Preserved Topic: Here's a little question for you. (Page 1 of 1)

I have three cups, upside-down, on the table in front of you. I tell you that one of them contains an object, and invite you to guess which one.

You choose one of them. We'll call the one you chose "cup A".

I overturn one of the remaining two (we'll call it "cup B") and show you that it's empty. I now give you the option to change your choice to the remaining cup "cup C", if you like, or to keep your choice as cup A.

If you decide to keep your choice of cup A, what is the probability that you will be correct?

[This message has been edited by Slime (edited 06-10-2002).]

well I know that magic trick,

50%, 1/2, .5

P(Cup A or Cup C)
P(CA)=1(ball)/2(cups)
P=1/2

If you stay with cup A, you only have a 1/3 chance of being right (its just like picking 1 out of 3, because you didn't change) But, if you switch, you have a better chance of getting it. Trust me



[This message has been edited by Dan (edited 06-10-2002).]

1 in 2 or 50%

Nope, its 1/3, I'll explain. If you have 3 choices, you have a 1 in 3 chance of getting it right. Which means a 2/3 chance of getting it wrong. Now, since he takes away one of the wrong answers, you now have a 2/3 chance of getting it right if you switch, and still a 1/3 chance of getting it right if you don't. Chances are you didn't choose right on your first pick, and since one of the other 2 are more likely the right cup, and he takes away the one that *isn't* the right cup..


[This message has been edited by Dan (edited 06-10-2002).]

yea dan, the process is quite simple

3 cups 1 ball, chance to get the right 1 is 1/3

but he was talking about with the 2 remaining so its 2 cups 1 ball

1/2


tri-eye

No, it doesn't change the chances. There were three cups. You picked one. No matter which one you choose, he can still take away one that isn't the right cup. So as long as you dont change, your odds are 1/3. Its just like he never took away the cup.

*sigh*

Alright dan you wanna do this the hard way?

Round 1: 3 cups 1 ball = 1/3
Round 2: 2 cups 1 ball = 1/2

1/3 ? 1/2 = P(slimes ?)

=1/6


___________________
tri-eye

But he's taken away one of the cups so now you only have two choices:

Stick with Cup A
Go with Cup C

No matter what you do, you have a 50% chance of being correct with your guess.

Put the math away kiddie. Heres the deal. When you are first offered the choice. You have a 1/3 chance of getting it right. That means that 2 out of 3 times, you'll get it wrong. SO....... theres a 2/3 chance, that the correct cup, is one of the unchosen ones.

Now.. assuming I haven't confused you, since theres a 2/3 chance of it being a cup you have not chosen, and since out of the 2 cups you haven't chosen, he shows you that one of them is not the correct cup, and takes it away. There is now a 2/3 chance that the Cup C is the correct one, which means theres a 1/3 chance that it is cup A.

Wanna argue more? how bout this. Go find 3 plastic cups, and a friend/relative. Then, get them to perform this experiment on you. You'll notice, when you switch, you'll usually get it right, and when you stick with your original choice, you'll usually get it wrong.


<Edit> Krets, its only a 1/2 chance if he only showed you 2 cups in the first place. Since you know that the cup he took away was *not* the correct cup, it changes everything. Now, if he took away the cup BEFORE you made your choice, than it would be 1/2, but since he didn't, my explainations stand.




[This message has been edited by Dan (edited 06-10-2002).]

As I understand it though (and I slept through most of my Elementary Stats class in college) you can't take previous events into account. There are two cups, you know one has to have a ball in it and one doesn't, so therefore you have a 50% chance.

Edit
Its like the question about coin tosses; if you toss a coin in the air 100 times and 60 of those times its heads, what is the chance of it landing heads on the next toss?

50%

Right?




[This message has been edited by kretsminky (edited 06-10-2002).]

http://www.geocities.com/danvenne/logic.html

I drew it for you, maybe this will explain better.

Haha! I made you all fight!

The answer, believe it or not, is 1/3. I know it's unintuitive, but it's true. Trust me, I went through the whole "it *has* to be 1/2!" thing myself, and now I realize the error of my ways.

Here's my explanation:

There is a 1/3 chance that the contestant picks the winning cup:
- one of the other cups is removed. There is a 100% chance that if the
contestant stays on their cup, they will win.

There is a 2/3 chance that the contestant picks the losing cup:
- one of the other cups is removed (the losing one). There is a 0% chance
that if the contestant stays on their cup, they will win.

So, the chances of them winning if they stay on their cup are 100% * 1/3 +
0% * 2/3 = 1/3.


Here's a better explanation that I found elsewhere:

First we need a random variable P that describes whichdoor the prize is behind (i.e., P = 1, 2, or 3, each with probability 1/3). Without lossof generality, we may assume that the contestant always picks door 1 (so that door 1 isdoor A). Now, if the prize is behind door 3, Monty Hall will open door 2 (so that door2 becomes door B and door 3 becomes door C). Similarly, if the prize is behind door 2,Monty Hall will open door 3. For these two cases Monty Hall had no choice.However, what if the prize is actually behind door 1? Now what is Monty going todo? Maybe he always shows door 2 in this case. Or maybe he always shows door 3. Ormaybe he uses some complicated secret algorithm for deciding which door to open. Inany case the contestant has no knowledge as to how Monty will behave in this situationand regards Monty's two possibilities as equally likely. Hence, we may as well assumethat Monty tosses a fair coin: if it comes up heads he opens door 2, while if it comes uptails he opens door 3. This random coin toss is independent of P . Therefore, our samplespace consists of six points as shown in the table below. If the contestant switches, thensuccess will correspond to the four sample points (H, 2), (H, 3), (T, 2) and (T, 3). Hencethe probability of success is 4/6 = 2/3. On the other hand, if the contestant holds on todoor A, then success will correspond to sample points (H, 1) and (T, 1) and the successprobability will be only 1/3.

[edit: BTW, this is called "monty's paradox."]

[This message has been edited by Slime (edited 06-10-2002).]

dude, you cant take chance it wont be picked and chance it will be picked and multiply them together to get the chance it will be picked

quote:

---

what is the probability that you will be correct?

---

its gotta be 1/3 time 1/2

put the math away old guy j/k

Slime, dude, your missing the point, 2 cups left and 100% chance to be right my ass


___________________
tri-eye

Should I mention that this has also been proven by approximation? =)

If you know any sort of language, carefully write a program to simulate it, and it'll come up with 1/3.

I do like Dan's one-sentence explanation, "When you are first offered the choice. You have a 1/3 chance of getting it right. That means that 2 out of 3 times, you'll get it wrong. SO....... theres a 2/3 chance, that the correct cup, is one of the unchosen ones."

That statement really sums it up.

See, in the end, there are three cases:
1. cup A is in fact correct, and either cup B or C have been overturned (doesn't matter which).
2. cup B is correct, and cup C has been overturned.
3. cup C is correct, and cup B has been overturned.

Cup A is correct only in one of these three equally likely cases.

This is why I hated statistics.

After doing a little research on the nut, I found an answer to this as well as some simulators that you can use yourself if you're skeptical (like me).
http://mathforum.org/dr.math/faq/faq.monty.hall.html

Its known more popularly as the Monty Hall problem but apparently it is better to change cups.

Even though when I did a simulator 100 times I can up with 45 out of 100 correct when I stayed with my original choice but hey, who am I to argue?

Just look at the picture! It explains.. everything.

45 out of 100? Interesting. Try more, though, like at least 10000 I'd say. I'm going to go write a simulator in JS myself...

hehe Riiiiight. I don't want to wear out my mouse or my finger.

The second time I tried I got 31 out of 63. Dead on.

You must have written your program incorrectly. Mine gives me .33561 (100,000 trials).

Interestingly, as I simplified the logic of the program, it got closer and closer to being a program that merely chooses a random number between one and three, and returns true if the number is one. 'Cause it turns out that it doesn't matter which cup is overturned, the probability that it's cup A stays the same.

a = Math.random();

	c1 = (a < 1/3);

	c2 = (a >= 1/3 && a < 2/3);

	c3 = (a >= 2/3);

	

	// choose c1

	if (c2) // if option 2 is correct...

	{

		// ...overturn c3

		c3 = false;

		//if it's c1, add one to prob

		if (c1) prob++;

	}

	else if (c3) // if option 3 is correct...

	{

		// ...overturn c2

		c2 = false;

		//if it's c1, add one to prob

		if (c1) prob++;

	}

	else

	{

		// it's c1, so add one to prob

		prob++;

	}

is simplified to

a = Math.random();

	c1 = (a < 1/3);

	c2 = (a >= 1/3 && a < 2/3);

	c3 = (a >= 2/3);

	

	// choose c1

	if (c2) // if option 2 is correct...

	{

		//if it's c1, add one to prob

		if (c1) prob++;

	}

	else if (c3) // if option 3 is correct...

	{

		//if it's c1, add one to prob

		if (c1) prob++;

	}

	else

	{

		it's c1, so add one to prob

		prob++;

	}

which can easily be simplified to

a = Math.random();

	c1 = (a < 1/3);

	c2 = (a >= 1/3 && a < 2/3);

	c3 = (a >= 2/3);

	

	//if it's c1, add one to prob

	if (c1) prob++;

[This message has been edited by Slime (edited 06-10-2002).]

Ummmm, my program?

I didn't write the program, it was a Java simulator built by someone else.

Oh. Strange.

Hmm You don't add the probabilty together you multiply it.

Ok everyone I wrote it down click here.


___________________
tri-eye

Jackass, read the fucking article.

God you're an idiot sometimes. Its a known paradox that's been proven time and again by people much smarter than you or me.

Im not gonna click that link. I know the answer, and have provided solutions, slime knows the answer, and has provided fantastic solutions, Krets posted a whole site with links dedicated to the answer. If you're still going to argue.. You're wrong get it?

Uhh I didnt even bother to read the other posts, I assumed none were made...

Oh well I tried, question though.

Slime why did you post this in the first place?


___________________
tri-eye

I think what you're forgetting, Insider, is that the odds *before* one of the cups is overturned *really do* have an effect on the odds afterwards.

See, let's say you have two choices. One choice is right and one is wrong. This does not *necessarily* mean that there is a 50-50 chance between them. It's possible that one of them is more likely than the other.

So, in this case, the question is what happens that causes it to be less likely that cup A isn't correct?

See, originally, before any cup is overturned, cup A has a 1 in 3 chance of being correct. Now, one of the other two cups are overturned. The key here is that *it's not a purely random choice* which cup is overturned. This gives the cup that wasn't overturned an advantage over cup A. This is what changes the probability in the only-two-cups-left situation.

If it *were* a truly random choice which cup was overturned, then 1/3 of the time, cup A would be right, 1/3 of the time, the non-overturned cup would be right, and 1/3 of the time, the overturned cup would be right. But we're not letting that last situation happen. We take that last situation, and we turn it into the second situation, by making sure we never overturn the winning cup. So those last two 1/3s combine into a single 2/3, leaving a 1/3 chance that cup A is right and a 2/3 chance that the non-overturned cup is right.

Now, that said, I could go on explaining this over and over in different ways, but you won't be convinced until you open your mind up to the idea that you might be wrong; as long as you're convinced you're right, you won't see why you aren't. So try to open your mind and really consider these arguments.

Whoops, 3 posts were made while I was typing that =)

Guys, don't be so harsh on him about this, it took *you* some convincing too =) (except for Dan)

Anyway, I posted this 'cause I saw it in the POV-Ray newsgroups and it interested me, and I suppose we all have that desire to point out to someone else that they're wrong, don't we? *grin*

I too needed some convincing originally.

I had this problem in math in 4th grade and the answer was:

1/3 first round
1/2 second roung

lol yea dan knew it strieght foward, gj dan

Poor slime... even after all of that, there are still those that don't bother to read all the posts...

Took me a while, but I eventually got it

njuice42
icq 957255

Nope, fooled you all. The ball is behind...

...Slime's ear! Ta-da!

I just thought of another way of looking at this.

Lets pretend, instead of taking away one of the cups, slime adds 2 more empty cups after you choose one, and asks if you'd like to switch now. If you don't switch, are your odds now 1/5 of having the right cup? No.. its one out of three, because it was one out of three when you picked at the start, and you never switched.