well known proba problem

[shapenaji]

HINT:

so, the way to solve that last one,

Box A contains xa black
Box B contains xb black and yb White
Box C contains yc white

If we express the final solution in terms of these, then sum over all possibilities, we should be gold.

[perceval]

HINT:

so, the way to solve that last one,

Box A contains xa black
Box B contains xb black and yb White
Box C contains yc white

If we express the final solution in terms of these, then sum over all possibilities, we should be gold.

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

[Bill Spight]

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]

Agreed, the problem as posed is ambiguously worded.

[tj86430]

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?

1/3

There are four options:

Oldest = Girl, Youngest = Girl
Oldest = Girl, Youngest = Boy
Oldest = Boy, Youngest = Girl
Oldest = Boy, Youngest = Boy

Since I have observed you with one girl, I can eliminate the fourth option. Of the three remaining options, the other child is a boy in two cases, a girl in one case, hence there is a 1/3 chance the other child is a girl.

If we are going to differentiate the children by age, then we have to include that in the enumeration of possible observations. The following four options fit the observation and are equally likely:

Oldest = Girl, Youngest = Girl, Observed = Oldest
Oldest = Girl, Youngest = Girl, Observed = Youngest
Oldest = Girl, Youngest = Boy, Observed = Oldest
Oldest = Boy, Youngest = Girl, Observed = Youngest

I haven't looked at the wikipedia article yet, but this (mitsun's) is my analysis, too.

In this case the intuitive answer is correct.

[tj86430]

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?

Ok, I'll bite:

Let A be the number of (black) stones in first box
let B be the number of black stones in second box
let C be the number of white stones in second box

the probability is A/(A+B) + B/(A+B)*(B-1)/(B+C-1)

[tj86430]

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:

There are seven cubes (dice, if you will). The sides of the cubes are plain, and painted either white (W) or black (B). All the cubes have different number of white and black sides:
- 6W, 0B
- 5W, 1B
- 4W, 2B
- 3W, 3B
- 2W, 4B
- 1W, 5B
- 0W, 6B
The cubes are in a container, so that we can't see them. You pick a random cube from the container with your eyes shut and hold the cube so that only one side is visible (assume that it is possible to hold it in such a manner). Then you show the cube to me and open your own eyes. Now we both see that the visible side is black. Neither you nor me know nothing else of this cube. I bet, that the cube has at most one white side. Would you accept?

Second bet proposal:

Same cubes as before. No one picks anything yet, all the cubes are in the container. This time I propose: You will pick a random cube from the box and throw it like a die (assume that they all behave like perfect dice). After it has settled, we will observe the color of the side facing up. I bet, that the cube has at most one side of the opposite color (i.e. if black is up, there is at most one white side, and if white is up, there is at most one black side). Would you accept?

[schawipp]

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

Code: Select all

#!/bin/perl
use strict;

# We assume that bowl 0 contains WW, bowl 1 BB, bowl 2 WB

my $size=100000;
my $count=0;
my $hit=0;

while($count<$size) {
  my $bowl = int(rand()*3);
  if ($bowl==0) {         
    # Bowl 0 with WW stones picked. We will get a second W stone in any case
    ++$count; ++$hit;
  } elsif ($bowl==2) {
    # Bowl 2 with WB picked. The probability for initially picking W is 50%
    # otherwise this draw will not count
    # (This is the point, which I didn't consider on first thought...!)
    if(rand()<0.5) { 
      ++$count;
    }
  }
}

my $result=$hit/$count*100;
print "Result = $result %\n";
ap@sim037:~$ perl bowls.pl 
Result = 66.521 %

Now, I'm kinda convinced

[perceval]

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:

There are seven cubes (dice, if you will). The sides of the cubes are plain, and painted either white (W) or black (B). All the cubes have different number of white and black sides:
- 6W, 0B
- 5W, 1B
- 4W, 2B
- 3W, 3B
- 2W, 4B
- 1W, 5B
- 0W, 6B
The cubes are in a container, so that we can't see them. You pick a random cube from the container with your eyes shut and hold the cube so that only one side is visible (assume that it is possible to hold it in such a manner). Then you show the cube to me and open your own eyes. Now we both see that the visible side is black. Neither you nor me know nothing else of this cube. I bet, that the cube has at most one white side. Would you accept?

Second bet proposal:

Same cubes as before. No one picks anything yet, all the cubes are in the container. This time I propose: You will pick a random cube from the box and throw it like a die (assume that they all behave like perfect dice). After it has settled, we will observe the color of the side facing up. I bet, that the cube has at most one side of the opposite color (i.e. if black is up, there is at most one white side, and if white is up, there is at most one black side). Would you accept?

First bet: i do not take it

again we have picked a random B side amongst 21. you win if the face belong to either the 1W/5B or 0W/6B side so in 11/21 case slightly over 0.5

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

rolling the dice is an illusion, whatever the side it land on i win if the dice have at least one side of each color, so 5/7 (15/21), good odds

[tj86430]

@perceval:

rolling the dice is an illusion, whatever the side it land on i win if the dice have at least one side of each color, so 5/7 (15/21), good odds

I guess I either explained badly or you misunderstood: you dont win with any two-colored cube/die. You only win if:
- the cube/die has 2B/4W, 3B/3W or 4B/2W
- the cube/die has 1W/5B AND the white side is up
- the cube/die has 1B/5W AND the black side is up

If the cube/die has 1W/5B with the black side up, then the phrase "has at most one side of the oppisite color (this case white)" is true (and vice versa).

The second bet is in fact exactly the same as the first.

The purpose of this experiment is to show that (I suspect) many people will accept the first bet (thinking that they have 2/3 chance to win), but not the second one (if they can calculcate probabilities correctly). In both cases the odds for your win are 10/21 (and 11/21 for my win), as you correctly calculated in the first bet.

[perceval]

@perceval:

rolling the dice is an illusion, whatever the side it land on i win if the dice have at least one side of each color, so 5/7 (15/21), good odds

I guess I either explained badly or you misunderstood: you dont win with any two-colored cube/die. You only win if:
- the cube/die has 2B/4W, 3B/3W or 4B/2W
- the cube/die has 1W/5B AND the white side is up
- the cube/die has 1B/5W AND the black side is up

If the cube/die has 1W/5B with the black side up, then the phrase "has at most one side of the oppisite color (this case white)" is true (and vice versa).

The second bet is in fact exactly the same as the first.

The purpose of this experiment is to show that (I suspect) many people will accept the first bet (thinking that they have 2/3 chance to win), but not the second one (if they can calculcate probabilities correctly). In both cases the odds for your win are 10/21 (and 11/21 for my win), as you correctly calculated in the first bet.

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

[shapenaji]

HINT:

so, the way to solve that last one,

Box A contains xa black
Box B contains xb black and yb White
Box C contains yc white

If we express the final solution in terms of these, then sum over all possibilities, we should be gold.

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

My bad, for some reason I just overlooked that post, yeah seems right. Although I agree that my wording of the original problem was a bit ambiguous.

If I had to reword it would be:

Box A contains an unknown number of black stones but no stones of any other color
likewise for B and C.

[jts]

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 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 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]

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]

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?