A brain teasing probability puzzle

BrutulTM

BrutulTM

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Here's a puzzle for you.

3 boxes, one containing a car, two containing goats. You choose one at random. You are then given the choice of keeping your one box, or trading it for both of the remaining boxes. Which do you choose? Is it 50/50 that the car will be in your box and not the other two?
 
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The Master

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Out of curiosity Ara, do you often argue with experts in their field? Like if a physicist explained to you that the reason things heat up upon entry into the atmosphere isn't because of friction, but because of the compression of a column of air, would you argue with him?

I'm wondering if the obstinate stupidity is specialized to statistics or informs your entire approach to the world.
 
Araysar

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BrutulTM_sl said:
Here's a puzzle for you.

3 boxes, one containing a car, two containing goats. You choose one at random. You are then given the choice of keeping your one box, or trading it for both of the remaining boxes. Which do you choose? Is it 50/50 that the car will be in your box and not the other two?
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Its not the same problem because both boxes are unknown.

here, let me redo it for you.

3 boxes, one you selected (unknown), the host has 2 other boxes , one which is opened and is a goat and the other is unknown

is your choice 33/66?
 
ZyyzYzzy

ZyyzYzzy

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Stats, how do they work?

You flip a coin 5 times, what are the odds you get heads 5 times in a row? Now you flip a coin 100 times, odds of getting heads 5 times in a row?
 
Araysar

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Siddar_sl said:
The odds of you having the right door is .1% the odds of Monty having the right door are 99.9%.
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Since Monty always removes 998 doors and always reduces the proposition to a binary proposition at the end, your true odds are always 50%. You can do whatever the fuck you want in 1st round. have a blind man pick it for you. have someone who knows whats behind the doors pick it for you, throw darts at them, spin a roulette wheel. It doesnt matter what you do, Your final choice, the one that matter is always 2 doors. 1 car 1 goat, 50/50
 
Araysar

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The Master_sl said:
Out of curiosity Ara, do you often argue with experts in their field? Like if a physicist explained to you that the reason things heat up upon entry into the atmosphere isn't because of friction, but because of the compression of a column of air, would you argue with him?

I'm wondering if the obstinate stupidity is specialized to statistics or informs your entire approach to the world.
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Who are experts? you morons?

Or the thousands upon thousands of Math and Stats PhDs who dispute the "accepted" answer of the Monty Hall problem because they see the inherent sleight of hand?
 
T

The Master

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Find me one person who has a PhD in statistics and disputes the answer to the basic Monty Hall problem, correctly stated.
 
BrutulTM

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Araysar_sl said:
Its not the same problem because both boxes are unknown.

here, let me redo it for you.

3 boxes, one you selected (unknown), the host has 2 other boxes , one which is opened and is a goat and the other is unknown

is your choice 33/66?
Click to expand...
Nonsense.

Put your box on one side and the other two boxes on the other side. Odds are 1/3 that it's on your side, 2/3 that it's on the other side. You KNOW that at least one of the boxes on the other side doesn't have the car, so removing one that doesn't have the car doesn't change the odds that the car is on the other side.
 
Araysar

Loser Araysar

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BrutulTM_sl said:
Nonsense.

Put your box on one side and the other two boxes on the other side. Odds are 1/3 that it's on your side, 2/3 that it's on the other side. You KNOW that one of the boxes on the other side doesn't have the car, so removing the one that doesn't have the car doesn't change the odds that the car is on the other side.
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Exactly. Since you know that one of the boxes on the other side doesnt have a car, it means that either the box on your side does, or the remaining 1 box on host's side does. That's a 50/50 proposition.

Now you you're starting to get it.
 
Swagdaddy

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finding out the 1$ doesnt change anything

you either have the 100,000 already or its in the other box, since you don't know which it's just a 50/50 cut and swapping doesnt increase your chances

i dont even see where the riddle is here
 
Tuco

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Araysar_sl said:
But Tuco, why wouldnt your odds be 1 in 1000? Why is it 0?
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Because you're required to complete two rounds. The second round forces you to chose to stay the same or you choose to switch. It's an irrelevant point I made to agree with some dumb shit you said earlier, pay no attention to it.


Araysar_sl said:
How can that be when you are presented with a binary option at the end?
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That's the crux of the issue, it's been explained to you in enough ways that a rational person would understand the answer to your question.
 
BrutulTM

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Araysar_sl said:
Exactly. Since you know that one of the boxes on the other side doesnt have a car, it means that either the box on your side does, or the remaining 1 box on host's side does. That's a 50/50 proposition.

Now you you're starting to get it.
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Why in the fuck am I still talking to you?
 
Siddar

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Araysar_sl said:
Since Monty always removes 998 doors and always reduces the proposition to a binary proposition at the end, your true odds are always 50%. You can do whatever the fuck you want in 1st round. have a blind man pick it for you. have someone who knows whats behind the doors pick it for you, throw darts at them, spin a roulette wheel. It doesnt matter what you do, Your final choice, the one that matter is always 2 doors. 1 car 1 goat, 50/50
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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.
 
Tuco

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The last time someone was this dense his name was Merlin and I rrp'd him for a few days. Choose your next posts carefully Araysar because you'll find yourself in his position. Who knows maybe you'll learn from it and take some stats classes in community college.
 
Araysar

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Tuco_sl said:
Because you're required to complete two rounds. The second round forces you to chose to stay the same or you choose to switch. It's an irrelevant point I made to agree with some dumb shit you said earlier, pay no attention to it.
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Oh so its completely irrelevant as to what you choose in 1st round? The 1st round is just theater to set up the second round???? HMM SOUNDS LIKE SOMETHING I HEARD BEFORE IN THIS THREAD.

And in the second round you're only given 1 choice? Stay with original door or switch to the second door?

SOUNDS LIKE A 50/50 PROPOSITION TO ME, BRO.
 
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felldoh

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felldoh_sl said:
The second round offers you a choice, but the odds never changed. There is still a (1/x) chance that your initial selection was the car, and a (x-1)/(x) chance that the car is the host's group of boxes. If you deny this to be true, then you believe that you have a 50% chance of initially selecting the car from a pool of that car + an infiniate number of goats.
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