| Home |
| Search |
| Today's Posts |
|
|
|
#1
Posted to sci.electronics.repair,rec.audio.pro
|
|||
|
|||
another puzzler
On May 18, 6:44*am, "David" wrote:
"Don Pearce" *wrote in message ... The host acts as a leak of information. It might help to imagine an alternate game, where the host does not know the contents of the doors, and the game is void if the host reveals the car. This version puts you back to 50/50 when the host reveals a goat, whether you switch doors or not. *** Not true. When the host reveals a goat whether he guessed or knew it was there makes absolutely no difference. You should still switch doors. David If the host does not know, he might quite as easily reveal the car. You then can't win it. Do you guarantee yourself 2/3 odds by switching then? No. If the host reveals a goat by chance, the odds do indeed drop to 50/50. d *** Sorry, I disagree. *Yes the host could reveal a car if he is unaware of the situation. If this happens, the game was defined as void. If the host instead reveals a goat, there is no difference whether he guessed or knew the goat was there. Let's look at the case of the ignorant host. There are three possibilities at the start of the game. The probability of each is 1/3 _1 2 3_ aCGG bGCG cGGC Let us say door 1 represents the contestant's pick. The host can pick either door 2 or door 3 Case a: Host picks Door 2. Result: Goat. Contestant switches to Door 3, loses. ...............Host picks Door 3 Result Goat. Contestant switches to Door 2, loses. Case b: Host picks Door 2. Result Car. Contestant loses ...............Host picks Door 3. Result Goat. Contestant switches to Door 2, wins Case c: Host picks Door 2. Result Goat. Contestant switches to Door 3, wins ...............Host picks Door 3 Result Car. Contestant loses. Of the six possible scenarios, the contestant loses four times. If the contestant does not switch after the ignorant host opens a door, the contestant loses four times. If we discard the times the host opens a door with a car behind it, the contestant wins two out of four times when he switches, and two out of four times when he doesn't switch. Therefore, switching picks has no effect on the odds when the host randomly opens one of the other doors. |
|
#2
Posted to sci.electronics.repair,rec.audio.pro
|
|||
|
|||
|
another puzzler
|
|
#3
Posted to sci.electronics.repair,rec.audio.pro
|
|||
|
|||
|
another puzzler
On May 19, 7:54*am, Carey Carlan wrote:
spamtrap1888 wrote in news:88df2861-f695-449b- : Let's look at the case of the ignorant host. There are three possibilities at the start of the game. The probability of each is 1/3 _1 2 3_ aCGG bGCG cGGC Let us say door 1 represents the contestant's pick. The host can pick either door 2 or door 3 Case a: Host picks Door 2. Result: Goat. Contestant switches to Door 3, loses. ..............Host picks Door 3 *Result Goat. Contestant switches to Door 2, loses. Case b: Host picks Door 2. Result Car. Contestant loses ..............Host picks Door 3. Result Goat. Contestant switches to Door 2, wins Case c: Host picks Door 2. Result Goat. Contestant switches to Door 3, wins ..............Host picks Door 3 *Result Car. Contestant loses. Of the six possible scenarios, the contestant loses four times. If the contestant does not switch after the ignorant host opens a door, the contestant loses four times. If we discard the times the host opens a door with a car behind it, the contestant wins two out of four times when he switches, and two out of four times when he doesn't switch. Therefore, switching picks has no effect on the odds when the host randomly opens one of the other doors. Then go back to the original where the host knows where the car is and the contestant switches. Case a: Host picks Door 2. Result: Goat. Contestant switches to Door 3, loses. ..............Host picks Door 3 *Result Goat. Contestant switches to Door 2, loses. Case b: Host picks Door 3. Result Goat. Contestant switches to Door 2, wins Case c: Host picks Door 2. Result Goat. Contestant switches to Door 3, wins Or the contestant doesn't switch. Case a: Host picks Door 2. Result: Goat. Contestant keeps Door 1, wins. ..............Host picks Door 3 *Result Goat. Contestant keeps Door 1, wins. Case b: Host picks Door 3. Result Goat. Contestant keeps Door 1, loses Case c: Host picks Door 2. Result Goat. Contestant keeps Door 1, loses After the Host opens the door the odds are even. *Makes no difference if the contestant changes doors or not. *This is the same as there only being two doors. The original claim was that the odds remained 1 in 3 even after the Host opened the door. *I still don't see it. Without switching, the contestant has a 1/3 chance of winning: Case a: Contestant picked door with car. Host can open either door, his choice, to reveal goat. Case b: Contestant picked door with goat. Host must open Door 3 to reveal goat. Case c: Contestant picked door with goat. Host must open Door 2 to reveal goat. With switching, the contestant now has a 2/3 chance of winning: Case a: Host can open either door, his choice. Contestant switches to unopened door, loses. Case b: Host opens Door 3, contestant switches to Door 2, wins. Case c: Host opens Door 2, contestant switches to Door 3, wins. |
|
#5
Posted to sci.electronics.repair,rec.audio.pro
|
|||
|
|||
|
another puzzler
"Bill Graham" wrote in
: I claqim there are two games. In the first game, you go to the studio, pick a door, and then go home to wait and see if they call you and tell you that you either won or lost. Your odds are only 1/3 of winning this game. But if you play the second game, then you go to the studio and mess around until the host opens up a door and shown you the donkey behind it. then you can play the game with 50-50 odds of winning. The only thing I have trouble explaining is why, in order to play this second game with the better odds, you have to switch doors. But, in fact, you do have to switch in order to switch games and take advantage of the better odds. It's not just 50/50. It's 67/33 in your favor. Here's another super-simple explanation: As Cecil Adams puts it (Adams 1990), "Monty is saying in effect: you can keep your one door or you can have the other two doors." |
|
#6
Posted to sci.electronics.repair,rec.audio.pro
|
|||
|
|||
|
another puzzler
In article ,
Bill Graham wrote: I claqim there are two games. In the first game, you go to the studio, pick a door, and then go home to wait and see if they call you and tell you that you either won or lost. Your odds are only 1/3 of winning this game. But if you play the second game, then you go to the studio and mess around until the host opens up a door and shown you the donkey behind it. then you can play the game with 50-50 odds of winning. The only thing I have trouble explaining is why, in order to play this second game with the better odds, you have to switch doors. But, in fact, you do have to switch in order to switch games and take advantage of the better odds. Yup. The distinction between the two games is based on the amount of information available to you at the moment you make your final decision. In the first game, the only information you have is that every door available to you to choose has a 1/3 chance of being correct. In the second game, additional information is given to you by the host, after you make your initial decision and before you make your second one. "The prize is *not* behind that door over there." Actually, you don't *have* to switch doors to "play the second game". You're playing it from the moment the host gives you this extra information. It's just that if you ignore this extra information, and *think* you're still playing the first game (as many people do), you are more likely than not to make a poor choice in the decision you make in this second game. -- Dave Platt AE6EO Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior I do _not_ wish to receive unsolicited commercial email, and I will boycott any company which has the gall to send me such ads! |
|
#7
Posted to sci.electronics.repair,rec.audio.pro
|
|||
|
|||
|
another puzzler
Dave Platt wrote:
In article , Bill Graham wrote: I claqim there are two games. In the first game, you go to the studio, pick a door, and then go home to wait and see if they call you and tell you that you either won or lost. Your odds are only 1/3 of winning this game. But if you play the second game, then you go to the studio and mess around until the host opens up a door and shown you the donkey behind it. then you can play the game with 50-50 odds of winning. The only thing I have trouble explaining is why, in order to play this second game with the better odds, you have to switch doors. But, in fact, you do have to switch in order to switch games and take advantage of the better odds. Yup. The distinction between the two games is based on the amount of information available to you at the moment you make your final decision. In the first game, the only information you have is that every door available to you to choose has a 1/3 chance of being correct. In the second game, additional information is given to you by the host, after you make your initial decision and before you make your second one. "The prize is *not* behind that door over there." Actually, you don't *have* to switch doors to "play the second game". You're playing it from the moment the host gives you this extra information. It's just that if you ignore this extra information, and *think* you're still playing the first game (as many people do), you are more likely than not to make a poor choice in the decision you make in this second game. Suppose for the moment that there are two contestants. One picks door two, and the other picks door one. Then the moderator opens door three and shows everyone that there is a donkey behind that door. Now, will it make any difference if the other two switch their initial picks or not? And, if they do swap doors, with they both enjoy a 2/3 chance of winning? |
|
#8
Posted to sci.electronics.repair,rec.audio.pro
|
|||
|
|||
|
another puzzler
In article ,
"Bill Graham" wrote: Suppose for the moment that there are two contestants. One picks door two, and the other picks door one. Then the moderator opens door three and shows everyone that there is a donkey behind that door. Now, will it make any difference if the other two switch their initial picks or not? And, if they do swap doors, with they both enjoy a 2/3 chance of winning? You can't just go changing the rules of the game willy nilly. The game is played with goats, not donkeys. Sheesh. |
|
#9
Posted to sci.electronics.repair,rec.audio.pro
|
|||
|
|||
|
another puzzler
Smitty Two wrote:
In article , "Bill Graham" wrote: Suppose for the moment that there are two contestants. One picks door two, and the other picks door one. Then the moderator opens door three and shows everyone that there is a donkey behind that door. Now, will it make any difference if the other two switch their initial picks or not? And, if they do swap doors, with they both enjoy a 2/3 chance of winning? You can't just go changing the rules of the game willy nilly. The game is played with goats, not donkeys. Sheesh. No sheep ? Jamie |
|
#10
Posted to sci.electronics.repair,rec.audio.pro
|
|||
|
|||
|
another puzzler
"Bill Graham" wrote in message ... Suppose for the moment that there are two contestants. One picks door two, and the other picks door one. Then the moderator opens door three and shows everyone that there is a donkey behind that door. Now, will it make any difference if the other two switch their initial picks or not? And, if they do swap doors, with they both enjoy a 2/3 chance of winning? It get's really silly now when both switch doors! :-) However neither gets 100% chance of winning and one still loses. They both can't win 66% of the time however. The problem is that the host can now only open one door if he is not to pick one already selected, and since his door now has a 1/3 chance of winning (rather than none as in the original game where he must always select one that has a goat) NO new information is provided if his is actually a goat/donkey. Therfore both contestentants now have a 50% chance of winning whether they switch or not. This in fact seems far more logical and thus obvious IMO. Trevor. |
|
#11
Posted to sci.electronics.repair,rec.audio.pro
|
|||
|
|||
|
another puzzler
On Fri, 20 May 2011 17:49:12 -0700, "Bill Graham"
wrote: Dave Platt wrote: In article , Bill Graham wrote: I claqim there are two games. In the first game, you go to the studio, pick a door, and then go home to wait and see if they call you and tell you that you either won or lost. Your odds are only 1/3 of winning this game. But if you play the second game, then you go to the studio and mess around until the host opens up a door and shown you the donkey behind it. then you can play the game with 50-50 odds of winning. The only thing I have trouble explaining is why, in order to play this second game with the better odds, you have to switch doors. But, in fact, you do have to switch in order to switch games and take advantage of the better odds. Yup. The distinction between the two games is based on the amount of information available to you at the moment you make your final decision. In the first game, the only information you have is that every door available to you to choose has a 1/3 chance of being correct. In the second game, additional information is given to you by the host, after you make your initial decision and before you make your second one. "The prize is *not* behind that door over there." Actually, you don't *have* to switch doors to "play the second game". You're playing it from the moment the host gives you this extra information. It's just that if you ignore this extra information, and *think* you're still playing the first game (as many people do), you are more likely than not to make a poor choice in the decision you make in this second game. Suppose for the moment that there are two contestants. One picks door two, and the other picks door one. Then the moderator opens door three and shows everyone that there is a donkey behind that door. Now, will it make any difference if the other two switch their initial picks or not? And, if they do swap doors, with they both enjoy a 2/3 chance of winning? In this scenario, once the third door is opened, they each have a 50/50 chance of winning, and there is no advantage in swapping choices. Before the revelation their chances of winning were 1/3. If it seems illogical that the odds are changed by this revelation, remember that one time in three the host will reveal, not a goat, but the car. d |
|
#12
Posted to sci.electronics.repair,rec.audio.pro
|
|||
|
|||
|
another puzzler
Don Pearce wrote:
On Fri, 20 May 2011 17:49:12 -0700, "Bill Graham" wrote: Dave Platt wrote: In article , Bill Graham wrote: I claqim there are two games. In the first game, you go to the studio, pick a door, and then go home to wait and see if they call you and tell you that you either won or lost. Your odds are only 1/3 of winning this game. But if you play the second game, then you go to the studio and mess around until the host opens up a door and shown you the donkey behind it. then you can play the game with 50-50 odds of winning. The only thing I have trouble explaining is why, in order to play this second game with the better odds, you have to switch doors. But, in fact, you do have to switch in order to switch games and take advantage of the better odds. Yup. The distinction between the two games is based on the amount of information available to you at the moment you make your final decision. In the first game, the only information you have is that every door available to you to choose has a 1/3 chance of being correct. In the second game, additional information is given to you by the host, after you make your initial decision and before you make your second one. "The prize is *not* behind that door over there." Actually, you don't *have* to switch doors to "play the second game". You're playing it from the moment the host gives you this extra information. It's just that if you ignore this extra information, and *think* you're still playing the first game (as many people do), you are more likely than not to make a poor choice in the decision you make in this second game. Suppose for the moment that there are two contestants. One picks door two, and the other picks door one. Then the moderator opens door three and shows everyone that there is a donkey behind that door. Now, will it make any difference if the other two switch their initial picks or not? And, if they do swap doors, with they both enjoy a 2/3 chance of winning? In this scenario, once the third door is opened, they each have a 50/50 chance of winning, and there is no advantage in swapping choices. Before the revelation their chances of winning were 1/3. If it seems illogical that the odds are changed by this revelation, remember that one time in three the host will reveal, not a goat, but the car. Only if the host opens a door at random, which isn't the case in the classic Monty Hall problem. The host knows where the car is, and always opens one of the other doors. -- Tciao for Now! John. |
|
#13
Posted to sci.electronics.repair,rec.audio.pro
|
|||
|
|||
|
another puzzler
On 5/20/2011 8:49 PM, Bill Graham wrote:
Suppose for the moment that there are two contestants. One picks door two, and the other picks door one. Then the moderator opens door three and shows everyone that there is a donkey behind that door. Now, will it make any difference if the other two switch their initial picks or not? And, if they do swap doors, with they both enjoy a 2/3 chance of winning? This is called the Monty Hall problem. And yes, switching does change the probabilities of winning, and this can be proved with a simple computer program. I use this in both my math classes when we cover probabilities, and in my programming classes. -- I'm never going to grow up. |
|
#14
Posted to sci.electronics.repair,rec.audio.pro
|
|||
|
|||
|
another puzzler
In article , PeterD
wrote: On 5/20/2011 8:49 PM, Bill Graham wrote: Suppose for the moment that there are two contestants. One picks door two, and the other picks door one. Then the moderator opens door three and shows everyone that there is a donkey behind that door. Now, will it make any difference if the other two switch their initial picks or not? And, if they do swap doors, with they both enjoy a 2/3 chance of winning? This is called the Monty Hall problem. And yes, switching does change the probabilities of winning, and this can be proved with a simple computer program. I use this in both my math classes when we cover probabilities, and in my programming classes. Did you just wake up from a long winter's nap? |
| Reply |
| Thread Tools | |
| Display Modes | |
|
|
Similar Threads
|
||||
| Thread | Forum | |||
| Tall black people continue to push not tall black people around | Audio Opinions | |||
| people helping people | Pro Audio | |||
| FS:dbx, Valley People etcFor Sale 1 - dbx 586 Tube Mic Preamp•Perfect condition in original box $400.00 1-Valley People Dynamite •Very good condition$300.00 1 - Valley People 415 DeEsser$250.00 •Great DeEss | Marketplace | |||
| Note to the politically challenged in RAO | Audio Opinions | |||
| Some people have had enough... | Audio Opinions | |||
Hybrid Mode
