Monty Hall Problem: A Mathematical Analysis

Rationale:

Regardless of whether you realize or not, we are surrounded by probability. Consistently, we use probability to plan around the weather. Meteorologists can't predict precisely what the weather will be, so they use devices and instruments to decide the probability that it will rain, snow or hail.
When the doctor gives us chances to survive, its probability. According to Eliezer S. Yudkowsky, “Reality dishes out experiences using probability, not plausibility”. I’ve chosen to do my mathematics exploration on probability on account of its wide use and my enthusiasm for it and I’ve decided to find different probabilities in the famous Monty Hall Problem where there is a car behind one of the doors and the other two doors have goats

Toward the start of the game one object will be allotted behind each door, two goats and a car.
III. The candidate will choose a door without knowing which door has a car allotted behind it.
IV. Knowing the fact that which door has a car and goat behind it, the host will open a door.
V. After opening the door, the host will ask the contestant if as to whether he might want to remain on his choice or change it.

Table-3 and Table-4 in the Appendix demonstrates the spreadsheet simulations for switching initial decision and staying on the same correspondingly.

Table-3 shows that when the candidate switches the initial decision he wins 128 out of 200 times, which implies that the candidate winning percentage is 64%.

Whereas Table-4 represents that when the candidate switches the initial decision, he wins 69 out of 200 times, which implies that the candidate winning percentage is 34.5%.

The above data represents that it is superior to switch the initial selection of the door than to stay on the same one.

Although this data can fluctuate in each simulation. But in any case, unfailingly, the probability of winning the car when you switch your initial decision will be more. It can be tested by more simulations in the