Are you actually asking? I'd be happy to explain it if you're not going to be obstinate about it.
A brain teasing probability puzzle
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Loser Araysar
Chief Russia Correspondent / Stock Pals CEO
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it doesnt matter whether you're doing it with cards, coins, simulators, etc.
it all works off the wrong premise.
Loser Araysar
Chief Russia Correspondent / Stock Pals CEO
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the simulator is accurate but that is because it incorporates the first choice as if it meant something, but its working off a wrong premise.
it doesnt. the first choice is meaningless because in no way does it affect the options presented to you in second choice. it is all pure theater.
your final choice is always 50/50 and thats the only real choice thats presented to you. the first choice is meaningless.
Hachima
Molten Core Raider
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The easy way to think about it for me is you have a 2/3 chance of picking a goat. If you pick a goat, the host has to show you the other goat. This means switching is a guaranteed win if you pick a goat. So by switching your odds are based on the 2/3 chance of picking a goat.
Loser Araysar
Chief Russia Correspondent / Stock Pals CEO
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how is that easy? your first choice means nothing because your second set of choices (REGARDLESS OF WHAT YOU PICK) is always the same:
1. door you picked (unknown)
2. door (unknown)
3. door with goat that host opened
regardless of what you pick the first time, the set of options above is always your ONLY choice for your second choice. your first choice is purely meaningless. and the second set of options presents you with 50/50
McCheese
SW: Sean, CW: Crone, GW: Wizardhawk
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They probably don't have statistics in Chechnya so it's a difficult concept for him.
The chance of picking the car is 1/3.
The host eliminates a goat.
You don't switch.
The chance that you picked the car remains the same - 1/3.
It's not an independent event. The way you're looking at it the doors are randomized again.
The host eliminates a goat.
You don't switch.
The chance that you picked the car remains the same - 1/3.
It's not an independent event. The way you're looking at it the doors are randomized again.
Your first choice matters because what you already picked is still on the table. You either picked the goat or the car on your first pick. You cannot possibly argue that you didn't pick either a goat or the car. If you picked a goat, switching wins. If you did not pick a goat, switching loses. But the odds of picking a goat are 2/3. It isn't a new pick, it is switch or don't switch from your your first pick. Your first choice is included in the problem.
McCheese
SW: Sean, CW: Crone, GW: Wizardhawk
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Where'd you hear that? There's a reason their nuclear power plants, submarines, spaceships, and anything else requiring more than high school algebra have routinely exploded.Russians are supposed to be good at math.
Noodleface
A Mod Real Quick
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I took Probability for Engineers last year and this was the first problem we went over. I won't keep you in suspense, I got a B-. I DONT KNOW THE FUCKING ANSWER
Noodleface
A Mod Real Quick
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OneofOne
Silver Baronet of the Realm
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I choose to switch my answer. Cause I was just trolling ya'll bro!11!!!
=/
Damn fine piece of statistical trivia though!
Lejina
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Well, I didn't do statistics since I did my BSc but I have to do a fair amount rearranging of formulas on a regular basis at work. So yeah, whatever about the art degree.
The whole thing is more of a gotcha than an actual test if you can do math gud. Unless you know the trick already or just happen to have been thru an example during a stat class, the instinct is to say the odds are down to 50/50 once you take out a bad door.
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.
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.
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
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
Downhammer
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In a true monty hall problem you would increase your odds by switching. What the OP presented is not technically a monty hall though. No where does he state that the host will only open a $1 box. He states that the host knows what's inside the box and that he opens a box, two unrelated statements. Your odds of winning improve only if the host "knows what's inside the boxes" AND "opens another boxthat contains $1" As stated it's just dumb luck the host didn't reveal $100,000 by opening box 3.