A brain teasing probability puzzle

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khalid

khalid

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Long discussed puzzle and a good example of how humans just aren't good at conditional probability. However, stupid also because on an actual gameshow you aren't dealing with unbiased chance. They are setting it up and can manipulate if he opens up a box or just takes your first choice based on whether you picked wrong or not.
 
T

The Master

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McCheese_sl said:
I think this was discussed at length a long time ago on FoH, and despite reading all of that discussion and the wikipedia entry below, I still don't understand it. fuck statistics, probability, math, and generally anything involving numbers!

http://en.wikipedia.org/wiki/Monty_Hall_problem
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There are two bad boxes and one good box. If you pick the bad box (2/3 chance) the host opens the other bad box. There is now a bad box (which you picked already) and a good box. Switch wins in this scenario. You pick the good box. Host opens either bad box. Leaving one good box which you picked and a bad box. You switch, you lose. That happens 1/3 of the time. So switching wins 2/3 of the time, based purely on your initial choice.

Some of the variants require more thought, it really comes down to the host's expected behavior.
 
Araysar

Loser Araysar

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The Master_sl said:
There are two bad boxes and one good box. If you pick the bad box (2/3 chance) the host opens the other bad box. There is now a bad box (which you picked already) and a good box. Switch wins in this scenario. You pick the good box. Host opens either bad box. Leaving one good box which you picked and a bad box. You switch, you lose. That happens 1/3 of the time. So switching wins 2/3 of the time, based purely on your initial choice.
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thats bullshit. it doesnt matter what your initial choice was.

what matters is that one box gets opened afterwards (a bad one) and you get to make a choice again between 2 boxes which is 50/50

it doesnt matter if you stick with the original box or choose another one, that is still an implicit 50/50 choice being made
 
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Araysar_sl said:
thats bullshit. it doesnt matter what your initial choice was.

what matters is that one box gets opened afterwards (a bad one) and you get to make a choice again between 2 boxes which is 50/50

it doesnt matter if you stick with the original box or choose another one, that is still an implicit 50/50 choice being made
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You're really not understanding the role the host is playing. Which is the most common mistake in failing to understand the Monty Hall problem. This is really easy to demonstrate iteratively, just go play each scenario. You won't get 50/50. It'll be 66/33.

http://www.youtube.com/watch?v=cXqDIFUB7YU
 
Araysar

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The Master_sl said:
You're really not understanding the role the host is playing. Which is the most common mistake in failing to understand the Monty Hall problem. This is really easy to demonstrate iteratively, just go play each scenario. You won't get 50/50. It'll be 66/33.

http://www.youtube.com/watch?v=cXqDIFUB7YU
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that didnt explain anything. it offers the same convoluted logic as people posting here before.

your initial 1 in 3 choice means nothing. its totally irrelevant

your final choice which is 50/50 means everything.

you only think your odds are improving because you consider the first 1 in 3 choice to mean anything at all.
 
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Because it did. The host's behavior matters, you don't seem to be understanding that. It is an essential part of the problem. It isn't some magical coincidence that if you switch, you win precisely 2/3 of the time. If you were right, it'd come out 50/50. It doesn't, so you aren't.
 
Araysar

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the host is offering you a 50/50 choice. at the end. thats all that matters.

what you are saying makes no sense AT ALL. I'm pretty sure you're just trying to parrot the wikipedia article to me without understanding it yourself.
 
Araysar

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the simulator is going off the wrong premise, assuming that the initial choice matters.

thats the flaw in this problem, the initial choice is actually meaningless.

after your first choice, your options will always be exactly the same, meaning your initial choice is meaningless. regardless of whether you select the car or the goat, there will always be another door with a goat opened and you will always be presented with a 50/50 choice again
 
T

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Because I agree with the Wikipedia article, which happens to be correct, I must be parroting it without understanding it? Are you saying that simply because you don't understand it or...?

I mean clearly this makes no sense to you. But, again, iteratively proving it to yourself is easy. Go actually get three boxes and work out each scenario. There aren't that many. See how many scenarios actually result in winning vs not. It won't be 50/50.
 
Araysar

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The Master_sl said:
Because I agree with the Wikipedia article, which happens to be correct, I must be parroting it without understanding it? Are you saying that simply because you don't understand it or...?

I mean clearly this makes no sense to you. But, again, iteratively proving it to yourself is easy. Go actually get three boxes and work out each scenario. There aren't that many. See how many scenarios actually result in winning vs not. It won't be 50/50.
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Araysar_sl said:
the simulator is going off the wrong premise, assuming that the initial choice matters.

thats the flaw in this problem, the initial choice is actually meaningless.

after your first choice, your options will always be exactly the same, meaning your initial choice is meaningless. regardless of whether you select the car or the goat, there will always be another door with a goat opened and you will always be presented with a 50/50 choice again
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.
 
T

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You realize the simulator is simulating the scenario presented, right? You can't ignore details of the scenario and pretend your answer makes any sense to the scenario.
 
Araysar

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The Master_sl said:
See how many scenarios actually result in winning vs not. It won't be 50/50.
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of course it wont be, because they all assume the initial choice matters. it doesnt.

the first choice always presents you with exact same set of options for your second choice regardless of what you pick. the unknown door you picked, the door with a goat that the host opens and the third unknown door.

since the set of options is exactly the same, that means that the first set of options doesnt matter at all, its just theater.

your choice is 50/50 regardless.
 
Tuco

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It's ironic that the voting on this poll is very close to 66/33 split, which is the probability split of winning if you switch.
 
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Araysar_sl said:
of course it wont be, because they all assume the initial choice matters. it doesnt.

the first choice always presents you with exact same set of options for your second choice regardless of what you pick. the unknown door you picked, the door with a goat that the host opens and the third unknown door.

since the set of options is exactly the same, that means that the first set of options doesnt matter at all, its just theater.

your choice is 50/50 regardless.
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Those aren't the options. Really. Go write down each possible scenario. 2/3 of them result in winning if you switch. The initial choice matters.
 
Araysar

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The Master_sl said:
Those aren't the options. Really. Go write down each possible scenario. 2/3 of them result in winning if you switch. The initial choice matters.
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how can the first choice matter when you are presented with exact same set of options for your second choice regardless of what you choose??

it's the definition of being irrelevant.
 
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Tea_sl

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Araysar_sl said:
the simulator is going off the wrong premise, assuming that the initial choice matters.

thats the flaw in this problem, the initial choice is actually meaningless.

after your first choice, your options will always be exactly the same, meaning your initial choice is meaningless. regardless of whether you select the car or the goat, there will always be another door with a goat opened and you will always be presented with a 50/50 choice again
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???? The simulator is the same as you doing it manually. If you don't believe it still then do it physically with 2 red playing cards and a black playing card with sufficient iterations the number will converge to 66%.
 
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