25-12-2004 15:47									#1
	Maths Problem Support Apolyton, buy GURPS/ Alpha Centauri

I have three non see through tea cups
I place them upside down on a table
Under one of the I place a prize
Under the other two cups I place nothing.
I know which cup the prize is under.
I invite somebody to pick a cup.
If it has the prize under it then they get the prize.
My first player picks a cup
At this point the player has 1 chance in 3 of wining the prize.
But before showing what is under the cup I tell him
I will show you what is under one of the unselected cups.
I then always turn over a cup with nothing under it.
I then tell the player that he can stay with the cup he chose earlier
or switch to the remaining cup.

Some eminient mathematics professors thought that it did not matter
which cup he chose from the two. Other eminent maths professors said it did matter.

What is his chance of winning if he sticks to his original choice?

What is his chance of winning if he switches.

	25-12-2004 15:54									#2
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Flip the table, grab the prize, run like hell.

	25-12-2004 15:55									#3
	Got spare money?

Well, there obviously some trick here, but I am going to go at it straightforward. Correct me if I am wrong in any of my assumptions.

A person chooses a cup. No matter what that cup has under it, the number of choices he has will drop to 2, by the elimination of a wrong answer. and then he can choose to stay with the cup, or not, which is basically the same as choosing one cup out of two all over again. This way, it's 50/50. Thus, I say: no difference. His chances are the same.

	25-12-2004 16:35									#4
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Switch.

If you don't pick the right cup, a 2 in 3 chance, then you guarantee that the remaining cup will be the one with the ball. Hence you have at least a 2 in 3 chance of winning if you switch.

Or put another way, if you don't switch, your chances of winning remain 1 in 3 regardless. Hence there must be a 2 in 3 chance of winning if you switch.

Last edited by Dauphin on 25-12-2004 at 16:40

	25-12-2004 17:40									#5
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quote: Or put another way, if you don't switch, your chances of winning remain 1 in 3 regardless.

How so?

	25-12-2004 18:05									#6
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I know the answer , so I'll put it in spoiler tags .

spoiler(highlight to read): Draw a table : Choice one : You pick junk . If you switch , you get the prize . Choice two : You pick junk . If you switch , you get the prize . Choice three : You pick the prize . If you switch , you get junk .

	25-12-2004 18:36									#7
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I'm thinking the same as Dauphin but struggling to prove this, here is my attempt :


Three cups, name them A,B & C

Each probability of it having the prize is 1/3

A+B+C=1

A=B=C

Player picks A,
at this point

A= 1/3
B+C= 2/3

B&C are assesed with the wrong option removed, lets say this is C

B now equals 2/3 as C = 0

A= 1/3
B= 2/3

therefore he is better to switch??

I may do some actual tests

	25-12-2004 18:38									#8
	Avatar Enlargement: We've got the solution

3 choices are an illusion. One of them is certain to be eliminated through the process.

	25-12-2004 18:44									#9
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Az: if you switch always, you have a 2/3 chance of winning each time. If you don't ever switch, you have a 1/3 chance.

	25-12-2004 18:48									#10
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quote: Az: if you switch always, you have a 2/3 chance of winning each time. If you don't ever switch, you have a 1/3 chance.

Once again, how come?
whatever you choose at first, you're left with two cups: one of them has a prize, the other has no prize. "switching or not switching" is just like saying "choosing this cup, or choosing that cup". The fact that you chose some cup before that is irrelevant.

	25-12-2004 19:04									#11
	Increase the size of your Attachments

Azazel - ckeck my spoiler for the table .

	25-12-2004 19:07									#12
	Lose 30 kilos (of popups)

Let us expand this scenario a bit . Imagine there are a hundred cups , but only one with a prize . You pick one . 98 false options are eliminated . Now , does it make sense to switch ?

	25-12-2004 19:07									#13
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сhecked it. It completely ignores the removal of a necessarily empty cup.

	25-12-2004 19:11									#14
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quote: Let us expand this scenario a bit . Imagine there are a hundred cups , but only one with a prize . You pick one . 98 false options are eliminated . Now , does it make sense to switch ?

Again, it won't make any difference.

Answer me this: You have 3 cups. One of them has a prize under it. Then, an empty one is removed.
A) what's your chance of picking the right one?
B) how is that scenario different?

	25-12-2004 19:52									#15
	Put an end to popups!

quote: Originally posted by Azazel quote: Or put another way, if you don't switch, your chances of winning remain 1 in 3 regardless. How so?

You have a one in three chance at stage 1, yes? If you refuse to change your choice your chances do not increase as you are told before hand that selection of non-balled cups will be made. ONLY if there was random selection would your odds increase as ONLY then will the extra information be provided and so ONLY then will conditional probabilities apply.

	25-12-2004 19:54									#16
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quote: Originally posted by Azazel quote: Let us expand this scenario a bit . Imagine there are a hundred cups , but only one with a prize . You pick one . 98 false options are eliminated . Now , does it make sense to switch ? Again, it won't make any difference. Answer me this: You have 3 cups. One of them has a prize under it. Then, an empty one is removed. A) what's your chance of picking the right one? B) how is that scenario different?

The point is you make the selection before an empty cup is removed.

	25-12-2004 19:57									#17
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quote: The point is you make the selection before an empty cup is removed.

No, not really. Since switching or not switching, is like making a choice all over again.

	25-12-2004 20:57									#18
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The second choice you make is not symmetrical though.

	25-12-2004 21:00									#19
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Ball is under Cup A

Case 1:

You choose the Cup A. Either Cup B or C is eliminated, it doesn't matter which. You switch and you lose. You stick and you win

Case 2:

You choose Cup B. Cup C is eliminated. You switch to A and you win. You stick and you lose

Case 3:

You choose Cup C. Cup B is eliminated. You switch to A and you win. You stick and you lose.

The fact that a cup is eliminated provides no useful information as Cup A will not be eliminated whatever happens.

	25-12-2004 21:17									#20
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quote: The fact that a cup is eliminated provides no useful information as Cup A will not be eliminated whatever happens.

Well, duh, of course the cup that has the ball won't be eliminated. This explains nothing. There is one certain thing: Before you choose to "stick or change", you'll have two cups. One of them will house the ball, the other will be empty. What cup you held to earlier is meaningless.

I think that an experiment is in order. Replace random choice with a roll of a dice

	25-12-2004 21:18									#21
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EDIT: I just suck

Last edited by VJ on 25-12-2004 at 21:43

	25-12-2004 21:40									#22
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It seems that I have to concede defeat. I've repeated the experiment, and the results are quite discouraging, to say the least. (Btw, thanks mom for helping me with this one )

Good going, guys

	25-12-2004 22:01									#23
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There's a 2/3 chance that you picked an empty cup.
The empty cup that is turned over changes nothing about the fact that there was a 2/3 chance that you picked an empty cup! It simply shows you that that other one is definitely empty.
In other words, there's still a 2/3 chance that your cup is empty.

If you're still not following...
If you were to repeat this experiment many times you would end up picking an empty cup in 2 out of every 3 cases.
In the case that you have picked an empty cup, with the other empty cup revealed, you should obviously switch to get the prize.
In other words, in 2 out of 3 cases, you have to switch to get the prize.

	25-12-2004 22:03									#24
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This puzzler is known as the Monty Hall problem, named after the host of the TV show Let's Make a Deal. Google for "Monty Hall Problem" for more info.

	25-12-2004 22:51									#25
	Remove this text

quote: Originally posted by Mercator There's a 2/3 chance that you picked an empty cup. The empty cup that is turned over changes nothing about the fact that there was a 2/3 chance that you picked an empty cup! It simply shows you that that other one is definitely empty. In other words, there's still a 2/3 chance that your cup is empty.

But only a 1/2 chance when you are given the second choice of switching your cup.



There are two choices here, each with a different chance, but the first choice does not really matter because the result is always the same.

You are given three choices.

You chose one.

One of the other choices, which is wrong, is removed.

You are then asked if you are sure of your original choice, or if you want to change it.

You now only have two choices.

	25-12-2004 23:20									#26
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Always switch.

A woman has two kids. At least one of them is a boy. What are the chances she has two boys?

	25-12-2004 23:33									#27
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Ah, good stuff bods,

well, I will attempt to answer,
There are 4 permutations normally

Woman has 2 girls = 1/4
Woman has 1 each = 1/2 (boy+girl and girl+boy)
Woman has 2 boys = 1/4

With the information given only 3 can be possible, as two girls cannot be possible therefore we are left with 3/4, Woman having 2 boys now assumes the probability of 1 in 3, or 33.3%

Let me know how wrong I am

	25-12-2004 23:41									#28
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Very interesting Worthy's...

	26-12-2004 01:07									#29
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quote: Originally posted by Azazel Well, there obviously some trick here, but I am going to go at it straightforward. Correct me if I am wrong in any of my assumptions. A person chooses a cup. No matter what that cup has under it, the number of choices he has will drop to 2, by the elimination of a wrong answer. and then he can choose to stay with the cup, or not, which is basically the same as choosing one cup out of two all over again. This way, it's 50/50. Thus, I say: no difference. His chances are the same.

Incorrect.Since you always turn over a cup with nothing under it (deliberately) the odds for the original cup do not improve. The odds for the other cup do.

	26-12-2004 01:21									#30
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quote: Originally posted by General Ludd There are two choices here, each with a different chance, but the first choice does not really matter because the result is always the same.

The first choice does matter, because 2 out of 3 times you would have picked an empty cup, seeing as how 2 out of the 3 cups are empty. So obviously your first guess would be an empty cup 2 out of 3 times even when there are only two cups left to choose from. Consequently, 2 out of 3 times, you should switch.