You are on a game show, and you're given the choice of three doors. Behind one door is a car; behind the others are goats. You pick a door, say A, and the host, who knows what's behind the doors, opens another door, say C, which has a goat. He then says to you, "Do you want to pick door B?"

Is it to your advantage to switch your choice to B?

We can solve this problem without getting deeply into the mathematics by drawing a tree diagram showing all possible sequences of events that can happen.

Event name | Event description |

A | The car is hidden behind door A |

B | The car is hidden behind door B |

C | The car is hidden behind door C |

Ga | The guest chooses door A |

Gb | The guest chooses door B |

Gc | The guest chooses door C |

Ma | Monty opens door A |

Mb | Monty opens door B |

Mc | Monty opens door C |

The list of events tabulates all the events that can occur.

The car can be behind one of three doors (events A,B and C). The Guest can then choose one of three doors (events Ga, Gb and Gc) and finally the game host (Monty Hall) can open a door to reveal a goat (events Ma, Mb and MC).

We have put all these events together in the "Full Event Diagram". Line A represents event A occurring, the car is behind door A. Then one of the events Ga, Gb or Gc can follow, the Guest can choose any door. Then Monty has to open a door, and this is where we are somewhat restricted. If A is followed by Ga then Ma cannot occur, so A followed by Ga must be concluded with Mb or Mc. If A is followed by Gb or Gc Monty cannot open door A (as the car is behind it) or the door chosen by the Guest and therefore he only has one choice, Mc after Gb and Mb after Gc.

Let us think about what additional information we have as the game progresses. We know that the Guest has chosen door A and we know Monty has chosen door C. We therefore know that Ga and Mc have occurred. Reflect this in our updated diagram below.

The new diagram shows the only two sequences on the diagram that fit the facts that we now have and we have coloured these blue and magenta. The question is are the blue and magenta sequences of equal probability?

Let the game run say 1,000,000 times. Follow the blue route. There is an equal probability of A, B or C happening. We would therefore expect that 1/3 (i.e. 333,333 times) we would set off on the blue line. Now we come to the Guest choice.Having set out along the blue line we get to the point where Ga, Gb or Gc happen. The guest chooses randomly, so on our 333,333 journeys down the blue route 111,111 journeys will continue on the blue route, on the other 222,222 occasions we move on to Gb (111,111) or Gc (111,111). We are now at the point when Monty chooses a door. We have travelled on the blue route and the car is behind door A and the Guest has chosen door A. Monty can now choose to open door B or door C, and as he will do this with equal probability we complete the blue journey through to Mc on 1/2 of occasions (55,555) and we follow through to Mb (the yellow route) on the other 55,555 occasions.

Now travel along the magenta route. The route B followed by Ga is, as with blue line, expected to be followed 111,111 times. However on this occasion Monty can only open door C as door B has the car and door and the Guest has chosen door A. Therefore all 111,111 magenta routes through Ga end up at Mc.

We therefore conclude that when we play the game 1,000,000 times 55,555 times we will travel along the blue route and 111,111 times we proceed along the magenta route. We do now know that we must be on one of these routes as we know Ga and Mc have taken place. The probability of magenta to blue is 111,111 to 55,555 or 2 to 1 and the Guest is advised to assume it is more likely he is on the magenta route (which started with B) than the blue route (which started with A). Therefore the Guest is strongly advised to change his original choice.

The game show is run 1,000,000 times. On each run we

- randomly select a door to keep the car behind
- randomly select a door which is the guest's choice
- if the two doors selected are different the third door becomes the host's choice (magenta path - guest has to change choice to win)
- if the two doors selected are the same the host will choose from the other two doors. The host's choice will be the lower numbered door 50% of the time and the higher numbered door the other 50% of the time (ABC being numbered 0, 1, and 2 for this purpose)(blue path - guest will win if choice kept)

On each of 1,000,000 occasions we have the identity of the door with the car behind it, the guest's original choice and the host's choice. We can now declare the results

Wins if the guest switches from the original choice | |

Wins if the guest sticks with the original choice | |

( Change : Stick ) win ratio |