Search results
Results from the WOW.Com Content Network
The Monty Hall problem is a brain teaser, in the form of a probability puzzle, based nominally on the American television game show Let's Make a Deal and named after its original host, Monty Hall. The problem was originally posed (and solved) in a letter by Steve Selvin to the American Statistician in 1975.
Monty_Hall_game_possibilities,_when_player_picks_door_1.png (653 × 272 pixels, file size: 12 KB, MIME type: image/png) This is a file from the Wikimedia Commons . Information from its description page there is shown below.
The following 16 pages use this file: Monty Hall problem; Talk:Monty Hall problem/Archive 13; Talk:Monty Hall problem/Archive 15; Talk:Monty Hall problem/Archive 21
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 probability problem arises from asking if the player should switch to the unrevealed door. Mathematically, the problem shows that a player switching to the other door has a 2 ⁄ 3 chance of winning under standard conditions, but this is a counterintuitive effect of switching one's choice of doors, and the problem gained wide attention due ...
The Monty Hall problem is a puzzle involving probability similar to the American game show Let's Make a Deal.The name comes from the show's host, Monty Hall.A widely known, but problematic (see below) statement of the problem is from Craig F. Whitaker of Columbia, Maryland in a letter to Marilyn vos Savant's September 9, 1990, column in Parade Magazine (as quoted by Bohl, Liberatore, and Nydick).
Whatever Marilyn wrote, whatever she intended, the Monty Hall problem today is about the problem in which the host is guaranteed to open a different door to the door chosen by the player and reveal a goat (which he can do because he knows where the car is). Richard Gill 17:30, 19 January 2013 (UTC) Others may score your arguments.--
For the standard version of the Monty Hall problem, that means in turn: Given that the host opens door 2, the car is behind door 3 (1/3)/(1/2) = 2/3 of the time. So in the standard Monty Hall problem the odds for switching versus staying are 2:1, irrespective of any particular actual door numbers.