# Monty Hall Puzzle

## Puzzle on Monty Hall

The premise of Monty Hall Puzzle:-

You are playing a game where the host gives you 3 boxes. One of the boxes contains an expensive diamond whereas, the other 2 are empty. You are available with the option to pick any one of the three boxes. You picked one box of the three. Now among the two boxes, one of them is sure to be empty. The host opens one of the remaining boxes and reveals it to be empty. Now he gives you a choice, where you can either open the box you had picked initially or pick the unopened box on the table. Whichever box you choose, you can open that box.

Question:-

• Which box will you pick? Will you stay with your initial choice or switch? Why?

## Puzzle on Monty Hall

The premise of Monty Hall Puzzle:-

You are playing a game where the host gives you 3 boxes. One of the boxes contains an expensive diamond whereas, the other 2 are empty. You are available with the option to pick any one of the three boxes. You picked one box of the three. Now among the two boxes, one of them is sure to be empty. The host opens one of the remaining boxes and reveals it to be empty. Now he gives you a choice, where you can either open the box you had picked initially or pick the unopened box on the table. Whichever box you choose, you can open that box.

Question:-

• Which box will you pick? Will you stay with your initial choice or switch? Why?

## Solution

• This puzzle is also known as the Monty Hall Puzzle.
• I will pick the box on the table to open as it increases the probability of getting the diamond from $\frac{1}{3}$ to $\frac{2}{3}$
• There are 3 boxes, and on picking one of the three boxes, the probability of winning is$\frac{1}{3}$.
• While the probability of the diamond not being in the selected box and being in one of the other two boxes is $\frac{2}{3}$.
• Once the empty box is open – the $\frac{2}{3}$rd probability of the diamond being in the unselected box shifts over to the remaining box.
• Thus the unopened box that was previously not selected now has a probability of $\frac{1}{3}$ containing the diamond.
• Hence, it is always advisable to switch in the scenario, because there is always a chance of $\frac{2}{3}rd$ for the diamond to be in the other box and not in the selected box.
• If we see this problem on a greater scale and imagine that instead of three were given a hundred boxes, and one of the boxes contains the diamond.
• Here the selected box has a probability of $\frac{1}{100}$ , for containing the diamond while that the remaining 99 boxes have a probability of $\frac{99}{100}$.
• Once the remaining 98 empty boxes are revealed, the last unopened box gets a probability of$\frac{99}{100}$ while the initially selected box has a probability of $\frac{1}{100}$
• This makes the box that is on the table have a better chance of containing the diamond then the one that was selected initially.