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Author | Topic: Three Curtains | |||||||||||||||||||||||||||||||||
rstrats Member (Idle past 102 days) Posts: 138 Joined: |
What would you do if after your initial pick of curtain #1, and before any curtain was opened, the host told you that you could stay with #1 or switch to BOTH #2 and #3?
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PaulK Member Posts: 17822 Joined: Member Rating: 2.2 |
That's just another variation that makes no difference.
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rstrats Member (Idle past 102 days) Posts: 138 Joined: |
PaulK,
re: "That's just another variation that makes no difference." Why do you think that having 2 curtains to look behind isn't better than having only 1?
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Modulous Member Posts: 7801 From: Manchester, UK Joined: |
Switching has to be better or worse or the same as not switching. If it's the same (worst case for the contestant) then it's still a 50-50 guess. Actually depending on the strategy in play, switching can be worse than 50/50 for the contestant. At the extreme and boring end, imagine the host only ever offers the switch in the cases where you picked the car. Now switching is a losing proposition every time. If the host is randomly opening one of the three doors, then the strategy becomes:1/9 host opens the door you picked AND its a winner (stick) 2/9 he opens your door and its a loser (switch) 1/9 he opens door 2 AND it's a winner (switch) 2/9 he opens door 2 AND it's a loser (odds go up, but switching is irrelevant.) 1/9 he opens 3 AND winner (switch) 2/9 he opens 3 and loser (doesnt matter) here, you win 1/9 (x3) of the time = 1/3 of the time outright. And you win 50% of the time the remaining. Your general odds going in are better than 50/50. If he randomly opens the two non picked doors1/3 he opens a door and it is a winner (1/6 for each door) 2/3 he opens a door and it is a loser. So in this example your odds go up because 1/3 of the time you win outright and 2/3 you have a 50/50 shot, meaning your odds are better than 50/50 And we know if he deliberately selects one of the remaining losing doors each time - you win 2/3 of the time by switching 100% of the time. Therefore, if you adopt the correct strategy you do a bit better than 50/50 if any switch is offered. Unless the host can choose to not offer a switch. In the cases where he does this, you lose 2/3 of the time if he chooses at random. If he selects to do this at whim he can make you lose 2/3 of the time or less (ie the times he offers a switch he improves your odds). We also note, that in complete random selections - switching is sometimes worse than 50/50 (1/9 of possibilities where he opens your door and it is a winner).
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PaulK Member Posts: 17822 Joined: Member Rating: 2.2 |
quote: That's what I said. And that is why I've been saying that you need to know the strategy. Simply by choosing when to offer the chance to switch the host can trivially guarantee that a switch will always win or always lose.
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PaulK Member Posts: 17822 Joined: Member Rating: 2.2 |
Oh dear. The point is that offering the chance to choose the other two is no different from revealing one of the other two as losing and offering the chance to switch to the remaining one. So it doesn't change the analysis at all.
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rstrats Member (Idle past 102 days) Posts: 138 Joined: |
PaulK,
re: "... offering the chance to choose the other two is no different from revealing one of the other two as losing and offering the chance to switch to the remaining one. So it doesn't change the analysis at all." And the analysis says that it is better to have 2 curtains to look behind than to having only 1.
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PaulK Member Posts: 17822 Joined: Member Rating: 2.2 |
No, it depends on the host's strategy as I keep telling you.
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Modulous Member Posts: 7801 From: Manchester, UK Joined: |
Actually depending on the strategy in play, switching can be worse than 50/50 for the contestant. That's what I said. I'm also trying to put the different strategies out there and in numbers, just to make it more interesting than a standard Monty Hall discussion. 1/3 of the time you pick right first time.But Monty offers you to switch 2/3 of the time (2/9) in this case and you auto win the rest (1/9) of the time 2/3 of the time you pick wrong. Monty offers to switch 2/9 of these times, but not the other 4/9ths. Switch option: 2/9 + 2/9 = 4/9ths.No switch: 5/9ths. If you always switch then 50% of the time you picked right first time and lose.If you never switch then 50% of the time you picked wrong first time and you lose. That is to say, switching makes no difference,you win 2/9ths you lose 2/9ths When you don't get a choice, you win 1/9, and lose 4/9ths. Thus you win 3/9ths (1/3), and you lose 6/9ths (2/3). I think that's the best strategy for the host which involves occasional revelation of non-chosen losing options coupled with the choice to switch(which is no different than just picking a random curtain/door) .
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Larni Member (Idle past 164 days) Posts: 4000 From: Liverpool Joined: |
Oh, I see now.
The above ontological example models the zero premise to BB theory. It does so by applying the relative uniformity assumption that the alleged zero event eventually ontologically progressed from the compressed alleged sub-microscopic chaos to bloom/expand into all of the present observable order, more than it models the Biblical record evidence for the existence of Jehovah, the maximal Biblical god designer. -Attributed to Buzsaw Message 53 The explain to them any scientific investigation that explains the existence of things qualifies as science and as an explanation-Attributed to Dawn Bertot Message 286 Does a query (thats a question Stile) that uses this physical reality, to look for an answer to its existence and properties become theoretical, considering its deductive conclusions are based against objective verifiable realities.-Attributed to Dawn Bertot Message 134
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Larni Member (Idle past 164 days) Posts: 4000 From: Liverpool Joined: |
That game uses doors, whereas the OP game uses curtains. Pure genius. You made my day with that.The above ontological example models the zero premise to BB theory. It does so by applying the relative uniformity assumption that the alleged zero event eventually ontologically progressed from the compressed alleged sub-microscopic chaos to bloom/expand into all of the present observable order, more than it models the Biblical record evidence for the existence of Jehovah, the maximal Biblical god designer. -Attributed to Buzsaw Message 53 The explain to them any scientific investigation that explains the existence of things qualifies as science and as an explanation-Attributed to Dawn Bertot Message 286 Does a query (thats a question Stile) that uses this physical reality, to look for an answer to its existence and properties become theoretical, considering its deductive conclusions are based against objective verifiable realities.-Attributed to Dawn Bertot Message 134
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PurpleYouko Member Posts: 714 From: Columbia Missouri Joined:
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Ok so somebody further up the thread said that I could replace the host with a computer program so I just did that.
The rules I used are these.1) car is placed randomly in one of three places 2) The first guess is made as 1, 2 or 3 3) Host chooses one of the unchosen places and eliminates it 4) contestant either sticks or changes 5) results are evaluated and spit out as count rates Once put into code form it becomes pretty obvious what the result will be without even running the simulation.The logic is inescapable. First of all there is one thing to check 1) Was my guess correct? (remember there are only 2 choices now since the host already opened a curtain/door that wasn't the car)YES (1 in 3 chance) --> Sticking wins, changing loses so add 1 to StickCount NO (2 in 3 chance) --> Sticking loses, changing wins so add 1 to ChangeCount The results of 1 million games comes out as followsStickCount = 332654 ChangeCount = 667346 Until putting this into code form I would have sworn that the chances were 50:50 but as soon as I wrote the problem in code form it's obvious that changing is going to win twice as often as sticking
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rstrats Member (Idle past 102 days) Posts: 138 Joined: |
PurpleYouko wins a cigar.
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PurpleYouko Member Posts: 714 From: Columbia Missouri Joined: |
meh.. I don't smoke.
How about you make it a nice pina collada that i can sip while sitting in a swimming pool
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Minnemooseus Member Posts: 3941 From: Duluth, Minnesota, U.S. (West end of Lake Superior) Joined: Member Rating: 10.0 |
From Car Talk:
http://www.cartalk.com/content/lets-hear-it-boys-0?question RAY: Here’s the first half of the puzzler: If a mother has two kids and the older one is a boy, what are the chances the younger is a boy? TOM: I know this puzzler. This is a killer. RAY: And the answer is 50-50. So I'll give you that part. Now, suppose a different mother has two kids and one of them is a boy. What are the chances that the other one is a boy? Well, it's got to be 50-50, right? TOM: You would think. RAY: Wrong. TOM: Of course. RAY: So, what are those chances? ----- http://www.cartalk.com/content/lets-hear-it-boys-0?answer (hidden in white block, drag mouse cursor over to see)
RAY: There are four possible scenarios: Older boy, younger boy. Older boy, younger girl. Older girl, younger boy. And older girl, younger girl. Now, in the first case, when I say the older one is a boy, that immediately leaves out the last two possibilities. It can only be boy-boy or boy-girl. So, in order for the other one to be a boy, it's a 50-50 chance. TOM: That's right. RAY: Now,when I say that one of them is a boy, it becomes harder for the other one to become a boy. And the chances are one in three. If you look at the scenarios, you have boy-boy, boy-girl and girl-boy. For the other one to be a boy, it's gotta be choice #1 which is boy-boy. There's one chance in three. Do we have a winner this week? Moose
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