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LobowolfXXX
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I will take on faith that one day, you'll be able to wrap your head around this admittedly counter-intuitive problem, which has puzzled a great many good minds for a great many years.

Until then, I assure you, you know less about it than you think you do. Given that Monte knows where the goats are, and was always going to show you one, your chances of finding the car double when you switch.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley.

"...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us."
S2000magician
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Quote:
On 2012-10-15 02:48, Kevinh5 wrote:
There is only one goat left, correct? There are two doors. The probability is exactly equal that either door hides a goat.

Again, it isn't.
S2000magician
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Kevin:

Consider this: Suppose that Monty's biased: he always shows you the lowest numbered door than he can. Always.

You pick door #1, and he shows you a goat behind door #3.

Where's the car?
Kevinh5
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I promise I will read each of the above posts from last night and give them serious consideration. But today is the deadline for folks who have filed tax extensions, so I will be too busy to do it at least until tonight, maybe later if I have a drink after work.

And yes, goats kept appearing behind doors in my sleep last night. Sometimes they'd tell me to switch my choice.
Michael Daniels
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For an interactive card trick based on the Monty Hall Principle, see

http://www.mindmagician.org/monty2.aspx

This may (or may not) help in understanding the logic of the original dilemma.

Mike
S2000magician
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Quote:
On 2012-10-15 08:26, Kevinh5 wrote:
I promise I will read each of the above posts from last night and give them serious consideration. But today is the deadline for folks who have filed tax extensions, so I will be too busy to do it at least until tonight, maybe later if I have a drink after work.

And yes, goats kept appearing behind doors in my sleep last night. Sometimes they'd tell me to switch my choice.

So, you're a CPA? Amongst other things, I have a degree in accounting (though I've never taken the CPA exams).

(I avoided the rush: I e-filed mine yesterday.)

Best of luck!
Kevinh5
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No, my license is granted at the Federal level by the United States Treasury - I'm an Enrolled Agent. CPAs are licensed by individual states and must seek reciprocity to do work in other states. Also, they ARE accounting experts, but may choose to specialize in taxation. Enrolled Agents (EAs) may practice worldwide before all administrative levels of the IRS, regardless of their state (or country) of residence. And all EAs are tax experts, albeit they may know little about accounting. Slightly different, but similar fields of work.


OK, it is my lunch time, so I'm allowed to play, and now I agree that if Monty knows where the car is, we can have only 1 out of 6 possible outcomes:


D1 D2 D3
Outcome 1 C G G - Monty will show D3=G
Outcome 2 C G G - Monty will show D2=G
Outcome 3 G C G - Monty will show D3=G
Outcome 4 G C G - Monty will show D3=G
Outcome 5 G G C - Monty will show D2=G
Outcome 6 G G C - Monty will show D2=G

I can now see what you wrote late last night/early this AM: If Monty shows D3=G, then we know we can't have outcomes 2, 5 or 6. I like referring to these as 'outcomes' and not as 'games'; the terminology 'games' was confusing me at 2:30 in the AM.

Now I've already chosen D1. I now agree, of the remaining possible outcomes, I have only a 1 in 3 chance of being correct, even though there are only two doors. The other 2 possible outcomes are that C is behind Door 2.

Thank you everyone, for being so patient with me while I put this into a framework that I could understand. No, I have never had a 'conditional probability' class before, I was a business major in undergrad, and an International Business major for my MBA. We only had one regular probability & statistics class in the masters program.



All of the things I've read so far are based on the assumption that Monty knows where the CAR is. Do we have any empirical evidence of this? Has someone associated with the show actually come out and said that he knew where the CAR was? Or if he was as unknowing as the contestant, did he reveal the car inadvertently 1 out of every 3 times (or with a high degree of confidence that it approached .3333) when he showed the first non-selected door?



Here is another hypothesis:

What about if Monty did not know where the CAR was, but was told where ONE GOAT was?

Contestant always chooses D1, same as before.


D1 D2 D3

C G G - Monty knows D2 = G, so he reveals D2 1 time every 6
C G G - Monty knows D3 = G, so he reveals D3 1 time every 6
G C G - Monty knows D1 = G, he doesn't know whether to reveal D2 or D3, he picks one or the
other half the time or 1 time every 12
G C G - Monty knows D3 = G, so he reveals D3 1 time every 6
G G C - Monty knows D1 = G, he doesn't know whether to reveal D2 or D3, he picks one or the
other half the time, or 1 time every 12
G G C - Monty knows D2 = G, so he reveals D2 1 time every 6


I think the probabilities change based on partial knowledge instead of perfect knowledge.


What say you math-a-magicians?




I'd really like to know what the frequency of D2 reveals was compared to D3 reveals when D1 was chosen, as well as how many times the D2 or D3 reveal was the car. Has anyone read anything about that to find out what REALLY happened? Or are we all just curious/content enough to prove that conditional probability is tricky stuff?
Kevinh5
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I can't get the proper formatting to make the tables appear correctly.
Kevinh5
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If Monty knows nothing at all, then my logic from last night applies. And an examination of the actual outcomes from the show should reveal statistically what, if anything, he knew, assuming his knowledge was consistent throughout the show's run.

I was too young at the time to even think about trying to remember each show.
S2000magician
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Quote:
On 2012-10-15 12:58, Kevinh5 wrote:
No, my license is granted at the Federal level by the United States Treasury - I'm an Enrolled Agent. CPAs are licensed by individual states and must seek reciprocity to do work in other states. Also, they ARE accounting experts, but may choose to specialize in taxation. Enrolled Agents (EAs) may practice worldwide before all administrative levels of the IRS, regardless of their state (or country) of residence. And all EAs are tax experts, albeit they may know little about accounting. Slightly different, but similar fields of work.

We have an EA do our taxes for us.

Don't work too hard (on taxes) today!
S2000magician
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Quote:
On 2012-10-15 12:58, Kevinh5 wrote:
All of the things I've read so far are based on the assumption that Monty knows where the CAR is.

Yup, Monty knows.

Quote:
On 2012-10-15 12:58, Kevinh5 wrote:
Do we have any empirical evidence of this?

Ten million shows and Monty never once opened the door to reveal the car. Either he's (incredibly, amazingly, miraculously) lucky, or else he knows. I'd bet on the latter.

Quote:
On 2012-10-15 12:58, Kevinh5 wrote:
Has someone associated with the show actually come out and said that he knew where the CAR was?

I don't know. But the empirical evidence is enough to convict.

Quote:
On 2012-10-15 12:58, Kevinh5 wrote:
Or if he was as unknowing as the contestant, did he reveal the car inadvertently 1 out of every 3 times (or with a high degree of confidence that it approached .3333) when he showed the first non-selected door?

Nope, he never revealed the car.

He knew.
S2000magician
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Quote:
On 2012-10-15 13:00, Kevinh5 wrote:
I can't get the proper formatting to make the tables appear correctly.

That's the Café's fault, not yours.
S2000magician
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Quote:
On 2012-10-15 13:09, Kevinh5 wrote:
If Monty knows nothing at all, then my logic from last night applies.

Agreed. But that's not the Monty Hall Problem. In the MHP, he knows. (That's explicit in the problem statement, if not in the TV show.)
LobowolfXXX
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It would really be a bad show if you picked D1, then he revealed the car behind D2 and asked if you wanted to switch.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley.

"...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us."
LobowolfXXX
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Cute and related: You roll 2 dice behind a screen. By agreement (and nobody's cheating), if neither die shows a 1, the roll is cancelled, and you roll again. If you DO roll a 1 (at least one), then (without rotating it to change the face), I slide a die showing a 1 from behind the screen, so you can see it, and leave the second die behind the screen.

What are the chances that the second die is showing a 1?
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley.

"...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us."
Kevinh5
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And it would be a very boring show if they started with 100 doors and revealed 98 goats one at a time too. After the first reveal of one goat, it is anti-climatic to reveal 97 more one at a time. Might as well reveal them 5, 10, or heck, even 97 at a time and get to the chase: switch for the last choice, or keep your original choice.



S-2000, if your tax pro is an EA, there is a high probability that he/she knows me or has taken one of my classes. I'm big in the professional continuing education circuit.
Kevinh5
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If Monty never revealed the car but always revealed a goat, I agree: he had to know.



Who wants to take my challenge and figure out how much fun it would(n't) be if Monty only knew where ONE GOAT was?
Kevinh5
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Interesting question, LoboXXX. We all know that a die has 6 sides, but we're not asking the probability of one die being a one-spot. We're asking the probability of both die being one-spots. And in light of what we just learned, we're asking the probability of the 2nd die being a one-spot given that die number 1 was a 1 spot.

I'll think about your puzzle if you'll think about mine.
S2000magician
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Quote:
On 2012-10-15 13:46, LobowolfXXX wrote:
It would really be a bad show if you picked D1, then he revealed the car behind D2 and asked if you wanted to switch . . .

. . . to door 3.
Kevinh5
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There are 11 ways to have at least 1 one-dot showing, but only 1 way to have 2 one-dots showing.

Given that 1 one-dot has been shown, what is the probability that the other die is also a 1? I think this is the question.
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