A brain teasing probability puzzle

Szlia

Szlia

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Araysar: if one uses your logic and apply it to the 1,000,000 doors versions, would you still say that the first choice does not matter because you always end up with

1 door picked (unkown)
1 door closed (unkown)
999,998 doors opened (goats)

making it a 0.5 chance of winning by switching?


Melvin: if the host had a hat and could chose to leave it on or remove it when the player's initial pick is the door with the car behind it, would it suddenly make it a better strategy for the player to always stay with his initial pick?
 
Neki

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Also, how is the original question was worded poorly?

The original question remains the same. Once one of the three boxes is revealed that is not the jackpot/desired box, you should switch to the other unrevealed box because statistically, your odds has just doubled from 33.3% to 66.6% to making the correct choice. I see nothing wrong in the wording of the OP.
 
khalid

khalid

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Well, this is more about a "gotcha" than a probability question. It isn't too shocking that otherwise smart people act completely deranged when their intuition turns out wrong. People hate to have been "gotten". It has nothing to do with intelligence or even math background, both which often fool people on this question.
 
Sabbat

Sabbat

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Original question and the poll question are fine, and the answer is yes, switching will increase your chances.

Araysar is both right and wrong, which clearly shows how this teaser works. He is right that the final switch choice leaves the player with a perceived 50/50 chance, because it IS just a binary choice, pick one and win, or pick the other and lose...

Araysar is wrong because the question is not what chance you have of picking between two options, it's about picking between three options and then being given a chance to switch. It's a very large difference.
 
G

Gopher Yerguns_sl

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This is glorious. There's always one guy who keeps arguing about this even after being shown repeatedly why he is wrong, and I love that it's Araysar who is the forum tard.
 
Malakriss

Malakriss

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You pick A
[A] [B C]
33% vs 66%

Host turns [B C] into single choice [X]
[A] [X]
33% vs 66%

Do you want to stick with A or switch to X?
 
Araysar

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Malakriss_sl said:
You pick A
[A] [B C]
33% vs 66%

Host turns [B C] into single choice [X]
[A] [X]
33% vs 66%

Do you want to stick with A or switch to X?
Click to expand...
there is never a 33/66 choice in the original round, the choice is always 0/0

you never win anything in 1st round and you never lose. and the options presented to you in 2nd round are always the same. it is if the first round never existed to begin with, it affects nothing. it is pure theater to muddle up the problem. and you morons are all falling for it hook, line and sinker.
 
Tuco

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Araysar_sl said:
she offers a perfect example


its marvelously amazing how many of you are so fucking stupid, go get refunds all of you, you are fucking dumber than my cat's shit.

everyone of you, go get refunds on your "math" and "statistics" degrees.

your choice at first selection is never 1 in 3, or 2 in 3, or 3 in 3, or whatever the fuck you want to believe. your first choice is ALWAYS invalidated by the host because at second selection he ALWAYS presents you with a 1/2 choice of exactly same options regardless of what you picked in your first turn thus your choice at first selection is never 1/3, even if you pick the right door. your choice at first selection is actually 0/0.
the host never turns around and says, you picked the right door on your first try so you win, he always offers you an alternate. and the alternate is always. one door you selected, a goat door, and an unknown door, which means your selection is always 50/50

you are all so fucking stupid, how the fuck do you even live with yourselves.

holy shit, i am gonna edit this post again and ask how is it possible that so many of you are so fucking stupid.


i should start using as an interview question to weed out all the dumb motherfuckers such as yourselves
Click to expand...
Maybe if you throw some more insults in there you'll convince yourself you are right.
 
iannis

iannis

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Araysar_sl said:
there is never a 33/66 choice in the original round, the choice is always 0/0

you never win anything in 1st round and you never lose. and the options presented to you in 2nd round are always the same. it is if the first round never existed to begin with, it affects nothing. it is pure theater to muddle up the problem. and you morons are all falling for it hook, line and sinker.
Click to expand...
If the first round never existed it's an entirely different question.

So in that context... sure. I guess. 2+2=4!=17
 
Araysar

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Tuco_sl said:
Maybe if you throw some more insults in there you'll convince yourself you are right.
Click to expand...
sorry i was kinda drunk last night. you are still all idiots but i didnt need to repeat it 5 times.

my point stands however. the 1/3 choice is an illusion. there's never a 1/3 choice because you never win in 1st round regardless of what you select - and the set up for the 2nd round is always the same. the options are always the same regardless of your choices in first round. the choice has always been and always will be binary 50/50
 
hodj

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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
 
Tuco

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Araysar_sl said:
sorry i was kinda drunk last night. you are still all idiots but i didnt need to repeat it 5 times.

my point stands however. the 1/3 choice is an illusion. there's never a 1/3 choice because you never win in 1st round regardless of what you select - and the set up for the 2nd round is always the same. the options are always the same regardless of your choices in first round. the choice has always been and always will be binary 50/50
Click to expand...
Are the odds in Lejina's 1million-1 goat example also 50/50?
hodj_sl said:
Isn't the idea that chance is cumulative fallacious?
Click to expand...
Nah, the idea is that when you switch you're the getting the best of two doors. In Lejina's example you're getting the best of 999,999 doors.
 
Araysar

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Tuco_sl said:
Are the odds in Lejina's 1million-1 goat example also 50/50?
Click to expand...
dont know and dont care to do the math since now you are changing the problem

is 1+1=2?

oh ok, what about 1+1,000,000?

its silly.
 
Hachima

Hachima

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If the rules were changed so the remaining 2 choices were randomized after removing a goat, then it would be 50/50. The key is you change, not pick from from a new scenario.
 
Araysar

Loser Araysar

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Hachima_sl said:
If the rules were changed so the remaining 2 choices were randomized after removing a goat, then it would be 50/50. The key is you change, not pick from from a new scenario.
Click to expand...
the key is that the first choice is never really a choice and you all think it is. the first choice affects NOTHING. regardless of what you choose, it creates the exact same scenario for 2nd choice. in 2nd scenario you dont pick 1 out of 3 anymore, you make an implicit 50/50 selection. either stay with your original selection (which is a choice in itself) or switch to the remaining one door. the problem never presents to you a third option because the host always reveals to you the third option. the 3 choices are a pure illusion. there are only 2 choices.

the sooner you realize that the first choice is actually all theater and the odds are 0/0 instead of 1/3 the better off you will be.
 
Hachima

Hachima

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For the non trolls
smile.png

What are your chances of picking a car on the first round?
If you always change, is there any other scenario other than picking the car the first round that would result in you losing?
 
Araysar

Loser Araysar

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Hachima_sl said:
For the non trolls
smile.png

What are your chances of picking a car on the first round?
Click to expand...
0/0

you can never win in 1st round
and you can never lose in 1st round

all you can do is be presented with the illusion of choice.
 
hodj

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Hachima_sl said:
If the rules were changed so the remaining 2 choices were randomized after removing a goat, then it would be 50/50. The key is you change, not pick from from a new scenario.
Click to expand...
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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