balls in a bag

Originally posted by wolfgang59
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!

In this case it seems there was a 99% chance the first bag was chosen on the first pull.

So it's something like 1% chance the next ball is black, if I understnad the thought process here.

Originally posted by uzless
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.

The key is the words "we restart if it is white".

In this game, like in the problem, we have restarted half the time we get a mixed bag. We don't ever restart when we get a BB bag. So there is only half the probability of having a mixed bag bag that you think.

Counting balls doesn't work very well because it ignores the bags.

Originally posted by uzless
In this case it seems there was a 99% chance the first bag was chosen on the first pull.

So it's something like 1% chance the next ball is black, if I understnad the thought process here.

correct (roughly) .. so can you see why 1/3 is wrong for the original question?

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.

This is false. There are four black balls that can be selected on the first attempt. Out of these four black balls, two of them have a pair which is also black. 2/4 = 50%.

What you are missing is the fact that there are two ways to choose a black ball from the BB bag, which doubles the likelihood that the first black ball has been selected from that bag.

Awesome thread. There should be a mandatory 1 week probability course to graduate high school.

Originally posted by randolph
Awesome thread. There should be a mandatory 1 week probability course to graduate high school.

Would that be enough? I think it takes a lot of time to get rid of "some" statistical biases that almost everyone seems to have initially... and still they creep up on you sometimes. In some extreme cases it feels like you're fighting with yourself, like your "intuition" just doesn't want to accept the arguments.

this is of course 1/3.

You say whats the prob of picking a black ball of out the same bag.

so now we just look at probs now and not before picking a ball.

there are 3 cases:

you picked a ball from a 2x black
you picked a ball from a black white
you picked a ball from a black white

the other ball will be black in 1/3 cases

Originally posted by Klasker
this is of course 1/3.

You say whats the prob of picking a black ball of out the same bag.

so now we just look at probs now and not before picking a ball.

there are 3 cases:

you picked a ball from a 2x black
you picked a ball from a black white
you picked a ball from a black white

the other ball will be black in 1/3 cases

I bet you also think you have a 50% chance of winning the lottery. After all, there are only two cases. You either have the winning ticket or you don't.

🙂

Originally posted by Banana King
🙂

Throwing my 50 cents into the mix... a totally non-mathematical mind amongst you formula junkies, so go easy.

WW is eliminated with the first draw of B.

B could only come from BB, BW or BW. It has a 1/3 chance of being from any one of them.

If it came from BB, there is a 3/5's chance of drawing B again, since there are possibly three B's left to two W's.

If it came from either of the BW's, there is again a 3/5's chance of drawing B again, since the remaining bags have BB and BW, or 60% chance of being drawn again.

Flame on.

((3/4)x(1/4))+(3/4)x(1/2)=
that was black balls
heres white
(1/4)+((1/2)x(1/2))=
am i right?

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]


The question originally asked, clearly states, that a black ball has already been drawn. And it asks, what is the probability NOW to draw another black ball. That means, the bag with two white balls is already excluded and does not count anymore for probability.

The black ball can only come from the b/b or one of the two b/w bags.

So the chance is 1/3 for drawing another black ball from that bag.

Originally posted by tharkesh
So the chance is 1/3 for drawing another black ball from that bag.

Out of interest, have you read any of the various previous messages in the thread explaining why that's wrong?

Originally posted by mtthw
Out of interest, have you read any of the various previous messages in the thread explaining why that's wrong?

Actually I read all. And all of them neglect the point, that by having drawn already a black ball, the possibilities have been limited to three bags...

The way the riddle was asked, implies, that you already went down the probability tree along one arm.

To explain the difference:

If you have thrown a coin and it shows up heads, what are the chances, that the next throw is heads?

The answer is 50%, because in this case it doesnt matter what you threw before. The event is called 'unrelated' to previous event, leaving us an 50% chance to throw heads, after having thrown heads already (becuase there exist only two options: h or t).

If you ask differently, it will get a different result: what is the chance of throwing heads twice in a row? This probability is 25%, because of the four options (hh,ht,th,tt).

As some explained before, the correct answer would have to be 1/3 now, because the decision about the first draw was taken already. The drawing of the second ball is NOT unrelated to the drawing of the first ball. That means, IF you know, that you have drawn a black ball, you also know, that you have one of the three following bags in front of you: bb, bw or wb. Having drawn b, leaves you in two cases with w and in one with b.

Ok, you read them, but you didn't understand them.

"And all of them neglect the point, that by having drawn already a black ball, the possibilities have been limited to three bags..."

But the three bags do not have the same probability.

Try this one. One bag, containing 999 white and one black. A second bag, containing 999 black and one white. You pick a bag at random, and pick a ball at random. It's black. Do you really think it's still 50/50 which bag you picked? This is the same principle.

Failing that, try it out!