A brain teasing probability puzzle

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Araysar

Loser Araysar

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Tuco_sl said:
Ok now you're just fucking with us. Stop ruining the schaedenfreude I get from these threads with fake misunderstanding!
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how am i fucking with you? you're changing the fundamental structure of the problem and you want me to argue that instead. i'm sure hodj can tell you the specific fallacy for this type of argument.
 
BrutulTM

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Your first choice is more likely to be wrong than right. (2/3 chance that you choose a wrong box)

The other wrong box is eliminated.

The remaining box is more likely to be right than wrong BECAUSE your first choice was more likely to be wrong than right. (1/3 chance that the remaining box is wrong)

I love how Araysar thinks he has outsmarted the world's mathematicians on a classic problem. That is the kind of stupidity needed to keep things like the politics thread going year after year.
 
Araysar

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hodj_sl said:
Makes sense, without the new randomization of the choices, you are engaging in a cumulative action of sorts, versus the proverbial coin toss where each toss is a single random event in isolation.
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there's no cumulative action. thats an illusion.

regardless of whether you pick a car or a goat in 1st round, the host never acknowledges that you won or lost. he only presents you with exact same set of options for for 2nd round, regardless of your selection.

The host will always open a goat door (this will always be possible because there's 2 of them) and you are always, ALWAYS faced with exact same choice regardless of any prior actions: the 2 doors. one is always the open goat door that the host opened. one is always your selected door, and one is always the unknown door.

the choice is always 50/50

i see how a lot of people here get confused about it because they mistakenly incorporate the "choice" from the first round into the problem. but there is no probability involved in 1st round because it always artifically manufactures the exact same scenario for the 2nd round, it is pure theater.

Round 1 is The Pledge
Round 2 is The Turn
and then the supposed revelation of 2/3 is The Prestige.

i cant believe you guys are so gullible.
 
Araysar

Loser Araysar

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BrutulTM_sl said:
Your first choice is more likely to be wrong than right. (2/3 chance that you choose a wrong box)

The other wrong box is eliminated.

The remaining box is more likely to be right than wrong BECAUSE your first choice was more likely to be wrong than right. (1/3 chance that the remaining box is wrong)

I love how Araysar thinks he has outsmarted the world's mathematicians on a classic problem. That is the kind of stupidity needed to keep things like the politics thread going year after year.
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how can your first choice be "more likely to be wrong than right." when you NEVER get the result for your first choice? the host never acknowledges whether you won or lost in 1st round and simply reconfigures the problem for 2nd round.

tell you what, do this problem with me 100 times and tell me how many times you won in the 1st round. is it 33/100? NOPE. it is 0/100. you won zero times in 1st round out of 100 selections. but thats ok. because you probably also expected to lose 67/100 times as well. i got good news for you. you didn't. you lost 0/100 times as well.
 
hodj

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Araysar_sl said:
there's no cumulative action. thats an illusion.

regardless of whether you pick a car or a goat in 1st round, the host never acknowledges that you won or lost. he only presents you with exact same set of options for for 2nd round, regardless of your selection.
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I'm pretty sure that's why they're saying its cumulative, rather than two choices in isolation. Because there is no "winner" after the first pick, your choices are cumulative, whereas, if they did declare a winner after the first choice, and then provided you a second set of choices involving only two doors, that would be two separate random actions in "isolation" versus the cumulative nature of the exercise as originally written.

So the way its written, you get a 1/3 chance, then a 2/3 chance, but if they called a stop after the first pull, then you would have a 33% chance the first pull, and a 50% chance the second.

Does what I'm saying make sense? My brain is full of resonance structures right now so I might just be speaking alien and I wouldn't be able to tell.
 
Araysar

Loser Araysar

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Tuco_sl said:
Whether you get the result you're still picking a goat or a car.
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Exactly. a 1 in 2 choice.

tumblr_m230fmFNuF1rnax3yo1_500.gif
 
iannis

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Where are you getting this reconfigure shit from? When did 3 card monte come into this?

Nothing is reconfigured. It's not the shell game. This is a single ongoing instance, not two discrete events.
 
Araysar

Loser Araysar

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hodj_sl said:
I'm pretty sure that's why they're saying its cumulative, rather than two choices in isolation. Because there is no "winner" after the first pick, your choices are cumulative, whereas, if they did declare a winner after the first choice, and then provided you a second set of choices involving only two doors, that would be two separate random actions in "isolation" versus the cumulative nature of the exercise as originally written.


Does what I'm saying make sense? My brain is full of resonance structures right now so I might just be speaking alien and I wouldn't be able to tell.
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cumulative implies that your first set of choices somehow affects your second set of choices.

in this situation it doesnt. as a matter of fact, this situation is exactly opposite of cumulative, they are in fact 2 separate set of choices because the first one has ZERO bearing on the second.


hodj_sl said:
So the way its written, you get a 1/3 chance, then a 2/3 chance, but if they called a stop after the first pull, then you would have a 33% chance the first pull, and a 50% chance the second.
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But they dont say stop. they never call STOP. Thats The Turn.
 
iannis

iannis

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It honestly is pretty amazing though how long Ara can keep it going with semi-credible objections. Grain alcohol makes you strong!

I wouldn't be able to do it. That's fucking dedication.
 
hodj

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Araysar_sl said:
cumulative implies that your first set of choices somehow affects your second set of choices.
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Eh I see your point on this, actually. You're saying the first choice is completely illusory and therefore irrelevant. I don't disagree.

However, when it comes to the question as posed mathematically/statistically, the argument about 33%->66% probability is also valid

Yours is more a philosophical attack on the validity of the first choice, which I think is a valid objection to the exercise, whereas I think others are making a mathematical argument, hence the disagreement.
 
Araysar

Loser Araysar

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hodj_sl said:
Eh I see your point on this, actually. You're saying the first choice is completely illusory and therefore irrelevant. I don't disagree.

However, when it comes to the question as posed mathematically/statistically, the argument about 33%->66% probability is also valid

Yours is more a philosophical attack on the validity of the first choice, which I think is a valid objection to the exercise, whereas I think others are making a mathematical argument, hence the disagreement.
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Thank you.

But I think the 33/66 is only "valid" because it faultily incorporates the first choice into the equation. But how/why can it do that when we both agreed that the first choice is purely superficial and affects nothing?

I think I am very much right on the mathematical portion of this.
 
Noodleface

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hodj_sl said:
Its been a while since I had a stats class and logic class, and I know we covered this in one or the other, but I can't for the life of me recall.

Isn't the idea that chance is cumulative fallacious?

Like you flip a coin ten times, each time you flip the coin you didn't accumulate a higher chance of getting heads or tails, the probability remains essentially 50/50.

Gambler's fallacy I believe its called, or Monty Carlo fallacy. But maybe it doesn't apply in this scenario because you start with 3 choices...but then it wouldn't apply in casinos where you have things like the roulette wheel...

http://en.wikipedia.org/wiki/Gambler's_fallacy
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If you flip a coin ten times, the probability is 0.50 that you'll get either heads or tails (p(1-p)). In any given ten-coin flip situation, you won't get 5 heads and 5 tails, that isn't what probability suggests, but rather it is the ideal situation that will lead to 0.50 chance. The thought is that if you flipped that same coin 10 million times, the average will be much closer to 0.50 then if you flip it ten times (in reality). Sample sizes and some shit. Fuck if I know. I'm just some dude.
 
Tuco

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Araysar_sl said:
Thank you.

But I think the 33/66 is only "valid" because it faultily incorporates the first choice into the equation. But how/why can it do that when we both agreed that the first choice is purely superficial and affects nothing?

I think I am very much right on the mathematical portion of this.
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Maybe Araysar isn't trolling since now he's backpeddling. I like how he puts valid in quotations to soften the blow of admitting he's full of shit.
 
hodj

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The problem with objecting to the mathematical validity of the problem is that people have actually proofed it mathematically and with structural forms of logic to show its true.

http://math.ucr.edu/~jdp/Monty_Hall/Monty_Hall.html

Noodleface_sl said:
If you flip a coin ten times, the probability is 0.50 that you'll get either heads or tails (p(1-p)). In any given ten-coin flip situation, you won't get 5 heads and 5 tails, that isn't what probability suggests, but rather it is the ideal situation that will lead to 0.50 chance. The thought is that if you flipped that same coin 10 million times, the average will be much closer to 0.50 then if you flip it ten times (in reality). Sample sizes and some shit. Fuck if I know. I'm just some dude.
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Yeah I know that, sorry if the way I phrased what I said made it sound like I was saying that a 50/50 chance implies perfect 50/50 results. As you said, its related to sample sizes. Small sample sets are more subject to error and bias, so if you flip a coin ten times you very likely won't see a 50/50 ratio, but the more times you flip it and increase your data set, then the closer the ratio moves towards a 50/50 split.
 
Haast

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Yes, the misunderstanding in this problem is the circumstances of the second choice. If you look at the second choice in a vacuum, it is definitely 50/50. However, when you add in the circumstance that your first choice was more likely to be wrong, it becomes skewed in favor of the switch.

This screwed with my mind until I read theEconomist's explanation. For some reason, it made more sense.
 
Araysar

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Tuco_sl said:
Maybe Araysar isn't trolling since now he's backpeddling. I like how he puts valid in quotations to soften the blow of admitting he's full of shit.
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the premise of what I argued yesterday and today has not changed one iota. go ahead and look at my posts from yesterday to today. they argue the exact same thing

the only backpedalling i see here is from the rest of you - as my explanation is finally dawning on you. you guys have all been tricked and misled by The Turn

the_prestige-11537.jpg
 
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