Games of Chance Course: Marble Problem

I was doing the Games of Chance course (1st chapter, 3rd quiz (which was about misconceptions, so maybe that's why :) )) I think Brilliant has given an incomplete explanation to the following problem:

When I put two marbles in this bag, I flipped a coin twice to determine their colors. For each flip,

  • if it was heads -> I put in a red marble;
  • if it was tails -> I put in a blue marble.

You reach into my bag and randomly take out one of the two marbles. It is red. You put it back in. Then you reach into the bag again. What is the chance that, this time, you pull out a blue marble?

That's the problem, here's the explanation that Brilliant gave:

Initially, when the marbles are put into the bag, there are four equally likely possibilities:

> A) The first marble dropped in is red, and the second is also red.
> B) The first marble dropped in is red, and the second is blue.
> C) The first marble dropped in is blue, and the second is red.
> D) The first marble dropped in is blue, and the second is also blue.

When the first red marble is picked, it could be any of these possibilities, and all of them are equally likely:

> 1) The first red marble from A.
> 2) The second red marble from A.
> 3) The red marble from B.
> 4) The red marble from C.

(Bag D is now entirely out of consideration.)

This means in 2 out of the 4 cases, the bag is an all-red bag, and in the other 2 cases, it is a mixed blue-red bag. So, there is a 1/2 probability of the bag being mixed.

For the next pull to be blue, two things need to happen. One, the bag pick needs be a mixed color bag (a 1/2 probability), and two, the blue marble instead of red needs to be drawn (another 1/2 probability). So the overall probability is 1/2 x 1/2 = 1/4 of pulling a blue marble.

(The most common mistake people make with this problem is thinking the initial red marble pull only eliminates bag D, while still considering bags A, B, and C to be equally likely. The initial red marble pull does eliminate bag D, but it also tilts the probability towards the bag being A.)

That last bit in the parentheses, is what I'm concerned about. They say that the probability gets tilted slightly towards bag A. But why? I'm asking this question to you reading this, I genuinely don't know the answer!

Even more confusing: The next problem looks like this:

When I put two marbles in this bag, I flipped a coin twice to determine their colors. For each flip:

If it was Heads -> I put in a red marble. If it was Tails -> I put in a blue marble.

After I had the bag ready, I looked into the bag at both marbles and announced, "at least one of marbles in this bag is red." To prove it, I took a red marble out of the bag and set it aside. I then asked you to reach into the bag and remove the only remaining marble. What is the chance that it is blue?

Hint: The answer is not 1/2.

I think that problem is exactly the same as the 1st one, but instead of putting the 1st picked out marble back in, you put it aside.

The explanation to this problem:

Most people get this problem wrong, calculating or guessing that the probability of choosing a blue marble is either 1/2, 2/6 = 1/3, or 3/4. In actuality, the probability is 2/3, here's why:

Initially, when the marbles are put into the bag, there are four equally likely possibilities for what happens, each of which has a 1/4 chance of being what occurred:

Case A) The first marble dropped in is red, and the second is also red.
Case B) The first marble dropped in is red, and the second is blue.
Case C) The first marble dropped in is blue, and the second is red.
Case D) The first marble dropped in is blue, and the second is also blue.

However, when I announce (truthfully) that there is at least one red marble in the bag, I am ruling out Case D -- it could still be any of the 3 other cases and no information favoring one over the others has been given. And because all of the remaining cases, A-C, are still equally likely, each of them is now known to have a 1/3 chance of being what occurred.

Considering what remains in each of those three cases (given that a red marble is then removed) the three possibilities for what remains are as follows:

Case A) The remaining marble is red.
Case B) The remaining marble is blue.
Case C) The remaining marble is blue.

Therefore, there is a 2/3 chance that the remaining marble which you draw out of the bag is a blue one.

So this time, they didn't say that the probability gets tilted towards A! Why?

#Logic

Note by Simon Houben
2 years ago

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Comments

I have the same question, Simon.

I think these two explanations use contradicting logic, and at least one of them is spurious. I feel really perplexed, but there seems to be no clear explanation I can find elsewhere.

Chenzi Guo - 1 year, 4 months ago

It’s tilted to the bag with two red marbles, because there two ways to pick a red marble from this bag (there are two red marbles in this bag), whereas there is only one way to pick a red marble from any other bag.

Although, I am also confused about the statement that these bags are equally likely in the second exercise. The process seem to be exactly the same.

Gregoro Ricci - 6 months, 2 weeks ago
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