Editing Talk:1282: Monty Hall

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"In that scenario, if a goat is revealed, there is in fact an equal probability of winning by switching or keeping the initial door." I'm not sure of this, can anyone explain? It seems to be stating that simply because of the random chance of the host picking the goat door in this situation then the facts about probability change.This seems to be a very large stretch. Surely in this situation since the host is picking randomly then your probabilities of losing are increased but the moment he does randomly pick a goat door then your chances of winning by switching remain the same as they were in the original problem, 2/3. --[[User:Lackadaisical|Lackadaisical]] ([[User talk:Lackadaisical|talk]]) 17:29, 25 October 2013 (UTC)
 
"In that scenario, if a goat is revealed, there is in fact an equal probability of winning by switching or keeping the initial door." I'm not sure of this, can anyone explain? It seems to be stating that simply because of the random chance of the host picking the goat door in this situation then the facts about probability change.This seems to be a very large stretch. Surely in this situation since the host is picking randomly then your probabilities of losing are increased but the moment he does randomly pick a goat door then your chances of winning by switching remain the same as they were in the original problem, 2/3. --[[User:Lackadaisical|Lackadaisical]] ([[User talk:Lackadaisical|talk]]) 17:29, 25 October 2013 (UTC)
:Let's see... for a "random door opened by host", split it as follows: Initial choice is G (2/3rds goat), or C (1/3rd car).  If G, then host has even chance of getting a goat, GG (2/3 * 1/2 = 1/3) or a car, GC (2/3 * 1/2 = 1/3), whereas if C then host has 100% chance of a goat, CG (1/3 * 1 = 1/3).  A GG happens at a frequency of 1/3rd (best to swap), a CG happens at a relative frequency of 1/3rd (best to stick) and a "game ender" happens for the final 1/3rd (no choice, no win). Thus in any non-ended game (i.e. excluding the car-revealing third of possibilities), you have equal chance ''regardless'' of what you do.
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:You can also analyse it explicitly by noting down the full six combinations of "Goat 1", "Goat 2" and "Car" in the "First chosen", "Host chosen" and "Swap choice" doors.  "Swap choice" is the winning move in two circumstances (having sequentially had "G1 then G2" or "G2 then G1") and "First choice" is the winning move in two others ("C then G1" or "C then G2"), with the two remaining (G1,C or G2,C) being end-games that make the option moot.
 
:Note, the above frequencies only happen with the 'dumb host'.  Re-run it with a 'knowing host' and you get your proof for the "best to swap" scenario. (I found the above explanation by 184.66.160.91 to be the most succinct explanation of this problem that I've seen, BTW.  Although maybe I have the advantage of understanding it in the first place.) [[Special:Contributions/31.111.43.68|31.111.43.68]] 00:36, 26 October 2013 (UTC)
 
  
 
I agree with the various comments above.  While this comic does use a goat, it really isn't making use of "The Monty Hall Problem" at all.  That involves offering a player a chance to switch doors, and there's no indication that happened here.  In this comic, for all we know, Beret selected door B, won a goat and, unlike most players, was pleased with having won a goat.
 
I agree with the various comments above.  While this comic does use a goat, it really isn't making use of "The Monty Hall Problem" at all.  That involves offering a player a chance to switch doors, and there's no indication that happened here.  In this comic, for all we know, Beret selected door B, won a goat and, unlike most players, was pleased with having won a goat.

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