Monty Hall Problem

Here is an interesting problem called Monty Hall problem.
Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice and why? Give a proper mathematical explanation.
Now it is quiet easy to say whether probability would increase or not. the problem comes to give a proper mathematical. Try it, and leave your answer in the comments section. I will give my answer in tomorrow’s post. Please do not refer net. Ask your teachers (like I did), friends, solve in groups but do not refer internet.

2 thoughts on “Monty Hall Problem

  1. It is a very interesting problem. A sort of a choice paradox, which explains switching is beneficial. The underlying concept is the conditional probability.

    Probability of winning when we don’t switch = (1/2) * (2/3) = 1/3 (probability that car is behind gate 1 and gate 3 has a goat)

    Probability of winning when we switch = 1-(1/2)*(2/3) = 2/3

    So, switching is beneficial.

  2. When no one understands the answer repose the question in terms of looking for the ace of diamond in a pack of 52 cards.

    Pick a card at random and the probability is 1/52 that it is the ace of diamonds.

    Probability it is in the rest of the pack 51/52

    now discard 50 cards from the rest of the pack that are not the ace of diamonds.

    Which of the two cards that are left is most likely to be the ace?

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