Life In 19x19

shapenaji wrote: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.

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?

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?

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

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?

EdLee wrote:Suppose you have 3 Go bowls:
- One with 2 Shell stones
- One with 2 Slate stones
- One with 1 Shell and 1 Slate stone

#!/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 %

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:

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?

tj86430 wrote:@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 wrote:

shapenaji wrote: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 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?

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%.