```Date: May 14, 1998 11:32 AM
Author: Domenico Rosa
Subject: Monty Hall Problem: Tree Diagram

The Monty Hall problem has been discussed in numerous posts to this list. The e-mail newsletter CHANCE News 7.04 reported that this problem has resurfaced in The Independent (London). Two articles, written by William Harston on 28 Mar 1998 and 11 Apr 1998, can be obtained over Lexis-Nexis.I am taking the liberty of forwarding the following tree-diagram, which I constructed in December 1990 and is based on the assumptions made by Marilyn vos Savant. Namely: 1. The initial placement of the auto (car) is done at random.2. The contestant chooses a door at random.3. The host will not open the contestant's door and will not    open the door containing the auto. 4. If both remaining doors contain a goat, the host will open    one at random. The notation denotes the following sequence of activities:Ai = the auto is placed behind door i, for i = 1, 2, 3.Cj = the contestant chooses door j, for j = 1, 2, 3.Hk = the host opens door k, for k = 1, 2, 3.         Auto            Contestant     Host             Proba-        Placed             Chooses      Opens   Outcome  bility                                        * H2    A1C1H2    1/18                                 1/2 *                                  *                               C1                            *     *                   1/3   *       1/2 *                        *                 * H3    A1C1H3    1/18                   *                *      1/3           1            A1 --------------> C2 ------> H3    A1C2H3    1/9            *    *           *          *              1          *        1/3     *   C3 ------> H2    A1C3H2    1/9         *   1/3  *       *                             1      *                        C1 ------> H3    A2C1H3    1/9     *                       *    *                1/3  *   *                   *                * H1    A2C2H1    1/18  *                 *            1/2 * *    1/3        *     1/3        *O ---------> A2 -------------> C2 *               *                *  *                 *            1/2 *   *                   *                * H3    A2C2H3    1/18    *                1/3  *     *                       *       1      *                        C3 ------> H1    A2C3H1    1/9       *    1/3  *          *                           1          *        1/3     *   C1 ------> H2    A3C1H2    1/9           *          *            *    *                   1            A3 --------------> C2 ------> H1    A3C2H1    1/9                *      1/3                   *                      *                 * H1    A3C3H1    1/18                   1/3   *       1/2 *                            *     *                               C3                                   *                                 1/2 *                                        * H2    A3C3H2    1/18When this game is played under the vos Savant assumptions, there are 12 elementary outcomes. Each outcome where the auto is behind the contestant's door has probability 1/18. Each outcome where the auto is behind the remaining door has probability 1/9. It follows that the conditional probabilities are 1/3 and 2/3 of the auto's being behind the contestant's door and behind the remaining door, respectively. I would like to express my gratitude to Professor Dawson Fulton who taught me Probability and Statistics when I was a college junior in 1968-69. Bertrand's Box Paradox, which is similar to the above, is one of the homework problems that taught us the concepts of conditional probability and Bayes' Formula.Domenico Rosa=======================================================================The Advanced Placement Statistics ListTo UNSUBSCRIBE send a message to majordomo@etc.bc.ca containing:unsubscribe apstat-l <email address used to subscribe>Discussion archives are athttp://forum.swarthmore.edu/epigone/apstat-lProblems with the list or your subscription? mailto:jswift@sd70.bc.ca=======================================================================
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