posted
You have three bags with two marbles each: black/black, black/white, and white/white.

You reach into a random bag and pull out a white marble. What are the odds that the other marble in that bag is white?

The official answer is 2/3, because the white/white marbles are not interchangeable: You might have pulled white(1) and white(2) is still in the bag, or you might have pulled white(2) and white(1) is still in the bag, or you might have pulled white(3) and black is still in the bag.

Has anyone actually done this experiment, and does it use the same logic as the Monty Hall problem?
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posted
Letterip, your second case has 0% chance that the other marble will be white, but you give it a 1/3 chance.
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misread the question - for some reason read it as one bag with 2 white and one black without replacement. Where was my mind at
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posted
No, Brian's right with the 2/3. Something that helps people get past the Monty Hall problem is to look at the specific permutations possible.

White 3, Black 3 White 1, White 2 White 2, White 1 Black 1, Black 2 Black 2, Black 1 Black 3, White 3

If you pull white first, the only permutations possible are the first three.

Another way to look at it: if you have one bag with two white stones and one bag with a white stone and a black stone, the odds that the first stone you pull will be white favor the possibility that you picked the bag with two white stones.
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The way to think of this, is that each time you play the game, if you draw black on the first move you restart the game.

BB - eliminated immediatedly, so we always restart with the others if we choose this bag.

WB - if you draw the black first, it is eliminated, so we have to redraw each time we get the black. This biases us to pick the WW. Ie 1/2 the time we draw black if we pick the BW bag, which is the same as randomly choose again from between WB and WW again, each time you hit black redraw.

So first time

1/4 WB, 1/4 BW, 1/2 WW

second time 1/2 times 1/4 WB 1/4 BW, 1/2 WW

etc

so cummulative probability of drawing WW is 1/2 + 1/2*1/4 + 1/2*1/16 + 1/2*1/64 ...

posted
Actually brian is wrong, because the reason I posted it in the first place is because I believe it is 50%.

I get that the permutations lead to 2/3, but I don't understand why we are supposed to use Permutation rather than Combination.

Focusing on the white/white combo, I fail to see why the marbles are not interchangeable. They do not in fact have little numbers painted on them, so when you pull it out you have white(in hand) and white(in bag). It doesn't matter which one is which.

To simplify my thinking: Would it change the math if there were only two bags to start with? Pretend the black/black bag never existed. Are the odds the same?

Would it change the math if the person setting up the game pulled the first marble out for me, but actually tricked me about the setup of the game? Suppose there was actually only one marble in each bag, one black and one white, but he told me there were two and the jerk actually palmed the initial white marble to force the quandary? Would you be able to tell the difference between the two setups?

Because the tricky setup is obviously 50%, and visually indistiguishable from the straight forward setup. I think.

Added to address LR's new post: Maybe my problem is that I am thinking of it like 'you pull a marble. what are the odds that the other marble in that bag is the same color' as opposed to 'oh, a black one. start over' Obviously the odds are the same for pulling two black ones, but if you combine the two problems - holy crap I just solved it. If you think of it as same color vs opposite color, then there is a 2/3 chance of getting a bag with two of the same color. And I was just looking at half of the problem, but by my own theory, the two halves are indistinguishable, and I think I just took a very long way around to explaining to myself why we use permutation instead of combination.

Apparently I am some sort of idiot savant or something.

[ October 21, 2013, 02:51 PM: Message edited by: Brian ]
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quote:Focusing on the white/white combo, I fail to see why the marbles are not interchangeable.

That's why I closed my post with another way of looking at it. I'll restate:

On your first draw, you have drawn a white marble. The odds of drawing a white marble first are twice as high for the bag with two white marbles than for the bag with only one white marble, for the simple reason that there are two chances of drawing a white marble there versus the one chance in the other bag. (We can ignore the black-only bag altogether, for obvious reasons.)

In white-marble-first situations, you have a 66% chance of being in the two-white-marbles bag.

---------

To answer your specific questions:

1) No, it would not change the math if there were only two bags to start with, because the only effect of the first draw is to limit your available set to those two bags.

2) If both bags were identical, the odds would indeed be 50%. The reason they are not 50% in this case is that one of the two cases is much more likely than the other.

posted
Brian, do you agree that if you performed this experiment repeatedly, the results would approach this: 2 out of 3 times you pull a white marble first, it came from the white/white bag?
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what is going on is that if the game only starts if your first marble is white, it acts as if you restart the game if you draw a black on the first move. It is the 'restart' effect that biases the outcomes.
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quote:Focusing on the white/white combo, I fail to see why the marbles are not interchangeable. They do not in fact have little numbers painted on them, so when you pull it out you have white(in hand) and white(in bag). It doesn't matter which one is which.

They're not interchangeable because you draw them one by one. You can't pull the first marble, then say that it's the second marble.

And that's where the probabilistic trick comes in here- there's a much higher chance that the first marble you pulled will be white if you got the WW than if you get the WB bag- the fact that your first pull was white actively influences the probability of the second pull; the second pull is not independent of it.

(That's also where your palmed marble example comes up short- there was no actual first pull, in that example, you've got a 100% chance of getting white first, so it has no effect on future draws. All that it is is a good way to con someone that otherwise does understand real probability pretty well.)
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posted
I added to my long post up above, but the gist is that I am very bad a probability math.

The 2/3 chance of drawing w with the w/w bag explaination is - as well as being a very good, simple construction - similar to my convoluted rambling. I think. I guess I was over-thinking it.
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posted
I had to read the whole thread to get it, but LR's description seems simplest. Even if there are only two bags, you still get a "do-over" if you draw the black marble (which happens one time out of four). So you are not looking at the total number of double draws, but rather, at 3/4s of the them (and that 3/4s is skewed towards having chosen the all white bag), and *then* asking about the probability of the second draw.

I think the trick is revealed when you state it thus:

If you draw two marbles from a single bag, the odds of the second being white (or black) is 50%.

However...

The odds of the second marble being the same color as the first is 66.7% (2/3), since 2/3s of the bags contain two like-colored marbles.

Therefore, if your first draw is white, the odds that the second will also be white is 67%, with the same statement being true if you substitute the word black.

The trick lies in removing the "if", and saying: "you drew a white ball; what are the odds that the remaining ball being white?" A misdirection, of sorts.
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posted
There are not so many bags and rocks that you can't look at the total universe of first and second picks. There are, in fact, 6 possible first picks, 3 of which are black, 3 of which are white: 1st stone from B-B 2nd stone from B-B Black stone from B-W White stone from B-W 1st stone from W-W 2nd stone from W-W

We know the first 3 choices did not occur, that leaves 3 possibilities (in the same order as above): Black stone from B-W 2nd stone from W-W 1st stone from W-W
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posted
OK, this is the same premise as the Monte Hall problem. Since only one bag has two white marbles there was a 1/3 chance you got that bag on your first pick. Therefore, if you picked a white marble there's a 2/3 chance the other bag is the one you want.

This was confusing (to me) in that with the Monte Hall problem the person who makes the initial selection is not equivalent to someone who wanders in after the first pick and only sees two options (50-50). I can't tell you how many times I proved to skeptical friends that you should always switch on the MH scenario, and I still fell for the same trap in Brian's example.

It's not a case of committing a logical fallacy, but of being logically fallible.
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posted
There's a reason that there's a lot of money to be made in being a con artist or the House. The realities of probability are incredibly conterintuitive unless you've studied them deeply enough to internalize them.
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