I came across this puzzle yesterday which was quite puzzling. It is termed the famous Monty Hall problem. It goes like this.
There are 3 doors one of which contains a “car” and the other two have “goat”. The contestant has the option to choose a door. The host, before opening the opted door, opens another door which always has a goat. Remember, the host knows what lies behind all the 3 doors. Now, the host asks the contestant if he will be willing to switch his earlier decision about the opted door. The question is this. What is the probability that the contestant will win by switching his earlier decision versus going with his original decision?
I will give you the solution. But, the more interesting part lies in understanding the solution. It looks like, by switching the earlier decision, the contestant has a 2/3 probability of winning when compared to going with his original decision which only has a 1/3 probability of winning…. Our intuition might suggest that, after the host revealed 1 door, we have 2 doors and the probability of winning, irrespective of switching should actually be 1/2. But, that is not true… Something to think about….or not….;)