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Now, since the player initially chose door 1, the chance that the host opens door 3 is 50% if the car is behind door 1, 100% if the car is behind door 2, 0% if the car is behind door 3. Thus the Bayes factor consists of the ratios 1 / 2 : 1 : 0 or equivalently 1 : 2 : 0 , while the prior odds were 1 : 1 : 1 .
This question is called the Monty Hall problem due to its resembling scenarios on the game show Let's Make a Deal, hosted by Monty Hall. It was a known logic problem before it was used in "Ask Marilyn". She said the selection should be switched to door #2 because it has a 2 ⁄ 3 probability of success, while door #1 has just 1 ⁄ 3.
The Monty Hall problem. I'm convinced I'm a lost cause so there's not really any point trying to explain further. ... idea that if Monty was an alien and the car was a speedboat and the goat was a ...
Either Monty Hall opens the second door and reveals a goat, in which case the contestant should switch to the last remaining door, or else Monty Hall opens a door and reveals the car, in which case the contestant should obviously switch to that door.--Lorenzo Traldi 10:41, 20 December 2006 (UTC)
Let's compare them. WLOG suppose the contestant picks door A. In the standard Monty Hall problem, there are four possible outcomes: Car is behind A, Monty opens B (prob 1/6) Car is behind A, Monty opens C (prob 1/6) Car is behind B, Monty opens C (prob 1/3) Car is behind C, Monty opens B (prob 1/3)
If they all show heads, then Monty puts the car behind door 1. If they all show tails, then Monty puts the car behind door 3. Otherwise, Monty puts the car behind door 2. In each case, Monty puts goats behind the other two doors. The player then comes on stage, and Monty opens door 3. , after the opening reveals a goat and
1 Problem description. 2 Solution. 3 Vos Savant's solution. 4 Aids to understanding. 5 Simulation. 6 Increasing the number of doors. 7 Sources of confusion.
From this tree it's clear if you've picked door 1 and the host has opened door 3, the probability the car is behind door 2 is twice the probability the car is behind door 1 (because the host must open door 3 if the car is behind door 2, but opens either door 2 or door 3 - presumably each one half the time - if the car is behind door 1).