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Message: <blockquote data-quote="Hawkman" data-source="post: 445577" data-attributes="member: 23715">Steve Selvin wrote a letter to the <a href="https://en.wikipedia.org/wiki/The_American_Statistician" target="_blank">American Statistician</a> in 1975, describing a problem based on the game show <a href="https://en.wikipedia.org/wiki/Let%27s_Make_a_Deal" target="_blank">Let's Make a Deal</a>, dubbing it the "Monty Hall problem" in a subsequent letter. The problem is equivalent mathematically to the <a href="https://en.wikipedia.org/wiki/Three_prisoners_problem" target="_blank">three prisoners problem</a> described in <a href="https://en.wikipedia.org/wiki/Martin_Gardner" target="_blank">Martin Gardner</a>'s "Mathematical Games" column in <a href="https://en.wikipedia.org/wiki/Scientific_American" target="_blank">Scientific American</a> in 1959 and the Three Shells Problem described in Gardner's book Aha Gotcha.<h3>Standard assumptions</h3>Ambiguities in the Parade version do not explicitly define the protocol of the host. However, Marilyn vos Savant's solution printed alongside Whitaker's question implies, and both Selvin and Savant explicitly define, the role of the host as follows:<ol> <li data-xf-list-type="ol">The host must always open a door that was not selected by the contestant.</li> <li data-xf-list-type="ol">The host must always open a door to reveal a goat and never the car.</li> <li data-xf-list-type="ol">The host must always offer the chance to switch between the door chosen originally and the closed door remaining.</li> </ol>It is also presumed that the car is placed (uniformly) at random behind one of the doors and that, if the player initially chooses the car, then the host's choice of which goat-hiding door to open is also made (uniformly) at random.Under these standard assumptions, the probability of winning the car after switching is 2/3. Varying the assumptions can change the probability of winning by switching doors as detailed in the <a href="https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_behaviors" target="_blank">section below</a>.These assumptions specify the mathematical model of host behavior and prize distribution being analyzed. In a decision-theoretic reading, whether switching is rational for the contestant also depends on the contestant knowing, or having reason to assume, that the host follows this protocol.<h3>Simple solutions</h3>1 and 2 below ,Switch and Win, 3 ,stay and win.<a href="https://en.wikipedia.org/wiki/File:Monty_Hall_Problem_-_Standard_probabilities.svg" target="_blank"><img src="https://thumb.wikimedia.org/wikipedia/commons/thumb/4/41/Monty_Hall_Problem_-_Standard_probabilities.svg/250px-Monty_Hall_Problem_-_Standard_probabilities.svg.png?utm_source=en.wikipedia.org&utm_campaign=parser&utm_content=thumbnail" alt="" class="fr-fic fr-dii fr-draggable " style="" /></a>Three initial configurations of the game. In two of them, the player wins by switching away from the choice made before a door was opened.The solution presented by Savant in Parade shows the three possible arrangements of one car and two goats behind three doors and the result of staying or switching after initially picking door 1 in each case:<table style='width: 100%'><tr><th>Behind door 1</th><th>Behind door 2</th><th>Behind door 3</th><th>Result if staying at door #1</th><th>Result if switching to the door offered</th></tr><tr><td>Goat</td><td>Goat</td><td>Car</td><td>Wins goat</td><td>Wins car</td></tr><tr><td>Goat</td><td>Car</td><td>Goat</td><td>Wins goat</td><td>Wins car</td></tr><tr><td>Car</td><td>Goat</td><td>Goat</td><td>Wins car</td><td>Wins goat</td></tr></table>A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three.An intuitive explanation is that, if the contestant initially picks a goat (2 of 3 doors), the contestant will win the car by switching because the other goat can no longer be picked – the host had to reveal its location – whereas if the contestant initially picks the car (1 of 3 doors), the contestant will not win the car by switching. Using the switching strategy, winning or losing thus only depends on whether the contestant has initially chosen a goat (2/3 probability) or the car (1/3 probability). The fact that the host subsequently reveals a goat in one of the unchosen doors changes nothing about the initial probability.<a href="https://en.wikipedia.org/wiki/File:Monty_Little_green_alien.svg" target="_blank"><img src="https://thumb.wikimedia.org/wikipedia/commons/thumb/8/87/Monty_Little_green_alien.svg/250px-Monty_Little_green_alien.svg.png?utm_source=en.wikipedia.org&utm_campaign=parser&utm_content=thumbnail" alt="" class="fr-fic fr-dii fr-draggable " style="" /></a>A different selection process, where the player chooses at random after any door has been opened, yields a different probability.Most people conclude that switching does not matter, because there would be a 50% chance of finding the car behind either of the two unopened doors. This would be true if the host selected a door to open at random, but this is not the case. The host-opened door depends on the player's initial choice, so the assumption of <a href="https://en.wikipedia.org/wiki/Independence_(probability_theory)" target="_blank">independence</a> does not hold. Before the host opens a door, there is a 1/3 probability that the car is behind each door. If the car is behind door 1, the host can open either door 2 or door 3, so the probability that the car is behind door 1 and the host opens door 3 is 1/3×1/2=1/6.<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/eb2104f9331c9511732af9bf47192eb4d356b9cc" alt="{\displaystyle \textstyle {\frac {1}{3}}\times {\frac {1}{2}}={\frac {1}{6}}.}" class="fr-fic fr-dii fr-draggable " style="" /> If the car is behind door 2 – with the player having picked door 1 – the host must open door 3, such the probability that the car is behind door 2 and the host opens door 3 is 1/3×1=1/3.<img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/53a7b383f333672cd79db36e4ea772f0765e843c" alt="{\displaystyle \textstyle {\frac {1}{3}}\times 1={\frac {1}{3}}.}" class="fr-fic fr-dii fr-draggable " style="" /> These are the only cases where the host opens door 3, so if the player has picked door 1 and the host opens door 3, the car is twice as likely to be behind door 2 as door 1. The key is that if the car is behind door 2 the host must open door 3, but if the car is behind door 1 the host can open either door.</blockquote>

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