DEAL OR NO DEAL?!?!!?
Imagine you are on a game show with only three sealed red boxes.
The three cash prizes are $1, $1 and $100,000. You pick a box, let's say box two.
The host, who knows what's inside the boxes, opens another box, say box three and then tempts you with an offer. Box three is opened in front of you revealing a $1 prize, and he offers you the chance to change your mind to choose box one. Does switching improve your chances of winning the $100,000?
Last edited by Neki; 09-12-2013 at 11:09 PM.
DEAL OR NO DEAL?!?!!?
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
Switch. You originally had a 33% chance of getting it right. By switching after the first pick, your chance of getting a win goes up to 66%.
K Spacey was my teacher, too.
Is your next thread going to be can planes take off on treadmills?
How is this brain teasing? Not only is it a well known example of statistical probability, it is simple stats.
I picked don't switch because I am not a bitch.
oderint dum metuant
It's a cool problem. The trick for me to understand it was knowing that the host would always open the door to a goat.
The question is dumb. Its only interesting if you're prone to over think and/or motivated to appear smart. The elimination of a $1 box leaves your chance at 50/50 no matter what box you choose. Three boxes are equal, one is revealed, and suddenly now one box is twice as likely as the other? Makes no sense
From the wiki, it seems to be a good read at my current BAC: http://probability.ca/jeff/writing/montyfall.pdf
Dis. Always switch.
They made an entire game show about this with the bald germ freak. And it was 100 boxes.
Edit: It's a math trick. Your chance of winning does go up because they are a continuous event and you do have 50/50 on the second pick. But because they are a continuous event the odds on your first pick does change. Specifically it goes up. It's one of those "the rock will never reach the tree" tricks.
Last edited by Iannis; 09-13-2013 at 02:33 AM.
I never watched that show. Did he ever pick the suitcases?
I think I watched the show for about 15 minutes once.
It was mostly just long pauses. The show was like 15% models in slinky dresses and 75% dramatic pausing. 10% commercial breaks.
your choice is still 50/50 whether you switch or not because one box has been revealed regardless. im not sure what makes this a brain teaser or a puzzle.
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.
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.
Last edited by The Master; 09-13-2013 at 02:55 AM.
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
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
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.
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.
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.
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
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.
http://www.grand-illusions.com/simulator/montysim.htm
Nice easy way to visualize how it does make a difference to switch. And someone already linked it =p
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.
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.
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.
Are you actually asking? I'd be happy to explain it if you're not going to be obstinate about it.
Araysar be trollin. It is a hard concept to grasp for some, but if you don't believe the simulator you are either trolling or obstinate.
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.
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.
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
Araysar once you accept that you're wrong you can start on the path of enlightenment and follow in the footsteps of Merlin.
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.
Last edited by Tea; 09-13-2013 at 03:36 AM.
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.
Russians are supposed to be good at math.
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
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Honestly I couldn't wrap my mind around the concept and was with Araysar: you guys are all fucking crazy.
However, I got around to visualize the concept with that bit from the wiki.
Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance. Here’s a good way to visualize what happened. Suppose there are a million doors, and you pick door #1. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You’d switch to that door pretty fast, wouldn’t you?
In this thread we will find out who has done maths in college and who has a liberal arts degree
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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.
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.
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
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 box that contains $1" As stated it's just dumb luck the host didn't reveal $100,000 by opening box 3.
Is it a prop plane or a jet?
This problem also depends entirely on the host being REQUIRED to show you a $1/goat box. If he is motivated to avoid paying out and can do whatever he wants, you're playing rock/paper/scissors with him
The problem also doesn't seem to apply to Deal or No Deal. I'm not too confident on that assertion though
Entire thing reeks of "let's present some silly scenario with unstated imperatives and sound clever by misleading people"
I agree with your assessment that "the wording of the problem is bullshit." It's exceptionally hard to use written words to explain multiple quantities and their relationships with very much precision at all, which is exactly why math exists.
But I offer a somewhat more detailed, precise, and accurate explanation of the scenario:
Always switching =
A is car, B & C are goats. Host shows B. You switch and lose. (1/4 of the time)
A is car, B & C are goats. Host shows C. You switch and lose. (1/4 of the 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. (1/4 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. (1/4 of the time)
Win 2/4, lose 2/4
Never switching =
A is car, B & C are goats. Host shows B. You win. (1/4 of the time)
A is car, B & C are goats. Host shows C. You win. (1/4 of the time)
A is goat, B is car, C is goat. Host shows C (he's forced to, it's the only goat option), you lose. (1/4 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 lose. (1/4 of the time)
Win 2/4, lose 2/4
What if I know the odds are better if I switch but I still don't want to switch? The prize is blood money and an innocent person will be killed if I win the 100,000 and I don't want to switch. But the poll says I can only answer no if I agree the odds don't change. This means I can't answer the poll!
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
Last edited by Araysar; 09-13-2013 at 07:27 AM.
I'm not even going to try to understand how some of these people remember to breathe. But it's pretty clear where they're being led astray about this problem. There's a false equivalence being made where both of the two different goats = you lose. The fallacy that "you lose" is only one single event leads to an incorrect enumeration of the possible outcomes.
If you replace the two goats in the original scenario with "the top half of a mtf transsexual" and "the bottom half of a ftm transsexual", and rephrase the rules so that players win whatever is behind the door they chose, then you'll start seeing people put enough effort in to correctly enumerate the possibilities.
I think we can dig deeper.
It's a continuous event, not two discrete ones. Your first choice informs your second. In itself the first choice is meaningless, in relation to the second choice the first choice is meaningful.
Pretty sure the problem is just the semantics of the original post, which most of us instantly recognized as a Monty variation and responded to as such. That would make it a hellish exam question -- a poorly stated monty hall problem. Khalid, don't do this shit to your poor hung-over half-retarded students.
perfect example of the group think fallacy because wikipedia said so
youre all operating from a faulty premise
If by faulty premise you mean we were responding to what he meant rather that what he wrote, then yes.
If by faulty premise you mean the logic of the 66% is bad, you're just dumb.
There's a 50/50 for you!
Big Dummy.
Last edited by Melvin; 09-13-2013 at 07:52 AM.
It doesn't matter. If your going to count each goat as a different goat,(Goat1 and Goat2), then you'll need to account for that in your winning. So for wins you'll have to add:
A is car, B & C are goats. Host shows B. You switch and lose. (1/6 of the time)
A is car, B & C are goats. Host shows C. You switch and lose. (1/6 of the time)
A is goat1, B is car, C is goat2. Host shows C (he's forced to, it's the only goat option), you switch to B, you win. (1/6 of the time)
A is goat2, B is car, C is goat1. Host shows C (he's forced to, it's the only goat option), you switch to B, you win. (1/6 of the time)
A is goat1, B is goat2, C is car. Host shows B (he's forced to, it's the only goat option), you switch to C, you win. (1/6 of the time)
A is goat2, B is goat1, C is car. Host shows B (he's forced to, it's the only goat option), you switch to C, you win. (1/6 of the time)
Which is surprise surprise.... 4/6... or 2/3.
Think of the concept like this, without probability. You have 10 doors. You choose door 1. You either chose the car, or the car is in one of the other 9 doors. The host opens 8 of the other doors as he cannot open the prize (its implied.). So the car is either behind your 1 door, or it is *still* behind one of the other 9 doors. But you know 8 of them are empty. By switching, your choosing "the other 9 doors". So which is more likely, the car is behind 1 chosen door, or its somewhere behind the other 9 chosen doors?
Now instead of 10 doors, you have 3 doors. So its either behind 1 door, or its behind the other 2 doors.
Hope that cleared it up.
Edit: If your just arguing over the wording of the original problem, then I guess we can all agree it was worded poorly. I'll leave the above as an answer to the intended Monty Hall problem if anyone was still curious.
Last edited by Zuuljin; 09-13-2013 at 08:03 AM.
Close, but no cigar.
Always switching =
A is car, B & C are goats 1 and 2, respectively. Host shows B. You switch and win goat 2.
A is car, B & C are goats 1 and 2, respectively. Host shows C. You switch and win goat 1.
A is car, B & C are goats 2 and 1, respectively. Host shows B. You switch and win goat 1.
A is car, B & C are goats 2 and 1, respectively. Host shows C. You switch and win goat 2.
A is goat1, B is car, C is goat2. Host shows C (he's forced to, it's the only goat option), you switch to B, you win a car.
A is goat2, B is car, C is goat1. Host shows C (he's forced to, it's the only goat option), you switch to B, you win a car.
A is goat1, B is goat2, C is car. Host shows B (he's forced to, it's the only goat option), you switch to C, you win a car.
A is goat2, B is goat1, C is car. Host shows B (he's forced to, it's the only goat option), you switch to C, you win a car.
So to rephrase the problem: Choose one of 3 doors. Behind it is the truth about Araysar (a)trolling hard or (b)really being obtuse. Then one door is opened for us, in which it is revealed Araysar is (b)really being obtuse. So we switch - because we hoping to find he's just (a)trolling hard - but since he IS trolling REAL hard, he instantly switches his stance (b)really being obtuse.
So the final door is opened and reveals... LUMIE. End of world ensues. The only thing that can now bring us salvation, is FAITH in the fact that Araysar is a rational person. And then it strikes us...
seatbelts.
/oblivion
I always felt that this was the root of the increase in odds for switching in the scenario. The setup for the poll can only exist if one potential outcome is implicitly removed from the results. You have a 100% chance of not losing the larger prize on your first choice because the host must show you $1 after your pick. The wording removes all relevance of the first pick since you can never miss the larger prize and the actual choice that matters happens after you know that. It works for four boxes still although it works for 4 in the same way it should work for three. There are never three boxes in the system this way.
What I'm really really saying is that I have an exceptionally well developed sense of smell. So well developed, in fact, that I can smell my own shit when someone points out that I've stepped in it. Looks like I completely misdiagnosed why the odds of the two different goats were halved.
How is it not a 'true' monty hall problem and how is it 'dumb luck' that he didn't reveal the $100,000 box if he knew what was inside the boxes?
Which he does both in the original question and thus increases the odds of winning if you switchYour odds of winning improve only if the host "knows what's inside the boxes" AND "opens another box that contains $1
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?
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.
Last edited by Neki; 09-13-2013 at 12:31 PM.
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.
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.
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.
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?
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.
Last edited by Araysar; 09-13-2013 at 01:07 PM.
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
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
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.
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.
For the non trolls
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?
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