A brain teasing probability puzzle

Siddar_sl said:

Monty only removes goats though. Monty will only end up holding a goat if you chose the car in a 1 in 1000 pick. Monty chooses second and auto holds the car if you didn't choose it. His odds of getting car are 99.9% and yours are .1%. Then you get to decide if you want your box are Monty's box in round two.

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Of course he always remove the goats.

Is the final proposition anything other than 1 goat and 1 car? Why would you think that any of your previous choices or actions even matter? Everything is predetermined for you, you're simply going through the motions in round 1 and monty is going through the motions in intermission.

The final choice is always 1 goat, 1 car, and nothing you or him choose or decide has any impact on what the setup is on the final choice unless you alter the entire rules of the game themselves.

Even if you selected a car in Round 1 out of 1000 doors, it doesnt mean anything. your choice is nullified when you're forcibly moved to round 2 and forced to make a choice again. a binary choice this time.

Why are you guys so upset? I'm simply trying to teach you all something new.

Araysar_sl said:

Why are you guys so upset? I'm simply trying to teach you all something new.

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Tea_sl said:

If you have a group of random people in a room how many do you need before there is a 50/50 chance that there is a shared birthday?

The answer may surprise you.

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No one answered. Can't be assed to cipher it out. What's the answer?

Dabamf_sl said:

The wording of the problem is stupid because it doesn't say "the host will always pick a goat." That's important and the reason I, and probably lots of others, misconceptualized it.



The second set of choices is not always the same.

Scenarios of switching, assuming you always pick A first (because the first choice is meaningless)
A is car, B/C are goats. You switch and lose (1/3 time)
A is goat, B is car, C is goat. Host shows C (he's forced to, it's the only goat option), you switch to B, you win (1/3 of the time)
A is goat, B is goat, C is car. Host shows B (he's forced to, it's the only goat option), you switch to C, you win (1/3 of the time)
Win 2/3, lose 1/3

Scenarios of not switching
A is car, you win
B is car, you lose
C is car, you lose
Win 1/3, lose 2/3

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The problem could just as well be set up as: There are 3 boxes, one containing a new car, 2 containing a $1 bill. The Host asks if you want a snack. No matter what your answer is, he eliminates one of the $1 bill boxes. The first choice has no real impact on the second, unless you wanted a specific $1 bill.

DiddleySquat_sl said:

Oh shit, Araysar isserious? That changes the premise totally.

Ara, I love you (non-homo way) but here you're wrong. The first pick is not meaningless, it changes the options for the game master in that he in some cases now is limited in his choice of which door to open. That in itself is information that changes the equation.

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How does it meaningfully change his options? You guys are acting like it matters which goat he eliminates. People keep saying the first choice informs the second choice, but how? Unless he has a tell, you don't have any more information.

Fawe_sl said:

There is no illusion.

1 door that you selected has 1/3 chance of being a car
1 door that is unknown has 2/3 chance of being a car

You now have an option: Keep your 1/3rd chance or switch to take the 2/3rd chance.

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No. The problem is that the 1/3 and 2/3 comes from the first round choice. Yeah, in the first round, you had a 1/3 chance of picking the car door, but it didn't matter. This is nothing more than a statistical fun fact. In the second round, you are making another choice, and this one is 50/50. Whether the choice is door 1 vs door 2, or same vs switch, it's a 50/50 choice.

I saw the mythbusters link where they verified this myth, but they are extremely suspect as a general rule. I saw in the same episode that they also think they busted the myth of being able to inflate a tire with starting fluid and a match. This is something I've seen done numerous times, so I know they must have been doing it wrong. What I'd like to know now, is what were the real life results from "Lets Make a Deal". It seems like he offered this choice often enough over the years that we should be able to see if there was an advantage one way over the other. Anyone got a link where someone has compiled those results?

In theory, there is no difference between theory and practice. But in practice, there is.

Tuco_sl said:

for fun:

Start:
ManA: $10
ManB: $10
ManC: $10
Hotel: $0
Bellhop: $0

Pay for Lodging:
ManA: $0
ManB: $0
ManC: $0
Hotel: $30
Bellhop: $0

Hotel Refund Lodging:
ManA: $1
ManB: $1
ManC: $1
Hotel: $25
Bellhop: $2

End delta:
ManA: -$9
ManB: -$9
ManC: -$9
Hotel: +$25
Bellhop: +$2
=
0

No problem.

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I'm shocked that you see the fallacy in this problem, but don't see it in the monty hall problem.

Hoss_sl said:

No one answered. Can't be assed to cipher it out. What's the answer? .

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for more:http://en.wikipedia.org/wiki/Birthday_problem

It's a sad day when forum groupthink leads to the independent thinker being cut from the herd and ostracized.