Hi All,

I found this interesting puzzle called the Monty Hall Puzzle.

You are the contestant on a Game Show. There are 3 doors – say A,B and C – and behind one door is the prize of a million dollars. The prize is randomly placed and the probability of finding the prize behind any door is 1/3.

You get to pick one door and if you’ve picked the right door, you win the prize.

There’s a catch. After you’ve chosen a door, the game show host will not immediately open that particular door. The host will open another door that you did not pick and which he knows does not contain the prize. Suppose you had picked door A, and the prize is behind door B, the host will open door C and let you know that there is nothing being door C.

At this point, you will be given a choice: Do you want to stick with your original choice (door A in the example above) or do you want to choose the other unopened door (door B in this example)?

Would you switch your choice?

Do you think that switching doors would have any impact on the probability of finding the prize?

I thought that it would not make any difference (probability would remain at 50-50 for the two open doors)

Surprisingly, it turns out that you SHOULD switch. You can read the proof of this here: https://codeground.in/blog/index.php/20 ... ll-puzzle/

Explaining the Monty Hall Puzzle

Let’s the divide the doors into two sets

Set 1: {The door you picked originally}

The Probability that the prize is in Set 1 is 1/3

Set 2: {The other two doors}

The Probability that the prize is in Set 2 is 2/3

The Host will open a door from Set 2 that does NOT have the prize, but the probability that the prize is in Set 2 still remains at 2/3.

Switching your choice to choose the remaining unopened door from Set 2 maximizes your odds of winning.

Was this useful? If you’re interested, you can read more about more such puzzles for Tech Interviews (https://codeground.in/blog/index.php/ca ... s-puzzles/) or you can take do some online programming challenges (https://codeground.in/screening-tests/c ... tests.html)