well known proba problem

shapenaji · **#61**

HINT:

perceval · **#62**

shapenaji wrote:

HINT:

that is what i tried above, anything wrong with my calculation ?

Bill Spight · **#63**

shapenaji wrote:

How about a variation on the original?

We have 3 boxes,
the first contains an unknown number of black stones
the second contains an unknown number of black and white stones,
the third contains an unknown number of white stones

We pick a box, pull a stone and it's black,

What is the probability that if we pull another stone from the box, that it will also be black?

Unknown.

TheBigH · **#64**

Agreed, the problem as posed is ambiguously worded.

tj86430 · **#65**

mitsun wrote:

HermanHiddema wrote:

Bill Spight wrote:

OK. Here is an similar problem. Suppose, first, that the number of boys born is equal to the number of girls born. Suppose also that you know that I have two children, but you do not know their genders. One afternoon you (randomly) run into my wife, along with one of my children, who is a girl. What is the probability that my other child is also a girl?

tj86430 · **#66**

shapenaji wrote:

How about a variation on the original?

We have 3 boxes,
the first contains an unknown number of black stones
the second contains an unknown number of black and white stones,
the third contains an unknown number of white stones

We pick a box, pull a stone and it's black,

What is the probability that if we pull another stone from the box, that it will also be black?

tj86430 · **#67**

While we are at it, I'd like to present one more (I thought of this yesterday evening before falling asleep):

I will propose you two bets (even money). Assumption is that you accept the bet, if you think that you are more likely to win than lose, setting aside all considerations about your financial situation, ethics, religion etc. The other assumption is that there is no cheating. The two bets are under the two hide-tags that follow. Please consider the first bet first, and only after you have decided how you would react to that, look at the second bet and make a decision on that as well.

First bet proposal:

Second bet proposal:

schawipp · **#68**

EdLee wrote:

Suppose you have 3 Go bowls:
- :white:

One with 2 Shell stones
- :black:

One with 2 Slate stones
- :white:

One with 1 Shell and 1 Slate stone

At the beginning, I had no idea why the result shouldn't be 50%. But then, I wrote a perl script:

Now, I'm kinda convinced

perceval · **#69**

tj86430 wrote:

While we are at it, I'd like to present one more (I thought of this yesterday evening before falling asleep):

I will propose you two bets (even money). Assumption is that you accept the bet, if you think that you are more likely to win than lose, setting aside all considerations about your financial situation, ethics, religion etc. The other assumption is that there is no cheating. The two bets are under the two hide-tags that follow. Please consider the first bet first, and only after you have decided how you would react to that, look at the second bet and make a decision on that as well.

First bet proposal:

Second bet proposal:

First bet: i do not take it

second bet bet: i take it EDIT: wrong i misread the 2nd bet

tj86430 · **#70**

@perceval:

perceval · **#71**

tj86430 wrote:

@perceval:

my bad, your explanation was clear , i thought it was *at least* instead of *at most* in the second bet

shapenaji · **#72**

perceval wrote:

shapenaji wrote:

HINT:

that is what i tried above, anything wrong with my calculation ?

jts · **#73**

A note for people who find the original problem unintuitive:

Before you draw any balls from the urn, you assume you have a 50% chance that stone 1 will be B and a 50% chance that stone 2 will be B. If, after you draw one stone and get information that rules out one of the urns (e.g., the :white:

urn), it would be strange if you still believed there was a 50% chance of drawing white (which is what you believed when you thought there was a 33% chance of picking from the :white:

urn).

Or to think about it another way -- assume you fall asleep on your bus and get out in a strange neighborhood. You aren't sure whether you got out in an English-speaking neighborhood, a Spanish-speaking neighborhood, or a neighborhood where half speak Spanish and half English. When you step off the bus, you know that there is a 50% chance that the first person ask for directions will speak English, 50% chance that the second person you ask for directions will speak English, etc.... but after you ask the first person and and he doesn't understand you, doesn't your heart immediately sink? Congratulations, you find probability intuitive!

Bantari · **#74**

Bill Spight wrote:

Bantari wrote:

So, the whole question, paraphrased, simplified, and without confusing and unnecessary details, boils down to this:

- You have TWO boxes, Bb and Bw.
- Pick one at random.
- What is the chance of having picked Bb? (since this is the only scenario in which the second/hidden stone is also B)

The answer is clearly 50%.

If you pick one of those two boxes at random, indeed, the probability is 50% that one of the boxes has two Black stones in it. But we have not used all the information that we got from drawing a Black stone from a box.

Suppose now that you draw a stone from the box that you picked, and it is Black. Now what is the probability that the box has another Black stone in it?

Yeah, sleep is a wonderful thing... I should not get into these types of discussions in the middle of the night.

Now I see that you can view the question in two different ways, maybe its the wording... and I think I got fixated on the wrong interpretation. Or maybe everybody else has... Anyhow... I think this is the bottom of the issue here:

Interpretation #1:
You can see it like the TJ's code - you have three boxes, draw a stone randomly, and out of the cases when B comes up first, what is the statistical probability that a second B will come second too? You have 3 possible cases in the results set: (B1,B2), (B2,B1), and (B3,W1) - and out of the 3 in 2 B will also be on second draw. Thus the answer in this case is 2/3, as demonstrated. Or:

Interpretation #2:
You FIX the situation so that B WILL COME FIRST ON EACH DRAW (the wording of the question MIGH SEEM TO stipulate this) - like you throw away the WW box and attach a string to A BLACK STONE in each remaining box ensuring it to be the first draw. In which case it is only the question which of the two boxes you pick and the answer is 50%.

I assume that interpretation #1 is perceived as 'correct' by the math types out there... although I don't really see why that should be without knowing more about the circumstances of the initial problem.

Anyways - does the above two interpretation seem like a likely source of the confusion?
It makes it a non-issue for me, mathematically... everything's clear other than the problem itself.

HermanHiddema · **#75**

Bantari wrote:

Interpretation #2:
You FIX the situation so that B WILL COME FIRST ON EACH DRAW (the wording of the question MIGH SEEM TO stipulate this) - like you throw away the WW box and attach a string to A BLACK STONE in each remaining box ensuring it to be the first draw. In which case it is only the question which of the two boxes you pick and the answer is 50%.

Well, it seems to me that this interpretation is a bit strange. The original question says nothing about attaching strings to stones or anything, it just describes what happens.

But anyway, I have another way to look at it:

Puzzle A: Given three bowls, one with two black stones, one with two white stones, one with one black and one white. You pick a bowl at random and draw a stone from it. It happens to be white. What is the chance the second stone in that bowl is also white?

Puzzle B: Given three bowls, one with two black stones, one with two white stones, one with one black and one white. You pick a bowl at random and draw a stone from it. It happens to be black. What is the chance the second stone in that bowl is also black?

Puzzle C: Given three bowls, one with two black stones, one with two white stones, one with one black and one white. You pick a bowl at random and draw a stone from it. What is the chance the second stone in that bowl is the same color as the first one?

In my opinion, given the symmetry of the situation, all three puzzles are really equivalent. I would also argue that the third one obviously has the answer 2/3, right?

Bantari · **#76**

HermanHiddema wrote:

Bantari wrote:

Interpretation #2:
You FIX the situation so that B WILL COME FIRST ON EACH DRAW (the wording of the question MIGH SEEM TO stipulate this) - like you throw away the WW box and attach a string to A BLACK STONE in each remaining box ensuring it to be the first draw. In which case it is only the question which of the two boxes you pick and the answer is 50%.

Well, it seems to me that this interpretation is a bit strange. The original question says nothing about attaching strings to stones or anything, it just describes what happens.

I know, the string is my idea... pretty clever, eh?

But seriously - its just one of the possible ways to ENSURE that the first stone picked is B. Another way would be to place A BLACK STONE on top of the box... Or whatever, I'm sure there are other ways Think in practical terms: If you were to conduct such experiment physically, with actual boxes and stones, how can you ENSURE that your first pick is a B stone?

HermanHiddema · **#77**

Bantari wrote:

HermanHiddema wrote:

Bantari wrote:

Interpretation #2:
You FIX the situation so that B WILL COME FIRST ON EACH DRAW (the wording of the question MIGH SEEM TO stipulate this) - like you throw away the WW box and attach a string to A BLACK STONE in each remaining box ensuring it to be the first draw. In which case it is only the question which of the two boxes you pick and the answer is 50%.

Well, it seems to me that this interpretation is a bit strange. The original question says nothing about attaching strings to stones or anything, it just describes what happens.

I know, the string is my idea... pretty clever, eh?

But seriously - its just one of the possible ways to ENSURE that the first stone picked is B. Another way would be to place A BLACK STONE on top of the box... Or whatever, I'm sure there are other ways Think in practical terms: If you were to conduct such experiment physically, with actual boxes and stones, how can you ENSURE that your first pick is a B stone?

Well, the thing is, why would you want to? The original problem uses the phrase: "This Coin turns out to be a Gold coin". It doesn't ensure anything, it just describes what happens. To me, the idea of "ensuring" looks like a far-fetched attempt to make the problem fit an intuitive but mistaken answer.

In practise, the color doesn't matter, due to symmetry. If you were to conduct the experiment, you would ask anyone grabbing a black stone to estimate the probability that the other stone in the bowl is also black, and you would ask anyone grabbing a white stone to estimate the probability that the other stone is also white. Both cases are exactly equivalent to the original problem, both have the same answer, and 100% of test subjects can be asked the question.

Bantari · **#78**

HermanHiddema wrote:

Bantari wrote:

HermanHiddema wrote:

Well, it seems to me that this interpretation is a bit strange. The original question says nothing about attaching strings to stones or anything, it just describes what happens.

I know, the string is my idea... pretty clever, eh?

But seriously - its just one of the possible ways to ENSURE that the first stone picked is B. Another way would be to place A BLACK STONE on top of the box... Or whatever, I'm sure there are other ways Think in practical terms: If you were to conduct such experiment physically, with actual boxes and stones, how can you ENSURE that your first pick is a B stone?

Well, the thing is, why would you want to?

Ugh... I don't know what to say to that. I guess I had it coming.
Anyways - nice talking to you, bud.

Bill Spight · **#79**

Bantari wrote:

Bill Spight wrote:

Bantari wrote:

So, the whole question, paraphrased, simplified, and without confusing and unnecessary details, boils down to this:

- You have TWO boxes, Bb and Bw.
- Pick one at random.
- What is the chance of having picked Bb? (since this is the only scenario in which the second/hidden stone is also B)

The answer is clearly 50%.

If you pick one of those two boxes at random, indeed, the probability is 50% that one of the boxes has two Black stones in it. But we have not used all the information that we got from drawing a Black stone from a box.

Suppose now that you draw a stone from the box that you picked, and it is Black. Now what is the probability that the box has another Black stone in it?

Yeah, sleep is a wonderful thing... I should not get into these types of discussions in the middle of the night.

Now I see that you can view the question in two different ways, maybe its the wording... and I think I got fixated on the wrong interpretation. Or maybe everybody else has... Anyhow... I think this is the bottom of the issue here:

Interpretation #1:
You can see it like the TJ's code - you have three boxes, draw a stone randomly, and out of the cases when B comes up first, what is the statistical probability that a second B will come second too? You have 3 possible cases in the results set: (B1,B2), (B2,B1), and (B3,W1) - and out of the 3 in 2 B will also be on second draw. Thus the answer in this case is 2/3, as demonstrated. Or:

Interpretation #2:
You FIX the situation so that B WILL COME FIRST ON EACH DRAW (the wording of the question MIGH SEEM TO stipulate this) - like you throw away the WW box and attach a string to A BLACK STONE in each remaining box ensuring it to be the first draw. In which case it is only the question which of the two boxes you pick and the answer is 50%.

I assume that interpretation #1 is perceived as 'correct' by the math types out there... although I don't really see why that should be without knowing more about the circumstances of the initial problem.

Anyways - does the above two interpretation seem like a likely source of the confusion?
It makes it a non-issue for me, mathematically... everything's clear other than the problem itself.

I went back and took a look at your first post in this thread. Here is what you were responding to.

Quote:

Choices at first:

- BB
- BW
- WW

I choose a box and take out a B stone, so the WW option has just been eliminated, hasn’t it? How should then a 2/3 chance remain? There is no /3 anymore, so it should be 2/2 -> 1:1 chance, no?

You are right that taking out a B stone is ambiguous.

But here is an earlier statement of the problem.

Quote:

Suppose you have 3 Go bowls:
- One with 2 Shell stones :white:

- One with 2 Slate stones :black:

- One with 1 Shell and 1 Slate stone :white:

Suppose you randomly pick a bowl, then take one of the 2 stones out from the bowl.
This stone turns out to be :white:

. What is the probability that the second stone in the bowl is also :white:

?

The color of the stone changed, but that is immaterial.

Here you did not deliberately pick a :white:

, that's just how it turned out.

Therefore an interpretation that guarantees that that the picked stone is :white:

is incorrect. (Sure, somebody may have rigged the choice, but probability is about what what we know. Somebody could have rigged the choice the other way, as well.

)

kivi · **#80**

Actually you can change the last line of the puzzle to be:

Quote:

This stone turns out to be a particular color. What is the probability that the second stone in the bowl is also same color?

in order to avoid the fixation about fixing the situation

well known proba problem

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