spamtrap1888 wrote in news: snipped-for-privacy@z15g2000prn.googlegroups.com:
Still trying to get my head around this.
How would shuffling unknown values affect my choice? If I didn't know before and you shuffle the choices, it's still a random choice on my part.
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.
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
I think, several years ago when I originally saw this, I argued as vehemently as you, Arny. It is extremely counter-intuitive, which goes to show intuition isn't always right!
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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
In my studies of this item, I found a statement that about 10% of
*everybody* never gets it, regardless of their intelligence or education.That suggests to me that some people learn things about problem solving that keep them from seeing certain solutions. The trick is to not do that, or if you do, somehow redirect how you approach these things.
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.David
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.
spamtrap1888 wrote in news:88df2861-f695-449b- snipped-for-privacy@34g2000pru.googlegroups.com:
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.
f
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.
Void is not one of the permitted outcomes. Suppose the host accidentally revealed the car - to be equivalent to the intentional goat revelation, he would then have to say "never mind, take the car anyway". That would leave you in the 1/3 2/3 situation. If he reveals a goat by chance the game degenerates to the simple situation - the host has chosen one of the three, and you get to pick between the remaining two, always assuming that he did not pick the car.
The point of the intentional revelation is that by switching you get - in effect - both doors, not just the one.
d*** Start at the beginning of this post and read all of the quoted stuff. The initial assumption is that 'void' IS a permitted outcome. If the void assumption is changed , I concede. David
I don't know why people make the Monty Hall Paradox so complex. I've explained it simply, twice.
All you have to do is understand why the initial probability of getting the big prize is 1/3 -- and everything else falls out in a completely straightforward manner.
OK, try this: blow the game up to 100 doors. Your chance of picking the winning door on the first try is 1 out of 100. Stick with that choice and each time a zonk is revealed the chance of the prize being in the remaining group increases. When you get down to two doors the chance of the prize being behind the other door is 99 out of 100. The chance of your first pick being correct is still 1 out of 100.
Later... Ron Capik
--
Ron Capik wrote in news:XqSdncQ0dq_JzkjQnZ2dnUVZ_q- snipped-for-privacy@giganews.com:
That may have penetrated.
It all hinges on the host knowing in advance which door hides the prize and revealing known (to him) bad choices. I still don't agree with the logic, but I think I follow it.
It's called the "Monty Hall problem"? I'll continue to research.
Thank you for your patience.
Ron Capik wrote in news:XqSdncQ0dq_JzkjQnZ2dnUVZ_q- snipped-for-privacy@giganews.com:
Here's the simple answer that I can understand:
"An even simpler solution is to reason that switching loses if and only if the player initially picks the car, which happens with probability 1/3, so switching must win with probability 2/3"
Eureka! That finally makes sense. And yes, everyone was trying to tell me that.
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.
"Bill Graham" wrote in news: snipped-for-privacy@giganews.com:
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."
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
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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?
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