balls in a bag

Originally posted by Tatanka Yotanka
25%

You picked a black ball and are asking what the chances of the other ball also being black. Since only 1 of the 4 bags had 2 black balls (which must be the case for the second ball to also be black) then the chances of having picked the Black+Black bag are 1 in 4: 25%

Ok, lets say Wolfgang pick bags 200 times, and does random ball picks.

In 50 he gets a "ww" bag, these times are excluded by the information in the question.

In 100 he gets a "bw" bag, in 50 of those he picks a black ball (the times he picked a white ball were excluded by the information)

In 50 he gets a "bb" bag.

So out of Wolfgang's 200 picks, 100 of the outcomes are excluded by the information in the question, leaving 100 possible outcomes.

Out of the 100 possible outcomes, in 50 of them he has the "bb" bag and there is another black ball in it.

50/100 = 50%

Originally posted by iamatiger
The wording of the question is not quite clear. If Wolfgang *looks* into a random bag, and takes out a black ball *if it is there*, then the odds are completely different to the situation when he randomly takes a ball out of a random bag and it happens to be black.

If it is the former method, then it is very like the Monty Hall problem, Wolfgang has on ...[text shortened]... e totally sure, given the wording of the question, that Wolfgang is picking the ball randomly...

Good point. I am picking the bag randomly and then randomly picking a ball from that bag.

The answer is 50%

Good explanations from Thomaster but I'm not sure I follow iamatiger's.

Originally posted by wolfgang59
Good point. I am picking the bag randomly and then randomly picking a ball from that bag.

The answer is 50%

Good explanations from Thomaster but I'm not sure I follow iamatiger's.

I'm sorry, but the answer is not 50%.

It is also dependent on when you want to calculate the odds of the second ball being black: If you calculate them before you know what the color of the first ball is then it's 25%. If you determine that the first ball is black and then calculate what the color of the second ball is going to be then it's 33% (1 in 3, since the White+White is no longer a viable possibility)

I don't see how you maintain your 50% assertion since only 25% of the bags were ever capable of producing the combination.

Originally posted by Tatanka Yotanka
I'm sorry, but the answer is not 50%.

Ok, then show us your calculations using math.

Originally posted by iamatiger
Ok, lets say Wolfgang pick bags 200 times, and does random ball picks.

In 50 he gets a "ww" bag, these times are excluded by the information in the question.

In 100 he gets a "bw" bag, in 50 of those he picks a black ball (the times he picked a white ball were excluded by the information)

In 50 he gets a "bb" bag.

So out of Wolfgang's 200 picks ...[text shortened]... n 50 of them he has the "bb" bag and there is another black ball in it.

50/100 = 50%

You are resorting to statistics, and you know what is said about those.

We are not talking about what happens over the course of 200 drawings, we are talking about him picking a bag one time and out of that bag picking a black marble and the chances of the second one also being black.

Originally posted by Tatanka Yotanka
I'm sorry, but the answer is not 50%.

It is also dependent on [b]when you want to calculate the odds of the second ball being black: If you calculate them before you know what the color of the first ball is then it's 25%. If you determine that the first ball is black and then calculate what the color of the second ball is going t ...[text shortened]... our 50% assertion since only 25% of the bags were ever capable of producing the combination.[/b]

The question is: ''What is the probability of the second ball being black, given that the first was black?''

That was what I have calculated. It is 50%.
Bayes:
p(A | black) = p{black | A) × p(A) / p(black)

p(black | A) = 1
p(A) = 1/4 = 0,25
p(black) = 0,5

Where do I go wrong, according to you?

Originally posted by Tatanka Yotanka
You are resorting to statistics, and you know what is said about those.

Usually by people that don't understand them. 🙂

Bayes' theorem is definitely the way to go. Here's my version of it.

P(2nd ball black | 1st ball black) = P(2nd ball black & 1st ball black)/P(1st ball black)

Both balls are black if and only if the bag with two black balls is chosen. So
P(2nd ball black & 1st ball black) = 1/4

P(1st ball black) = 1/2 - straightforward (just as likely to be either)

Therefore: P(2nd ball black | 1st ball black) = (1/4)/(1/2) = 1/2

So 50% - I agree with the others. You want to disagree, either point out the error in the calculation or disprove Bayes' theorem. And good luck with that!

Originally posted by mtthw
Usually by people that don't understand them. 🙂

You want to disagree, either point out the error in the calculation

We know the layout, but to recap: There are 8 marbles: 4 black; 4 white; divided and packaged as described.

A marble is chosen and there was a 50% chance that the marble would be black. Lucky us; it's black! That black marble could have come from one of 3 bags. Of the 3 possible bags only one of them can produce another black marble. Your chances are 1/3, or 33%.

However, through the magical/dubious use of statistics you declare that it's impossible for both of the B+W bags to produce a black marble first. Statistics dictate that only one of those bags can produce a black first because one of them has to produce a white first. So armed with the almighty statistical proof, you summarily discount one of the 3 possibilities and decide that there are really only 2 bags left so you say 50%.

Or worse, and closer to the truth: The statistics don't say it's impossible, just that it's improbable ... yet you discard the 3rd bag anyway.

Originally posted by Tatanka Yotanka
We know the layout, but to recap: There are 8 marbles: 4 black; 4 white; divided and packaged as described.

A marble is chosen and there was a 50% chance that the marble would be black. Lucky us; it's black! That black marble could have come from one of 3 bags. Of the 3 possible bags only one of them can produce another black marble. Your chances on't say it's impossible, just that it's improbable ... yet you discard the 3rd bag anyway.

If you still don't believe everyone else, simulate it in a popular spreadsheet!


labels:
A1 = rand1, B1 = "Bag", C1 = rand2, D1 = "Ball 1", E1 = "Ball 2"

In A2 we are going to have a random integer between 0 and 3 to decide which of the four bags we get, so the formula can be:
=FLOOR(RAND()*4,1)

In B2 we have the bag we chose:
=IF(A2 = 0, "ww", IF(A2 = 3, "bb", "bw" )
(note the subsequent formulas depend on those exact strings)

so we can see we will get mixed bags twice as often as either of the other two bags, which is what the question specifies.

In C2 we are going to have another random number which controls which of the two balls in the bag we get:
=FLOOR(RAND()*2,1)

In D2 we are going to have the colour of the first ball:
=MID(B2, C1+1, 1)
this will select the first or second character in the ball name depending on the value of C1.

and in E2 we are going to have the colour of the second ball
=MID(B2, 2-C1, 1)
this will always be the other character in the ball name

Now copy those formulas down about 1000 rows. We have done 1000 bag/ball picks
Copy everything and paste as values so the ball picks don't change on us. otherwise the autofiltering below won't work.

Now highlight row 1 and do data / filter to get filter dropdowns based on the data in row 1.

Select the filter in the "ball 1" column and set to show "b" only.

highlight a single column and excel will give you the count of visible cells (actually cells+1 because it counts the column heading too), it will be about 500. This is the number of times a black ball was picked. Remember this number.

Without changing the "ball 1" filter, select the filter on the "ball 2" column and set to show "b" only.

Now highlight a column and excel will again tell you how many cells are visible after the filter. The number will be about 250, proving that half of the times that the first ball is black the second ball is too.

Originally posted by iamatiger
If you still don't believe everyone else

It's not a matter of belief. It's a matter of knowing that there were 3 possible bags from which that first black marble could have come. One of those can produce another one. It's very simple. You can come up with a million formulas and theories in a vain attempt to obfuscate the simplicity of the problem; some people will even buy it.

I guess I'm just one of those sheep that refuse to recognize that statistically only 2 bags could have produced that initial black marble when I'm told that 3 bags had one to offer up.

Originally posted by Tatanka Yotanka
It's not a matter of belief. It's a matter of knowing that there were 3 possible bags from which that first black marble could have come. One of those can produce another one. It's very simple. You can come up with a million formulas and theories in a vain attempt to obfuscate the simplicity of the problem; some people will even buy it.

I guess I ...[text shortened]... ould have produced that initial black marble when I'm told that 3 bags had one to offer up.

You're just trolling, aren't you?

Originally posted by iamatiger
If you still don't believe everyone else, simulate it in a popular spreadsheet!


labels:
A1 = rand1, B1 = "Bag", C1 = rand2, D1 = "Ball 1", E1 = "Ball 2"

In A2 we are going to have a random integer between 0 and 3 to decide which of the four bags we get, so the formula can be:
=FLOOR(RAND()*4,1)

In B2 we have the bag we chose:
=IF(A2 = 0, "ww", ...[text shortened]... econd ball
=MID(B2, 2-C1, 1)
this will always be the other character in the ball name

Good grief! Get a module, would you! Honestly, who programs inside the spreadsheet cells! Yikes!

Originally posted by Palynka
You're just trolling, aren't you?

Hey, good job on shortening up that uncomfortable text. If the uncomfortable shoe fits, delete it in the name of intangible brevity and pretend it never happened. Meh.

Originally posted by Tatanka Yotanka
Hey, good job on shortening up that uncomfortable text. If the uncomfortable shoe fits, delete it in the name of intangible brevity and pretend it never happened. Meh.

You could be a little more subtle in the future, though. Pretty poor trolling display, to be honest.

As I see it, and i admit I do tend to see the math side of things incorrect, I see the first part of the question being almost irrelevant since we are told the first ball is black. We should really start the question here because we are given new information and must recalculate the odds of pulling a black ball again.

The question is really this: there are 3 bags with one ball in each. 2 bags have white balls, one bag has a black ball. What are the odds of pulling the black ball?

Of course, the answer is 1/3.

Tell me where i'm wrong because it seems you guys are answering a different question.