View Full Version : Test your logic

You are on a game show. The object of the game is to win a car. The host shows you three doors. Behind two of the doors there are goats. Behind the other door is a car. You select your door. The host says OK, but doesn't open your door. Instead, he opens one of the other doors to show you a goat (the host knows what's behind the doors and he opened a goat door on purpose). The host now says that you can still change your door if you want to. Or, you can stick with your original door.

What should you do, stay with your original door, or pick the other unopened door?

EDIT: Tell me the odds, don't just randomly say switch or stay.

BloodFruit

03-21-2008, 09:39 PM

Switch .

Switch .

What are the odds you pick the car?

Index

03-21-2008, 09:47 PM

Stay .

notmaggot

03-21-2008, 09:48 PM

What are the odds you pick the car?

50/50. this is a really stupid question.

Scarecrow

03-21-2008, 09:51 PM

This problem is olde.

You switch, there's a higher chance of you winning, I forget how though.

50/50. this is a really stupid question.

Really, you think so? This is a logic riddle. I would not have posted it if it was that easy.

It is definetely NOT a 50-50 chance.

BloodFruit

03-21-2008, 09:52 PM

There is one chance in three that the door you first picked is the correct one. This can't change just because you're shown there isn't a prize behind some other door. Since the prize has to be behind either the door you picked or the other closed door, and there is one chance in three that its the one you picked, there must be two chances in three its behind the other door. Switch, and you'll be right twice as often.

notmaggot

03-21-2008, 09:52 PM

Really, you think so? This is a logic riddle. I would not have posted it if it was that easy.

It is definetely NOT a 50-50 chance.

that doesnt make any sense. jjiosdfdfjjifodjioejiewosedsfhuhuesuhsfufhusdhudhj usdfjusdjudsfhju /seizure

how would it not be? there are only 2 doors left. one has a goat the other has a car. its the same as flipping a coin.

FizzlemanJ

03-21-2008, 09:54 PM

It's 50/50 right?

edit: Oh wait no, it's one out of three.

StrunG

03-21-2008, 10:01 PM

Isn't it 0? If you choose the right door he's going to just open one of the other ones. If you pick a wrong one, he'll either open it anyways or just open the other wrong one.

Exile

03-21-2008, 10:22 PM

This is really not that hard to understand, guys. You have a 2/3 chance of winning if you switch, every time.

There's a 1/3 chance of picking the car first, and the only time switching will cause you to pick a goat is if you choose the car.

There's a 2/3 chance of picking one of the goats first. Switching in this instance will get you the car.

Assuming you stay with your first picked choice, you only have a 1/3 chance of winning, because those are your odds of initially choosing the car. However, no matter what goat you pick (A or B), switching will always get you the car, because the other goat is eliminated from the odds. Switching from one goat can never get you the other goat. Staying is therefore illogical.

Think of it this way if this is confusing:

Options (Switching)

Pick the car first, switch, you lose

Pick goat A first, switch, you win

Pick goat B first, switch, you win

Options (Staying)

Pick the car first, stay, you win

Pick goat A first, switch, you lose

Pick goat B first, switch, you lose

You're given 2 options, each with 3 instances. One of the options allows you to win 2 out of 3 of those instances, while the other only allows you to win 1 out of 3 instance. Which would you rather go for?

It's not all that complicated.

As you can see in my graph, you have a 2/3 (75%) chance of getting the car if you switch.

http://img228.imageshack.us/img228/1976/ooooooooooommmmmmmmmmgglm2.png (http://imageshack.us)

http://img228.imageshack.us/my.php?image=ooooooooooommmmmmmmmmgglm2.png

With your original pick of the 3 doors, you have a 1/3 (25%) chance of picking the car. The odds say you picked wrong the first time, so you should switch.

EDIT: Exilement nailed it, good job.

Exile

03-21-2008, 10:32 PM

1/3 (25%)

lol wut

BloodFruit

03-21-2008, 10:36 PM

2/3 (75%)

lol wut.

Cereal Man

03-21-2008, 10:40 PM

Uh, what if you picked the car the first time?

lol wut

lol wut.

shuddap. im dumb.

Schwa

03-21-2008, 10:41 PM

shuddap. im dumb.

lol wut.

Exile

03-21-2008, 10:45 PM

Uh, what if you picked the car the first time?

Congratulations, you had a 1/3 chance of doing that.

You have a 2/3 chance of picking one of the goats.

That means switching every time will get you the car 2/3 times, while staying gets you it 1/3 of the time.

So if you picked it the first time, you're pretty much screwed, because mathematically you should switch, every time.

What? No. Just no. After he opens one door to show 1 goat, it's no longer a chance out of three, it's out of two. Get it? Got it? After one goat is eliminated, and you are choosing between door a and door b, there is NO door c. Therefore, it's a 50/50 chance at the last. Before the revealing of one goat and after are two seperate problems, not one.

What? No. Just no. After he opens one door to show 1 goat, it's no longer a chance out of three, it's out of two. Get it? Got it? After one goat is eliminated, and you are choosing between door a and door b, there is NO door c. Therefore, it's a 50/50 chance at the last. Before the revealing of one goat and after are two seperate problems, not one.

That is what most people think. However you are wrong. Your peanut sized brain apparently cannot comprehend the riddle.

Exile

03-21-2008, 11:29 PM

You can ignore the 3rd door, yes, but you can't ignore your first choice.

In the simplest form of this problem, you can assume this: if you pick the car, switching gets you a goat. if you pick one of the 2 goats, switching gets you the car.

Remember this:

Options (Switching)

Pick the car first, switch, you lose

Pick goat A first, switch, you win

Pick goat B first, switch, you win

Options (Staying)

Pick the car first, stay, you win

Pick goat A first, switch, you lose

Pick goat B first, switch, you lose

6 outcomes, that means each outcome has a 1 in 6 chance of occuring. There are 3 that cause you to win, and 3 that cause you to lose. You could say that this would constitute a 50/50 chance of winning, since 3/6 outcomes will cause you to win.

However, you only have 2 options (switch or stay). That means each option gives you 3 possible outcomes. 3 allow you to win, 3 allow you to lose. That means there are only two possible ways to divide those 6 outcomes between the 2 options:

One option has 3 possible chances to win, the other has 3 possible chances to lose.

OR

One option has 2 possible chances to win and 1 to lose, and the other has 1 possible chance to win and 2 to lose.

The first situation would be the only possibility for this to give you a 50/50 chance. However, that's simply a hypothetical. It's not how the problem goes at all.

There's no possible way for this to be reduced to a 50/50 chance, and there's no possible way for me to explain it any further.

Jeremy

03-21-2008, 11:30 PM

Your peanut sized brain

Irony .

Exile

03-21-2008, 11:34 PM

This also might help:

http://upload.wikimedia.org/wikipedia/commons/thumb/9/9e/Monty_tree.svg/520px-Monty_tree.svg.png

Vervanda

03-21-2008, 11:40 PM

still dont grasp the concept

Geez Exilement, you're doing a lot of work.

Eccentrix

03-22-2008, 05:45 PM

I'd rather ride the goat

Schwa

03-22-2008, 05:52 PM

I'd rather ride the goat

Are you thinking what I think you're thinking? *wink wink nudge nudge*

Eccentrix

03-23-2008, 09:55 AM

idk *wink wink nudge nudge*

Scarecrow

03-23-2008, 07:59 PM

I remember seeing a newspaper article about this a while back. I'll see what I can find.

*edit*

Proof of 2/3:

http://en.wikipedia.org/wiki/Monty_hall_problem

the afro ninja

03-26-2008, 01:04 AM

muttonhead, why do u post things tht puzzle your feeble mind?

every1 thinks....knows your a dumma$$

Index

03-26-2008, 01:10 AM

muttonhead, why do u post things tht puzzle your feeble mind?

every1 thinks....knows your a dumma$$

Shut up stupid noob. Spell things right and say "dumbass" not dumba$$, it makes you look like a douche bag.

the afro ninja

03-26-2008, 01:47 AM

trying not to get it cencorsed ****ing dumbass cunt bitch ****er kike dick blowing shit hole of cum, seriously you're pissing me off, leave me alone and ill leave u alone

Index

03-26-2008, 11:42 AM

trying not to get it cencorsed ****ing dumbass cunt bitch ****er kike dick blowing shit hole of cum, seriously you're pissing me off, leave me alone and ill leave u alone

AHAHAHAHAH HAHAHAHAH

Noobs these days, they think they have power over people.

muttonhead, why do u post things tht puzzle your feeble mind?

every1 thinks....knows your a dumma$$

haha. Somebody wants attention..

Dragonkof

03-29-2008, 11:34 PM

Bayes' theorem

An analysis of the problem using the formalism of Bayesian probability theory (Gill 2002) makes explicit the role of the assumptions underlying the problem. In Bayesian terms, probabilities are associated to propositions, and express a degree of belief in their truth, subject to whatever background information happens to be known. For this problem the background is the set of game rules, and the propositions of interest are:

http://upload.wikimedia.org/math/8/0/f/80f49ddbe630733c95dc5fa80c6f6ef8.png : The car is behind Door i, for i equal to 1, 2 or 3.

http://upload.wikimedia.org/math/5/8/3/583394e4a1e949eb3002666cd3475fb0.png : The host opens Door j after the player has picked Door i, for i and j equal to 1, 2 or 3.

For example, http://upload.wikimedia.org/math/5/6/6/566395fdb9f8ee7ad63964694596894a.png denotes the proposition the car is behind Door 1, and http://upload.wikimedia.org/math/0/c/8/0c8639b8694f08c77ca33b4b47ead860.png denotes the proposition the host opens Door 2 after the player has picked Door 1. Indicating the background information with http://upload.wikimedia.org/math/8/e/a/8eac61b98a5665772fe9372b60176a9b.png, the assumptions are formally stated as follows.

First, the car can be behind any door, and all doors are a priori equally likely to hide the car. In this context a priori means before the game is played, or before seeing the goat. Hence, the prior probability of a proposition http://upload.wikimedia.org/math/8/0/f/80f49ddbe630733c95dc5fa80c6f6ef8.png is:

http://upload.wikimedia.org/math/4/6/7/467a0ef3ff9497ad646e991c45c6cfd8.png

Second, the host will always open a door that has no car behind it, chosen from among the two not picked by the player. If two such doors are available, each one is equally likely to be opened. This rule determines the conditional probability of a proposition http://upload.wikimedia.org/math/5/8/3/583394e4a1e949eb3002666cd3475fb0.png subject to where the car is — i.e., conditioned on a proposition http://upload.wikimedia.org/math/f/3/f/f3fdbc8dac079e683f00479a307c8303.png. Specifically, it is:

http://upload.wikimedia.org/math/0/4/1/0414573a99b1d14920379cfbb72b1fed.png

The problem can now be solved by scoring each strategy with its associated posterior probability of winning, that is with its probability subject to the host's opening of one of the doors. Without loss of generality, assume, by re-numbering the doors if necessary, that the player picks Door 1, and that the host then opens Door 3, revealing a goat. In other words, the host makes proposition H_{13}\, true.

The posterior probability of winning by not switching doors, subject to the game rules and H_{13}\,, is then P(C_1 | H_{13},\,I). Using Bayes' theorem this is expressed as:

By the assumptions stated above, the numerator of the right-hand side is:

The normalizing constant at the denominator can be evaluated by expanding it using the definitions of marginal probability and conditional probability:

http://upload.wikimedia.org/math/c/c/7/cc7fcaf53f438e4d5e40e5fb0e99e1e9.png

Dividing the numerator by the normalizing constant yields:

Note that this is equal to the prior probability of the car's being behind the initially chosen door, meaning that the host's action has not contributed any novel information with regard to this eventuality. In fact, the following argument shows that the effect of the host's action consists entirely of redistributing the probabilities for the car's being behind either of the other two doors.

The probability of winning by switching the selection to Door 2, http://upload.wikimedia.org/math/9/5/6/9568196a6b9383dcc1798a285958117a.png can be evaluated by requiring that the posterior probabilities of all the http://upload.wikimedia.org/math/8/0/f/80f49ddbe630733c95dc5fa80c6f6ef8.pngpropositions add to 1. That is:

http://upload.wikimedia.org/math/7/d/0/7d0c144aff972c42c1da84831bc5c9b6.png

There is no car behind Door 3, since the host opened it, so the last term must be zero. This can be proven using Bayes' theorem and the previous results:

http://upload.wikimedia.org/math/f/6/6/f66730e204ff67264ed615f25f064618.png

Hence:

http://upload.wikimedia.org/math/1/8/5/18524cc0646e4c6510855462e0eb84ec.png

This shows that the winning strategy is to switch the selection to Door 2. It also makes clear that the host's showing of the goat behind Door 3 has the effect of transferring the 1/3 of winning probability a-priori associated with that door to the remaining unselected and unopened one, thus making it the most likely winning choice.

i cant be stuffed adding all the pictures, so get over it.

Scarecrow

03-30-2008, 05:38 AM

wiki get?

Shanto

03-30-2008, 09:03 AM

There was a game show like this.

Dragonkof

03-31-2008, 04:54 AM

wiki get?

indeed =P

Setmin

03-31-2008, 08:18 PM

AHAHAHAHAH HAHAHAHAH

Noobs these days, they think they have power over people.

you're a noob, **** off

pagan

03-31-2008, 08:20 PM

you're a noob, **** off

seth am i a noob

Setmin

03-31-2008, 08:24 PM

seth am i a noob

nope : )

NaTuRaL

03-31-2008, 08:26 PM

You get 2/3 if your counting consecutive. But if you have a car behind one door, and a goat behind another, its a 1/2 chance.

mutton, why the hell do you post this shit?

Apples

03-31-2008, 08:26 PM

Hi setmin :Smile:

Setmin

03-31-2008, 08:29 PM

Hi setmin :Smile:

hey dude sup long time no see

pagan

03-31-2008, 08:49 PM

You get 2/3 if your counting consecutive. But if you have a car behind one door, and a goat behind another, its a 1/2 chance.

mutton, why the hell do you post this shit?

sly elf

Apples

03-31-2008, 08:52 PM

hey dude sup long time no see

I kno.

how r u?

aurorai

03-31-2008, 08:57 PM

I seriously bet the host is a dick and theres a goat behind every door.

NaTuRaL

04-01-2008, 04:34 PM

sly elf

I no my username is gey, kay? pp is being gey. if he sees this, i request i name change

You rang?

InsertNameHere

04-03-2008, 09:40 PM

I seriously bet the host is a dick and theres a goat behind every door.

lol, you would have to be a really big jackass to do that...

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