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Kevinh5
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Also, Monty doesn't give the probability from D3 ONLY to D2. He gives that information equally to D1 and D2.

So he gives D3 = 1/3 half to D1 (1/6) and half to D2 (1/6), leaving P(D1=car) = 3/6 and P(D2 = car) = 3/6, exactly even.

It isn't possible for Monty to give all of D3's probability to any one other door. The probability, once revealed to not be a car, is divided equally among the remaining doors. That's how math works. Both sides of an equation. Which is why it cancels out, or is worthless, for each side of the equation.
Kevinh5
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Lobowolfxxx, it doesn't matter whether Monty shows bias or not. Of all of the times I choose D1, half the time he reveals D2 and half the time he reveals D3.
LobowolfXXX
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It matters quite a bit as to the likelihood of getting the car vs. getting the goat.
"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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If he shows no bias, he is guessing, the same as I. He will reveaal D2 1/2 of the time, and D3 1/2 of the time. (He will inadvertently reveal the car once out of 3 times and a goat 2 out of 3 times, as Tomas wrote.)

If he knows where the car is, thus has bias to show a goat, he will reveal D2 1/6 + 1/3 = 3/6ths of the time, and he will reveal D3 1/6 + 1/3 = 3/6th of the time.

3/6 = 1/2. Same. It matters not whether he shows bias.
Kevinh5
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Where the car is is independent of my guess, his guess, and he will never show the car if he knows where it is, but the goat information is a red herring and worthless.
LobowolfXXX
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Quote:
On 2012-10-15 02:17, Kevinh5 wrote:
Of all of the times I choose D1, half the time he reveals D2 and half the time he reveals D3.


The question (assuming a biased choice on Monte's part) isn't how often he reveals D2 vs. how often he reveals D3; it's how often he reveals D2 (given that D1 was the door initially selected) when D1 has the car vs. how often he reveals D2 when D3 has the car.

Assuming D1 was chosen, when D1 has the car, he reveals D2, which has the goat, 1/2 of the time.
Assuming D1 was chosen, when D3 has the car, he reveals D2 every time, because he has no alternative.

Thus, it's twice as likely that D3 has the car, given that D1 was chosen and D2 was revealed.
"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 00:57, Kevinh5 wrote:
Those who believe the odds increase to 2/3 by switching are making a serious flaw in logic.

Kevin: I'm sorry, but you're wrong. You're the one making the serious flaw in logic.

Quote:
On 2012-10-15 00:57, Kevinh5 wrote:
Let P(Door# = car) represent the probability that door number X contains the car.

In the beginning
P(1=car) = 1/3
P(2=car) = 1/3
and
P(3=car) = 1/3

So far, so good, but your statement is incomplete. Not only are all of those probabilities equal to 1/3 at the beginning (as you state), but they remain 1/3 throughout. Those probabilities are established by the people who hide the cars behind the doors; nothing that comes after the car is in place affects those probabilities.

This is the single, most important fact in the analysis.

Quote:
On 2012-10-15 00:57, Kevinh5 wrote:
I choose door number 1.

Good choice; I'd have done the same.

Quote:
On 2012-10-15 00:57, Kevinh5 wrote:
Door number 3 is revealed to be a goat. Monty asks "do you want to change your choice?"

Yup.

Quote:
On 2012-10-15 00:57, Kevinh5 wrote:
Here is where the logic gets flawed by so many. Because door 3 has been revealed, they believe that by changing their choice to Door 2, they get to add the probability from Door 3 to the probability of Door 2, but by keeping Door 1 choice, they are left only with the original probability of Door 1.

No, that's not what "they" think. (At least, that's not what I think, nor any other mathematician of my acquaintance.)

What we think is that we now have a new problem. We are not computing P(1 = car) and P(2 = car). We are computing P((1 = car) and (Monty showed us that 3 = goat)) and P((2 = car) and (Monty showed us that 3 = goat)). The original probabilities are no longer in play; we have a new problem with new probabilities.

It's not a matter of simply adding P(2=car) and P(3=car), although on the surface it looks as though that's what's happening.

Quote:
On 2012-10-15 00:57, Kevinh5 wrote:
Hogwash. If you're going to add P(3=car), you'd have to add to both choices, D1 and D2.

Here's where your logic breaks down. You say that they have to do this (add the probability to both), but you give no mathematical justification for it. Calling it hogwash isn't an argument, much less a proof.

Quote:
On 2012-10-15 00:57, Kevinh5 wrote:
Thus P(1 or 3 = car) = 2/3, and
P(2 or 3 = car) = 2/3.

Sorry, it just doesn't work this way.

Quote:
On 2012-10-15 00:57, Kevinh5 wrote:
Had Monty asked BEFORE revealing D3 = goat, whether you wanted to keep D1 or trade for BOTH D2 + D3, then I agree, switching gives you a 2/3 chance of winning. But this is because you get 2 doors instead of only 1.

True, but irrelevant.

Quote:
On 2012-10-15 00:57, Kevinh5 wrote:
Once the reveal is made that D3 = goat, D3 is worthless, thrown out of the equation, and all that matters is D1 and D2.

Absolutely true. Nobody is arguing that there are any choices but these two. That is not the flaw in your reasoning. The flaw in your reasoning is that you believe that because there are two choices, they have to have equal probability. In fact, they don't.

Quote:
On 2012-10-15 00:57, Kevinh5 wrote:
Sorry, but the experts got it wrong.

No, sir, we haven't. Seriously: you're better off switching, and if Monty isn't biased, you're better off by 2:1.

Quote:
On 2012-10-15 00:57, Kevinh5 wrote:
Anyways, one of the links above clearly stated Answer 2 was that both D1 and D2 had a 1 out of 2 chance of being correct.

If it did, it's wrong.

Quote:
On 2012-10-15 00:57, Kevinh5 wrote:
Which is what I concluded before reading everyone elses flawed logic.

Indulge me a bit, if you will; I'll give it one more try.

Suppose that you play this game six times. To make it simple, you always start by choosing door 1. Unbeknownst to you, the cars are distributed thus:

1. Car behind door #1.
2. Car behind door #2.
3. Car behind door #3.
4. Car behind door #1.
5. Car behind door #2.
6. Car behind door #3.

Each of these situations is equally probable: P(1) = P(2) = P(3) = 1/3.

What's Monty going to do?

In game 1 and game 4, he can show you either door #2 or door #3. He's unbiased, so he'll show you one one time and the other the other time; let's say that he shows you door #2 in game 1, and door #3 in game 4.

In game 2 and game 5, he has to show you door #3. In game 3 and game 6, he has to show you door #2.

Now, let's continue with your scenario: you picked door #1 (already covered), and Monty showed you a goat behind door #3. Thus, you weren't playing game 1, nor game 3, nor game 6: in those games, he shows you door #2, not door #3. Once he shows you door #3, you know that you're playing game 2, or game 4, or game 5, and all of those games are still equally likely. You win in game 4 by not switching, but you lose in game 2 or game 5 by not switching; you win 1/3 of the time by not switching. You lose game 4 by switching, but you win game 2 or game 5 by switching; you win 2/3 of the time by switching.

That's the proper analysis. If it doesn't persuade you, I've failed. But it's not faulty.
Kevinh5
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I'll study that. But not tonight. This goat is tired. Thank you.
LobowolfXXX
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Just grab a deck of cards and a friend...20 minutes ought to convince you. Take the ace of hearts as the car, and the twos of spades and clubs as the goats. Friend mixes them up, you pick one, and friend shows you a black two afterward. Do it 10 minutes keeping your initial choice, then 10 minutes switching.

You'll find (given unmarked cards and a friend with a poker face) than if you keep your initial choice every time, you'll win the 1/3 of the time that you were right the first time, and lose the 2/3 of the time you were wrong the first time.

You'll also find that if you switch every time, you'll win the 2/3 of the time that you were right the first time (since he kindly exposes the other losing card for you), and lose the 1/3 of the time you were right the first time.
"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 02:39, Kevinh5 wrote:
I'll study that. But not tonight. This goat is tired. Thank you.

Amen!

'Night!
LobowolfXXX
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Too late to edit: If you switch every time, you'll win the 2/3 of the time that you were WRONG the first time...
"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 02:41, LobowolfXXX wrote:
You'll also find that if you switch every time, you'll win the 2/3 of the time that you were wrong the first time (since he kindly exposes the other losing card for you), and lose the 1/3 of the time you were right the first time.

Fixed.
Kevinh5
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But a flaw I already see is that we could be playing game 1, and this is the time that Monty chose to reveal door 3 not door 2. Monty is indifferent to which he chooses. He doesn't HAVE to choose to reveal Door 2 just because you want to call this game 1.
LobowolfXXX
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Now, about the water level in a swimming pool if you drop a brick off of a raft and it sinks to the bottom of the pool...
"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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Smile
LobowolfXXX
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Quote:
On 2012-10-15 02:42, Kevinh5 wrote:
But a flaw I already see is that we could be playing game 1, and this is the time that Monty chose to reveal door 3 not door 2. Monty is indifferent to which he chooses. He doesn't HAVE to choose to reveal Door 2 just because you want to call this game 1.


If he had 100 doors and showed you 98 of the 99 goats, would you think your chances were 50-50 when it got down to the final 2 doors?
"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 02:42, Kevinh5 wrote:
But a flaw I already see is that we could be playing game 1, and this is the time that Monty chose to reveal door 3 not door 2. Monty is indifferent to which he chooses. He doesn't HAVE to choose to reveal Door 2 just because you want to call this game 1.

But he will choose door #2 some of the time, and that's the crux of the matter; he'll always show you door #3 when he has to (games 2 and 5), but he'll show you door #3 only sometimes when he has a choice (games 1 and 4). That's exactly the reason that

P((1=car) and (Monty shows you a goat behind door #3))
and
P((2=car) and (Monty shows you a goat behind door #3))

aren't equal, and that's the point: you're concluding that they are equal.
Kevinh5
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There is only one goat left, correct? There are two doors. The probability is exactly equal that either door hides a goat.
owen.daniel
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Kevin.

Let's move away from the description in terms of "giving the probability behind one door to another door"; when I first read that statement in an earlier post I thought that it was a nice way to visualise it, but there is no rigour in that description.

You say that in this case the experts have gotten it wrong, in fact they haven't, and the reason why it is generating so much debate here is because it is an excellent example of how we fail to appreciate conditional probabilities in day to day life. In general, conditional probabilities (as the names suggest) study the probability of a particular thing happening, given that you have information about something else. This is normally written P[A |B], and read "the probability of A given B".

If A and B have nothing to do with one another (i.e. they are independent) then P[A |B ] = P[A]. Take for example the probability I toss a coin and it lands heads, given that I previously rolled a 5 on a dice. The dicee has nothing to do with the coin, so P[head |5 on dice ] = P[head] 1/2. If instead the events are related then this is no longer true. Consider instead an experiment in which I toss two coins; now the probability P[ both are heads |the first coin is a head ] definitely depends on both coins. However, if you assume that the first coin is a head, then the probability that the second one is a head is a half, and the answer is P[ both are heads |first is a head ] = 1/2. Note how this differs from P[ both are heads ] = 1/4.

Now we are going to need a result about conditional probabilities. Suppose that we have a sequence B1, B2, B3,...Bn of events which cover all possible outcomes (so for instance if we are doing the two coin tossing experiment one such sequence would be B1 = both coins heads, B2 = both coins tails, B3 = one heads, one tails). Then there is a law in probability which says that:

P[A] = P[B1]*P[A |B1 ] + P[B2]*P[A|B2] + ... + P[Bn]*P[A|Bn]

Right, so what has this got to do with Monty Hall. In our case we'll let A be the event that you win, and let B be the event that you switch. So what the monty hall problem claims is that P[A |B ] = P[you win, given that you switch] = 2/3. Now I'm going to show why this is true from the probabilistic point of view, but beware, it becomes a bit confusing, because we are going to take a conditional probability of a conditional event... i.e. something of the form P[A|B|C]! But this works (this makes more sense if you see A|B as just one statement. So we are now going to define C1 be the event that you originally chose the car, and C2 the event that you chose a donkey. Note that these are the only two possible choices, and so these describe all outcomes, and we can use our rule above

P[A |B ] = P[C1]*P[A|B |C1] + P[C2]*P[A|B |C2]
= P[1st choice is the car ]*P[win given that you switch, and that you first chose the car]
+ P[1st choice is a donkey ]*P[win given that you switch, and that you first chose a donkey]

as you have already shown that you know, P[C1] = 1/3, whilst P[C2] = 2/3. You have also made it clear that you understand that once we have been shown what is behind on door switching either guarantees that we win the car, or guarantees that we don't, which correspond to probabilities 0 and 1 in the conditional statements.

= 1/3 * 0 + 2/3 * 1
= 2/3

Right. That was all quite long, and will probably take some time to read through properly. I suggest you look up conditional probability online and that'll help sort it... There are plenty of places where it will be described from a good level... wikipedia will tend to over complicate things with more general forms of conditional probabilities that you most likely wont be interested in.

Best of luck.

Owen
Kevinh5
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They ARE equal, as I wrote above. 3/6 times D2 and 3/6 times D3.
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