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  2. Monty Hall problem - Wikipedia

    en.wikipedia.org/wiki/Monty_Hall_problem

    As in the Monty Hall problem, the intuitive answer is ⁠ 1 / 2 ⁠, but the probability is actually ⁠ 2 / 3 ⁠. The Three Prisoners problem, published in Martin Gardner's Mathematical Games column in Scientific American in 1959 [7] [55] is equivalent to the Monty Hall problem. This problem involves three condemned prisoners, a random one of ...

  3. Talk:Monty Hall problem/Archive 1 - Wikipedia

    en.wikipedia.org/wiki/Talk:Monty_Hall_problem/...

    Text from another version, posted under the title Monte Hall Problem. Charles Matthews 20:57, 10 Jun 2004 (UTC) The “Monte Hall Problem” (Applied Probability) On the old game show “Let’s Make a Deal”, the final segment involved two contestants who had won the most money that day. There were 3 closed doors with prizes hidden behind ...

  4. Talk:Monty Hall problem/Arguments/Archive 8 - Wikipedia

    en.wikipedia.org/wiki/Talk:Monty_Hall_problem/...

    Probability 2/3 of missing the car since all doors are equally likely - according to his knowledge. The host opens a door and shows a goat, but the player doesn't take any notice of the number. Still probility 2/3 that there's no car behind door 1, because whether or not the car is behind door 1, the host is certain to do what he did.

  5. Talk:Monty Hall problem/Archive 2 - Wikipedia

    en.wikipedia.org/wiki/Talk:Monty_Hall_problem/...

    For n=3, both terms - (3-1)/3 x 1/(3-2) and (3-1)/3 - result to 2/3. So the change is a good choice for both general interpretations, as well for the one special case t=3. But it is illegal to refer to the alleged obviousness of just one general interpretation to illustrate the special case t=3.

  6. Talk:Monty Hall problem/Arguments - Wikipedia

    en.wikipedia.org/wiki/Talk:Monty_Hall_problem...

    I believe in constant refinement. 'Monty Hall problem', Wikipedia, There are 3!=6 sequences for arrangement of 3 prizes behind 3 doors. The prizes are {goat1, goat2, car}. For any 1 sequence there are 4 possible games (series of choices) by the player 'p' and host 'h' including an option for the player to change their choice. For {g1 g2 car}:

  7. Talk:Monty Hall problem/Arguments/Archive 3 - Wikipedia

    en.wikipedia.org/wiki/Talk:Monty_Hall_problem/...

    Start with an ace and two jokers. You're the dealer. Shuffle the 3 cards. Deal them face down left to right. The leftmost card corresponds to "door 1", the middle one "door 2" and

  8. Talk:Monty Hall problem/draft2 - Wikipedia

    en.wikipedia.org/wiki/Talk:Monty_Hall_problem/draft2

    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. 8 Other ...

  9. Talk:Monty Hall problem/Archive 35 - Wikipedia

    en.wikipedia.org/wiki/Talk:Monty_Hall_problem/...

    In my opinion many within the 2/3 fraction up to today don't see the difference between the blank fact that the host opens a not chosen door with a goat, and the enforcement by the rules of the game to do so, which is crucial for the 2/3 solution, especially too for "reenactments" and "computer proofs".--Albtal 12:38, 13 January 2013 (UTC)