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An irresistible probability paradox

I decided to share this probability paradox and have your say in which choice you should make.

"You are in a game show where there are 3 doors and you have to choose one of the doors. You are trying to win the million dollar prize which is behind one door. The other 2 doors have nothing in them so you want to make sure that you can win.

Let's call these doors \(D_1, D_2\) and \(D_3\). You pick out \(D_1\) and tell the host of the game show. The host, who knows which door has the million dollars shows you \(D_3\), which has nothing, and asks you if you want to change your mind."

How would you reply and give reasons.

Note by Sharky Kesa
3 years, 7 months ago

No vote yet
1 vote

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Comments

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This one is named as " Monty Hall problem" Also featured in the movie "21"

\(D_{2}\) is the probabilistically favorable answer. Here's the explanation:

Vos Savant's response was that the contestant should switch to the other door. (vos Savant 1990a)

The argument relies on assumptions, explicit in extended solution descriptions given by Selvin (1975b) and by vos Savant (1991a), that the host always opens a different door from the door chosen by the player and always reveals a goat by this action—because he knows where the car is hidden. Leonard Mlodinow stated: "The Monty Hall problem is hard to grasp, because unless you think about it carefully, the role of the host goes unappreciated." (Mlodinow 2008)

Contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their choice have only a 1/3 chance. One way to see this is to notice that 2/3 of the time, the initial choice of the player is a door hiding a goat. When that is the case, the host is forced to open the other goat door, and the remaining closed door hides the car. "Switching" only fails to give the car when the player picks the "right" door (the door hiding the car) to begin with, which only happens 1/3 of the time.

Source : wikipedia

Vishal Sharma - 3 years, 7 months ago

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Let's compute the probability that we will win if we switch:

 Total probability = Probabiilty that there's prize in D1 * Probability that we will win if we switch given there's prize in D1

                          + Probability that there isn't prize in D1 * Probability that we will win if we switch given there's no prize in D1

                          = (1/3) * (0) + (2/3) * (1)

                          = 2/3

In a formal way:

P (T) = P(S1) * P(A|S1) + P(S2) * P(A|S2)
     = (1/3) * (0) + (2/3) * (1)
     = 2/3

So its better to switch but it's too hard to think about this when you are in the game. Agree?

Lokesh Sharma - 3 years, 7 months ago

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Keep D1, its supposed to be hard to pick. There really is no way to know which so might as well just guess

Alexis Henderson - 3 years, 7 months ago

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Actually, it's better to switch.

Finn Hulse - 3 years, 7 months ago

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