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[Guest]

  On 1/21/2010 at 5:55 PM, dawh said:

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Like Chuck said, the boxes are a distraction. You aren't selecting boxes, you are choosing marbles and you have a 2/3 chance of having selected a marble from the double blue box once you find the first marble is blue. Your alternate method of viewing the problem is flawed. Yes, you have a 100% chance if both boxes have 2 blue marbles, but it doesn't make sense for the probability to go to 50% if you change 1 out of 4 marbles. In your example, you started with 4 possible marbles that you selected first. After you changed 1 of the marbles to orange, that still leaves 3 marbles that you could have selected in your first pick. Just because two have the same value from the same box doesn't make them the same event. You're twice as likely to have picked a marble from box 2 in the first place.

Chances are this puzzle already exists in some form on this forum as I've seen it with three cards marked with Xs and there is a Wikipedia article on the puzzle with a slightly different casting as well, though the problem itself is identical. The article explains why Chuck et al. have the correct interpretation of the problem and also recognizes that the problem is similar to the Monty Hall problem and the boy and girl problem, both of which have been discussed at length on this forum.

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It is true that the first box does not matter. But it is not equally likely that you will have chosen box 2 and box 3. Once you have chosen a blue marble, you must pick from the same box. It is twice as likely that you chose the first blue marble from the box with two blue marbles than from the box with one blue marble. Thus 2 out of 3 times you are choosing the second marble from the box with two blue marbles

[Guest]

Right on Chuck.

Often in mathematics (it seems especially so in statistics) the calculationally derived answer is contrary to our intuition, but is none the less true.

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Consider the example where

One box has 100 orange marbles

One box has 100 blue marbles

One box has 99 orange marbles and 1 blue marble.

After picking a random marble and discovering that it is blue, I ask, "What box did the marble come from?" You would of course say it wasn't the first. It could be the second or third, but my money is the third. You may say that it is the more likely (More probable) case. That being so, you would guess that the next marble would be blue as well. This is the scenario we are presented, but instead of 100/99 marbles, we are looking at 2/1.

The answer is, as Chuck and other have pointed out, 2/3.

[Guest]

The first box is irrelevant. The probability of choosing a blue marble is irrelevant, it has already happened. So now you have two boxes left and you have to assume that either one is missing a blue marble. So now there are only two possible options. One is that the next marble is blue and the other would be that the next marble is orange. 1/2.

[Guest]

No doubt, your mathematical calculations correct. However, often in mathematics we miss-read or miss-interpret the question. This particular question, in my opinion (which could be incorrect), does not ask about the probability of picking blue marbles out of a box.

Let’s examine the problem

Given that you just pulled a blue marble out of *one of these three boxes*.

What is the probability that the other marble *in this box* is also blue?

The first line indicates that the selection of a marble is directly tied to the boxes themselves. The statement also indicates that the marble being selected is blue. This eliminates the first box from the problem and leaves two boxes which contain at least one blue marble. Given that a blue marble is taken out of the box, each box then contains only one marble. The second line of the question asks about the color of the remaining marble as it relates to the box from which the first marble is drawn. We can easily deduce that if a blue marble is removed from a box containing two blue marbles, then one blue marble remains. Likewise, if a blue marble is removed from a box containing one blue marble and one orange marble, then a non-blue marble remains. Thus, the probability that the marble remaining from a set of two possible boxes after one blue marble is removed, is 50%.

[Guest]

"However, often in mathematics we miss-read or miss-interpret the question."

I agree that this is what is happening here. Consider the question:

What is the probability that the 1st blue marble picked came from box #2.

What is the probability that the 1st blue marble picked came from box #3.

The answer is clearly that there is a 2/3 probability that the 1st marble came from box 2 and 1/3 probability that it came from box 3 (and of course 0 probability that it came frombox 1).

This is effectively what the question is asking.

As in the case with 100 marbles described above, if you think the 1st blue marble is just as likely to have come from the box with 1 blue as opposed to the box with 100, then I'll bet 2:1 odds against your supposed 50% probability that the next marble will in fact be blue.

That is: given that you picked a blue marble 1st, you are more likely drawing from box 2 than box 3.

[Guest]

unfortunately, in this question, box two counts twice.

2/3

ignore box 1,

now if we start drawing random marbles from random boxes, i hope you will agree

box 2 will have a blue twice as often as box 3.

try it.. simulate it... stick your answer.. whatever

s

[Guest]

  On 1/21/2010 at 7:21 PM, Cantor said:

"However, often in mathematics we miss-read or miss-interpret the question."

I agree that this is what is happening here. Consider the question:

What is the probability that the 1st blue marble picked came from box #2.

What is the probability that the 1st blue marble picked came from box #3.

The answer is clearly that there is a 2/3 probability that the 1st marble came from box 2 and 1/3 probability that it came from box 3 (and of course 0 probability that it came frombox 1).

This is effectively what the question is asking.

As in the case with 100 marbles described above, if you think the 1st blue marble is just as likely to have come from the box with 1 blue as opposed to the box with 100, then I'll bet 2:1 odds against your supposed 50% probability that the next marble will in fact be blue.

That is: given that you picked a blue marble 1st, you are more likely drawing from box 2 than box 3.

I agree that you are more likely to draw a blue marble out of box two from the original condition of the boxes having two marbles. However, that is not what the question is asking. You must look at the boxes as if they only have one marble in them. And since there are two boxes that could have had a blue marble in them, there there are two possible marbles that can be drawn. One is blue and the other is not-blue, ergo a 50-50 chance of drawing a blue marble.

[Guest]

  On 1/21/2010 at 7:39 PM, hundleyj said:

I agree that you are more likely to draw a blue marble out of box two from the original condition of the boxes having two marbles. However, that is not what the question is asking. You must look at the boxes as if they only have one marble in them. And since there are two boxes that could have had a blue marble in them, there there are two possible marbles that can be drawn. One is blue and the other is not-blue, ergo a 50-50 chance of drawing a blue marble.

ok, hundleyj, i'm going to beat this into you with a stick.

you have just pulled a blue marble out of one of these boxes.

you have either marble 3,4 or,6

now

3 has another blue marble in the box

4 has another blue marble in the box

6 does not have a blue marble in the box.

2/3

[Guest]

Davis: What are the odds of a riot in Dog River?

Karen: 50/50

Davis: 50/50 !?

Karen: That's right. We'll either have one or we won't.

[Guest]

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2/3

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lets do some counting kids! How many different balls can be chosen such that we will look at the second ball) (in simple terms, how many blue balls are present?)

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3! Yeah! Good job! that is our base of probability! x/3. Now, if you choose the first blue ball, is it successful or a failure?

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success!

how about the second blue ball?

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success!

and the third?

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failure!

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so how many successes?

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2!

and what was our total?

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3!

so what is the probability?

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2/3!

I know that all the spoilers can get annoying, but I went step by step, proving my solution. Post if there is a mistake in my logic (but there isn't).

[Guest]

I think the answer is 1/3.

[Guest]

Good post Roark. The spoilers are justified. I'm just impressed you went through all the trouble.

Good problem Xandam. I enjoy these stat/probability problems that require some actual knowledge above and beyond intuition to solve. Just like the Monty Hall problem mentioned earlier.

Unfortunately, it is hard to reason with those who only use intuition as an argument. I understand them completely though. I've been in that boat (such as the first time I saw the Monty Hall problem). My gut told me right away that the answer was 1/2, but I recognized the conditional probability statement and decided to actually use the formula which of course led me to the correct answer. I think I enjoy trying to justify the correct answer more than actually solving for the correct answer.

I think if some of the individuals who insist on 1/2 would actually try the experiment, they would be amazed at the results even if they don't understand why they got them.

[Guest]

  On 1/21/2010 at 10:30 PM, Cantor said:

Good post Roark. The spoilers are justified. I'm just impressed you went through all the trouble.

Good problem Xandam. I enjoy these stat/probability problems that require some actual knowledge above and beyond intuition to solve. Just like the Monty Hall problem mentioned earlier.

Unfortunately, it is hard to reason with those who only use intuition as an argument. I understand them completely though. I've been in that boat (such as the first time I saw the Monty Hall problem). My gut told me right away that the answer was 1/2, but I recognized the conditional probability statement and decided to actually use the formula which of course led me to the correct answer. I think I enjoy trying to justify the correct answer more than actually solving for the correct answer.

I think if some of the individuals who insist on 1/2 would actually try the experiment, they would be amazed at the results even if they don't understand why they got them.

Yeah, I was on the other boat for the Monty Hall when we started it in psychology. This one seems easier. For example, this is the same as asking how many ways can you roll a seven with two dice? Every craps or settlers of catan player will tell you six. the people saying one half think 3/4 is the same as 4/3. EDIT: they are not the same thing, in case you were doubting that too.

Edited January 21, 2010 by roark13

[Guest]

No relevance to the problem, but I was just playing a game of Catan on my iPhone when I read your post. Neat coincidence. Now what's the probability of that?

[Guest]

Had to post to anything that mentions Catan. :-) And you're right that if you actually do the experiment (you only need the 2 boxes, 4 "marbles") the argument is over. I love this type of problem because intelligent people will see it completely differently. But if you start drawing marbles (for those of us who have lost our marbles, this can be a challenge) the logic becomes clear.

[Guest]

2/6 because there are 6 marbles total and 3 blue marbles. 1 blue marble was taken out, so 2 blue marbles ramained. simple!

[Guest]

  On 1/21/2010 at 11:24 PM, brain storm said:

2/6 because there are 6 marbles total and 3 blue marbles. 1 blue marble was taken out, so 2 blue marbles ramained. simple!

please read my solution. The first blue has to be in the second balls case. READ THE ACTUAL PROBLEM!

[bonanova]

The puzzle does on this forum.

Three cards: black-black, black-white, white-white.

Same question; page__view__findpost__p__6216.

Two thirds.

In comparing total and favorable outcomes, it's rather easy to forget about equal likelihood.

If that happens, we become unalterably certain of an incorrect result.

A simple, outrageous example asks: how many lottery tickets must I buy in order to have a 50% chance of winning?

The obvious answer is, only one ticket!

Why is it so clear that only one ticket is needed?

There are only two outcomes; one of them is favorable.

[Guest]

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1/2

[Guest]

I know it's lazy, but I don't want to trawl through ALL the previous posts for how people arrive at an answer other than 1/2. Can someone please explain to me? From my understanding, if you draw a blue ball, this eliminates all other events previous. So you have a blue ball in your hand that could have come from one of two boxes - one that had two blues originally, and one that had one blue and one orange. Because you have drawn a blue ball (and all other events previous have now been eliminated) the probability that the other ball is also blue is one in two. I can't see a fault in that logic. But I'm no mathematician, so can someone please explain? Cheers.

[Guest]

  On 1/21/2010 at 11:44 PM, bonanova said:

The puzzle does on this forum.

Three cards: black-black, black-white, white-white.

Same question; page__view__findpost__p__6216.

Two thirds.

In comparing total and favorable outcomes, it's rather easy to forget about equal likelihood.

If that happens, we become unalterably certain of an incorrect result.

A simple, outrageous example asks: how many lottery tickets must I buy in order to have a 50% chance of winning?

The obvious answer is, only one ticket!

Why is it so clear that only one ticket is needed?

There are only two outcomes; one of them is favorable.

exactly!!!! 2/3

[Guest]

I always get a kick out of how zealously people hold to their intuition in questions like this. And don't feel bad if you've argued against the correct solution. When early versions of the Monty Hall problem started showing up in papers and the like, many very upset academics (some statisticians themselves) wrote in to argue against the solution that was provided.

After all the explanations that have been provided, if you still aren't convinced, just try an experiment. It's not a hard problem to duplicate and shouldn't need a prohibitive number of trials to see the pattern.

[Guest]

  On 1/22/2010 at 9:43 AM, Tuckleton said:

I always get a kick out of how zealously people hold to their intuition in questions like this. And don't feel bad if you've argued against the correct solution. When early versions of the Monty Hall problem started showing up in papers and the like, many very upset academics (some statisticians themselves) wrote in to argue against the solution that was provided.

After all the explanations that have been provided, if you still aren't convinced, just try an experiment. It's not a hard problem to duplicate and shouldn't need a prohibitive number of trials to see the pattern.

  Patouie said:

Had to post to anything that mentions Catan. :-) And you're right that if you actually do the experiment (you only need the 2 boxes, 4 "marbles") the argument is over. I love this type of problem because intelligent people will see it completely differently. But if you start drawing marbles (for those of us who have lost our marbles, this can be a challenge) the logic becomes clear.

Well said Tuckleton and Patouie. I remember being convinced of answer A on this puzzle -- ready to stake my life on it. Then, I tried it, with a couple of boxes and tennis balls and realized why it was answer B. Soon I found myself surrounded by half-a-dozen math students in acollege parking lot, wanting to "beat me up" because I didn't agree with the "obvious" answer A. I should also mention that I gave the problem to two math professors and they both got answer A on their first try.

If you try it, count only the trials where you draw the blue ball first, keep a little tally and after 15 or 20 trials, it starts to make sense.

[Guest]

Hmmm...

I have been converted!

I could not, for the life of me, understand how anyone could reason a solution other than 50%. However, just 5 seconds ago, I finally understood. It absolutely DOES matter what the likelihood is of selecting a marble either from the blue/blue or from the blue/orange (and that choosing it from the blue/blue is twice as likely). The scales have been lifted from my eyes!

Thanks all for such a great mind exercise.

[bonanova]

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If you substitute orange for blue the same answer must obtain.

Which is simply the probability that the marbles are the same.

This is true in 2 of 3 cases.