • Test it – If you still think it is a 50-50 prop, then test/verify your assumptions or hypothesis using a simulation run over and over (Monte Carlo)
  • Focus on process rather than outcome (remember in the original example you won the car….it ended up being the right choice that time but not necessarily the smart choice).

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