Monty Hall MATLAB Code w graphs and figures Consider a game

Monty Hall

MATLAB Code w/ graphs and figures

Consider a game show in which the contestant is to choose among three doors, one of which conceals a new car, while the other two are empty. Regardless of the choice made, at lease one of the remaining doors is empty. The show hose opens one door to show lit empty. The contestant is given the opportunity to switch doors. Should the contestant switch? To experimentally determine the conestant\'s best strategy, first conduct the experiment 100 times using the strategy of the contestant not changing his/her original choice. Then, conduct the experiment 100 times with the contestant changing his/her original choice when given the opportunity by the show host. Estimate the probabilities that the contestant wins and car using both strategies. Can you provide a theoretical reason for your results?

Solution

To understand why it’s better to switch doors, let’s play out a few scenarios. Let’s see what will happen if you were to always stay with your original choice. We’ll play out three scenarios, one for each door that the car could be behind (door number 1, door number 2, or door number 3). And it doesn’t matter which door you start out with, so, to keep it simple, we’ll always start by choosing door number 1.

Stay strategy, scenario 1: the car is behind door number 1. You choose door number 1, then the host reveals a goat behind door number 2 and because you always stay, you stay with door number 1. You win the car! Stay strategy, scenario 2: the car is behind door number 2. You start by picking door number 1, the host reveals a goat behind door number 3, and you’re using the stay strategy so you stay with door number 1. You get a goat and don’t win the car. Stay strategy, scenario 3: the car is behind door number 3. You pick door number 1, the host opens door number 2 to reveal a goat, you stay with door number 1, and you get a goat. So, using the stay strategy, you won the car one out of three times. That means that in any one instance of playing the game, your chance of winning the car if you choose to stay is 1/3 or about 33%.

Now let’s try switching doors. Again, we’ll always start by picking door number 1. Switch strategy, scenario 1: the car is behind door number 1. You choose door number 1, the host opens door number 2 to reveal a goat, you are using the switch strategy so you switch to door number 3. You get a goat. Switch strategy, scenario 2: the car is behind door number 2. You start by picking door number 1, the host opens door number 3 to reveal a goat, you switch to door number 2 and win the car! Switch strategy, scenario 3: the car is behind door number 3. You pick door number 1, the host opens door number 2 to reveal a goat, you switch to door number 3 and win the car again! So, with the switch strategy you won the car 2 out of 3 times. That means, that in any one instance of the game, your chance of winning the car if you choose to switch doors is 2/3 or about 67%.

Therefore, if you play the game three times and stay, on average you’ll win the car once. But if you play the game three times and switch each time, on average you’ll win the car twice. That’s twice as many cars!