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MiszaBot I (talk | contribs) m Archiving 2 thread(s) from Talk:Monty Hall problem/Arguments. |
MiszaBot I (talk | contribs) m Archiving 2 thread(s) from Talk:Monty Hall problem/Arguments. |
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:The statement means that right at the start, before any doors have been opened, you have a 1 in 3 chance of picking the car. It does not refer to your probability of going on to win the game. If you are smart, this is 2/3. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 22:54, 6 May 2011 (UTC)
== Lundy's Solution ==
Mathematician Nicholas Lundy suggests numbering your goats or making them different colors to reveal that there are 6 possibilities.
<br /><br />
C -|- G1 -|- G2<br />
C -|- G2 -|- G1<br />
G1 -|- C -|- G2<br />
G1 -|- G2 -|- C<br />
G2 -|- C -|- G1<br />
G2 -|- G1 -|- C<br /><br />
Show a goat in any room and you remove two possibilities (the ones with cars) and therefore always have 1/2 chance of winning by switching or staying. <small><span class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Camtin|Camtin]] ([[User talk:Camtin|talk]] • [[Special:Contributions/Camtin|contribs]]) 04:05, 25 June 2011 (UTC)</span></small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->
:You remove three possibilities, as one of the solutions starting with the car will have the wrong goat in the revealed door. [[Special:Contributions/141.161.133.130|141.161.133.130]] ([[User talk:141.161.133.130|talk]]) 14:31, 20 July 2011 (UTC)
::Is this solution published anywhere? If so, where? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 19:21, 20 July 2011 (UTC)
:This explanation appears to ''assume'' that the host may sometimes open a door with a car and/or that he may sometimes open the already-selected door. A significant amount of confusion about the Monty Hall problem arises due to the ambiguity of the problem statement. I don't know what 141.161.133.130's "wrong goat in the revealed door" comment means, but he/she very likely has the "standard" version of the problem in mind, which instead assumes that the host ''always'' opens one of the non-selected doors, and ''never'' opens the door with the car. I believe that this illustrates why it is important to make such assumptions explicit. [[Special:Contributions/139.99.16.28|139.99.16.28]] ([[User talk:139.99.16.28|talk]]) 19:17, 18 October 2011 (UTC)
== Simple explanation! - Really ==
Ok so it seems to my lay self that there is a lot of over complication of a pretty basic problem.
I will now try to explain how the it works in the simplest possible manner.
1. Pick a door (chances of getting a goat = 2/3)
2. A goat gets revealed, you know that the door you chose to begin with most likely holds a goat, and that one door has the car
3. Switch, since you're pretty sure you have a goat (2/3 chance) you can be equally sure that the other door is a car.
4. Therefore after switching you have 2/3 chance that you have the car.
If that seems confusing, think about the problem with 1,000,000 doors and 999,999 goats and 1 car.
1. Pick a door, you almost definitely pick a goat
2. All goats but one are removed, leaving one door with the car and one with the goat.
3. You basically know that you chose a goat in the first round, so you obviously change your choice, making it almost guaranteed that you pick the car.
Easy Peasy :)
[[Special:Contributions/86.15.43.185|86.15.43.185]] ([[User talk:86.15.43.185|talk]]) 17:57, 25 July 2011 (UTC)
PiersyP
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