The Game Show Problem

Choose one of three doors, of which one has a prize behind it. After your initial selection, a wrong door is eliminated and you are given the opportunity to switch to the other door. Would you switch?

Suppose you are on a game show and the host asks you to choose one of three doors.  Of these doors, one has a prize behind and the other two have nothing.  Your goal is to choose the door with the prize, which at this point you only have a 1 in 3 shot of guessing correctly, not the best chance.  For this example I am going to choose door 1.  The host of the show then, instead of opening the door you chose decides to open a door without the prize (not one that you chose).  In this example the host opens up door number 3.  Next the host of the game show asks if you want to switch to the other door (in this example door 2).

The average person would now believe they have a fifty-fifty shot at winning the money depending on which door they choose, therefore they will stick with their original choice.  They would be correct if there were only two doors to choose from in the first place.  There is actually a 1 in 3 shot of winning if you stay and and 2 in 3 shot of winning if you switch to the other door.  Continue reading the next paragraphs if you want to read the explanations and reasons why there is a better chance of winning if you switch to the other doors.  Below is a table showing what would happen whether you switched or stayed if the prize was behind door number 2.

Chose Door 1

Chose Door 2

Chose Door 3

Staying

Loses

Wins

Loses

Switching

Wins

Loses

Wins

This game show problem, also called the Monty Hall, problem is loosely based on the Let’s Make a Deal show whose original host was Monty Hall.  There are many solutions to demonstrate why is odd outcome is very true.  One simple way is beside the door you chose the other two doors have a 2 in 3 shot of having the prize.  The host just takes away the door of these that is incorrect, thus you can either choose your original door of the other “two” doors.

P(A|B) = frac{P(B | A), P(A)}{P(B)}. ,

  A more complex solution can be found using Bayes’ Theorem giving relations between probabilities (shown above).  This can be extended to be (C is the door with prize, S is door selected, H is door host opens):

P(C=c|H=h, S=s),=frac{P(H=h|C=c, S=s)P(C=c|S=s)}{P(H=h|S=s)},

Simplifying this to numbers:

P(C=2|H=3, S=1) = frac{1timesfrac 13}{frac 12 times frac 13 + 1timesfrac 13 + 0 times frac 13}=tfrac 23.

Thus there is a 2 out of 3 chance in winning if you switch to the second door.  If you do not understand this just go with the simple solution or look up the Monty Hall problem of Bayes theorem on google to learn more.

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