balls in a bag

Originally posted by uzless
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: ther ...[text shortened]... s 1/3.

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

It is wrong because the black ball picked has given a hint, that this is more likely to be a bag with two black balls in.

Consider this:

You have two coins, one is fair and one always tosses heads. You don't know which is which.

You toss one of the coins, it comes up heads.
Could this be the fake coin? Do you know any more than you did before you tossed it?

Picking a ball out of the bag is a bit like tossing that coin.

Originally posted by Tatanka Yotanka
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.

My proof was not statistical. My proof was mathematical. There is no margin of error.

You are making a very common mistake when people try to do conditional probability by instinct. The fact that the first ball is black gives us more information. The probability of the three bags been chosen, given that we know the first ball was black, are not the same.

That's what the maths is for - Bayes' theorem allows us to calculate the difference.

Or you can look at it this way, which doesn't use any fancy maths.

All 8 balls are equally likely to be chosen first. We know a black ball was chosen, so it was equally likely to be any of the four black balls.

Two of the black balls are in a bag with a white ball. Two are in a bag with a black ball (each other). So the chance that the "other" ball is black is 2/4, or 50%.

Originally posted by uzless
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: ther ...[text shortened]... s 1/3.

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

Imagine you have a bag with 100k black and 2 white balls and another bag with 100k white balls and two black balls.

You choose one of the two bags randomly and pull out a black ball. Does your instinct still tell you that the probability of pulling a black ball from the same bag is (virtually) 50/50?

So the first ball gives you information about which bag you are likely to be holding and is not irrelevant.

Originally posted by uzless
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: ther ...[text shortened]... s 1/3.

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

An elegant (I think) way to disprove 1/3

IF the answer were 1/3 it would mean the chance of a white was 2/3 ... yes?

So if I pick a black ball first the chances of a white are 2/3
since the problem is symmetrical;
If I pick a white ball first the chances of a black are 2/3

therefore, regardless of whether I pick a black or white first the chances of getting a mixed pairing is 2/3

BUT WE KNOW THAT ONLY HALF THE BAGS HAVE A BLACK AND WHITE !!!!
AND THEREFORE THE ODDS OF GETTING B&W or W&B ARE 1/2

to calculate the probability of a black ball being drawn again from a bag randomly chosen, we should also consider the probability of each bag being chosen in random and factor it in.

as it is, the fact that the first ball drawn was black already eliminates one bag from the probabilities since that bag contains both white balls. that leaves us with three bags where the probability is that the bag with two black balls has been chosen is only one out of three, since in order to be able to draw another black ball, one must have picked the one with two black balls in it, otherwise, if either of the other two bags were picked instead, then the next ball is definitely white. therefore, this gives us only a 1 in three chance or a 33.33% probability!

Originally posted by roma45
my thinking is if you pick a black ball that means the bag with the two white balls is out of the equation. leaving 3 bags with 4 black and 2 white in total, since 1 black is picked that leaves 3 black aginst 2 white, hope i am right

the flaw in that equation however is its failure to consider that a black ball has already been drawn, which actually lowers the probability of black being the second ball into exactly half of it. while its true that there are still three black balls and two white balls left, we know that they are not in only one bag. in fact, two of those bags contain white balls which means only one bag is capable of producing another black since a black ball has already been drawn, giving us one out of three probability.

Originally posted by kyletoybits
as it is, the fact that the first ball drawn was black already eliminates one bag from the probabilities since that bag contains both white balls. that leaves us with three bags where the probability is that the bag with two black balls has been chosen is only one out of three...

BUT...knowing the first random ball was black modifies the probability of which bag it was drawn out of. You particially accept that - the probability of the bag with two balls is reduced to zero. Why assume the other bags are still all equally likely? Read Palynka's last post - I think that's a good example that shows this isn't true.

I still think the easiest way to think of it is to consider the balls, not the bags. All four of the black balls are equally likely to be the one chosen. Two out of the four are in a bag with another black ball, the other two are with a white one. 2/4 = 1/2. Simple!

Originally posted by wolfgang59
A different version of a puzzle I put up some time ago.


I have 4 identical bags each containing 2 balls.
1 bag has 2 white balls
1 bag has 2 black balls
2 bags have 1 black and 1 white

I pick a bag at random and take out a ball. It is black.

[b]What is the probability of the other ball being black?[/b]

help, every answer on here seems to make sense please tell me the answer, cant see why it can be 50%

Originally posted by roma45
help, every answer on here seems to make sense please tell me the answer, cant see why it can be 50%

SLOWLY

Bag 1 contains WW
Bag 2 contains BB
Bag 3 contains BW
Bag 4 contains BW

Randomly I pick a bag and [b]randomly[/]pick a ball from that bag.
It is black.

what are the chances I chose bag1? ...... zero
what are the chances I chose bag 2? ...... 1/2 (2 balls in 4)
what are the chances I chose bag 3? ...... 1/4 (1 ball in 4)
what are the chances I chose bag 4? ...... 1/4 (1 ball in 4)



Opponents to this say
what are the chances I chose bag 1? ...... zero
what are the chances I chose bag 2? ...... 1/3 (1 bag in 3)
what are the chances I chose bag 3? ...... 1/3 (1 bag in 3)
what are the chances I chose bag 4? ...... 1/3 (1 bag in 3)
BUT bag 2 is TWICE as likely to have been chosen !!!!! so this MUST be wrong

For those of you who think answer is 1/3 use same logic on this puzzle.

I have 4 bags each containing 1,000 balls

Bag 1 contains 1,000 white balls
Bag 2 contains 1,000 black balls
Bag 3 contains 999B and 1W
Bag 4 contains 999B and 1W

I pick a bag and take a white ball out.
What are the chances that the next ball is black?
2/3?
NO!

For those that think the probability is 1/3rd, let us play this game.

I take a random bag and take out a ball. We restart if it is white.

If it is black, I take another ball, if this new ball is black you pay me $3, if the new ball is white I pay you $2.

You will win: in 3 goes you pay me $3 on average and I pay you $4, how can you lose?

Reveal Hidden Content

Why didnt I think of that?
rec

Originally posted by wolfgang59
A different version of a puzzle I put up some time ago.


I have 4 identical bags each containing 2 balls.
1 bag has 2 white balls
1 bag has 2 black balls
2 bags have 1 black and 1 white

I pick a bag at random and take out a ball. It is black.

[b]What is the probability of the other ball being black?[/b]

It is 100% likely the "other ball" is black and that it is white because the "other ball" can be any of the remaining balls in any of the bags of my choosing. 😛

Originally posted by iamatiger
For those that think the probability is 1/3rd, let us play this game.

I take a random bag and take out a ball. We restart if it is white.

If it is black, I take another ball, if this new ball is black you pay me $3, if the new ball is white I pay you $2.

You will win: in 3 goes you pay me $3 on average and I pay you $4, how can you lose?

[hidden] bankrupcy looms [/hidden]

I'm still not getting it. I see it like this.

you now have 3 bags. one bag has a black ball. the other 2 bags have 1 white ball each.

The odds of picking the black ball are clearly 1 in 3.


You guys seem to be calculating what the odds are BEFORE you pick the first ball. I'm calculating the odds AFTER you pick the first ball, as is asked by the OP question.