marbles again

Hang on a while Whim, I am 2/3 the way to solving this problem. However, it is not so black and white.

50/50 its either black or white the first marble has no bearing on the second

50/50?

Oh! I dont think thats right. However, I have been known to be wrong, but not this time...I hope.

The first marble out could :-
the marble you put in
the other marble white
the other marble black
2/3 in favor of white

The second time not knowing which was pulled first time it only leaves an either or i.e 50/50

I see where you are coming from.

However, there are two marbles we can label PUTIN and MYSTERY, but since the MYSTERY marble may be either black or white, we can call these alternatives MYSTERY black and MYSTERY white. (Since PUTIN is known to be white, it can simply be PUTin.) There are four possible outcomes of our first drawing:

1) Marble PUTin is drawn, leaving MYSTERY white in the bag;
2) Marble PUTin is drawn, leaving MYSTERY black in the bag;
3) Marble MYSTERY white is drawn, leaving PUTin in the bag;
4) Marble MYSTERY black is drawn, leaving PUTin in the bag;

However, since a white marble is drawn, (4) did not occur. So there are three other possible outcomes, two of which have a white marble left, so the odds are 2/3. Which coincidently is the same number as my first post.

To err, I am told is human, therefore I hope you are right so I may be proved so.

If the PUTin marble is drawn first there are only two possibilities left , the MYSTERY marble black or white.
If the MYSTERY marble is drawn there are still only two possibilities left ,the PUTin marble but as we do not know which marble we have drawn it is still only two possibilities black or white. The question was what are the chances on the final draw. The odds are different if you ask the odds from the start of the drawing.

Whims question:
"What is the probability that the other marble in the bag is also white?"

Oh no. Now I am beginning to side with you. Except for the fact, that if the first marble out is the Mystery, then there is a 1:1 that the remaining is white.

Now I am really confused.

Tryagain, I think it's 2/3 too. But wasn't sure. Perhaps mark can act as referee?

It comes down to one marble so it can only be white or black.

2/3 (assuming that the marble in the bag is black or white with equal probabilities)

If you don't believe it, let's make it a betting game.

A neutral third party will put a black or white marble in the bag (each with probability .5).

You will add a white marble, then draw a marble from the bag.

If the marble is black, no money is exchanged.
If the marble is white then
- you pay me $1 if the remaining marble is white
- I pay you $1.25 if the remaining marble is black

How can you pass up those odds?

Easy, since you win only half the time there was a black marble in the bag, but I win every time there was a white marble in the bag.

On average, you'll lose 25 cents per game in spite of the fact that you think the odds are in your favor.

In a bag is a marble that is either white or black. You put a white marble into the bag. Then someone blindfolded takes out one marble, it is white.
What is the probability that the other marble in the bag is also white?

Is this right?

The chance you pick a white one if two whites are in the bag is 1.
The chance you pick a white one if one black and one white marble are in the bag is 1/2

So the general chance you pick a white one is then 1/2*1+1/2*1/2 = 3/4.

The chance the marble still in the bag is also white is, 1/2 / 3/4 = 2/3

Half of the time (when white was in the bag), the probability is 1 that another white is in the bag.
One quarter of the time (when black was in the bag), the probability is 0 that another white is in the bag.

(.5*1 + .25*0)/(.5+.25) = .5/.75 = 2/3

Yes, Whim, you are right.

In general, if the probability is p that a white marble was in the bag, then the probability is
p / [p + (1 - p) / 2]
that a white marble is still in the bag.

I vote for 50/50.
Here's an easy way to see why.

You are asked for probabilities AFTER you know the drawn marble was white (not before). So it is a fact, white marble is drawn, so what are the chances that next to come out is white?

Let's start with the original bag. There's one marble, either black or white. You put in a white, and take out a white. Doesn't matter if you took out the one you just put in, or the mystery marble. What does matter, that the marble remaining in bag is the SAME COLOR as was the original marble in the bag. Therefore it is 50/50 chance to be white or black (just as the original marble's chance was 50/50)...

If you were asked for chances to draw two whites in a row, it would be different... But here... white in - white out, leaves us with the original bag content. And that's when we are asked what are the chances... If black came out first - no question would have been posed.

Thank you lem, just as I said ,it is what was asked.

White in - white out doesn't necessarily leave us with the original bag content. It may leave us with a marble of the same color in the bag, but it's not necessarily the same marble.



So the bet I offered is equivalent to the problem. We can ignore games where a black marble is drawn since we're only concerned with probabilities after a white marble is drawn, correct?

Ok, thinking of it after a nights sleep. You are right. I re-read the original problem, and since the person drawing marble is blindfolded as opposed to always drawing a white - you would be correct. We disregard 1 game out of every 4, so It's 2/3 chance for white to come out next.