# Math Help Needed: Calculating Probabilities

But you wrote 66% despite what your brain said. That’s interesting!

I think that this problem is similar to the Monty Hall problem, yes.

Would you say then that the answer is 50%?

No, because there is a chance you are holding the original marble. Which means the marble in the box could be the second marble which we know is white. Or, it could be the original marble in the box, which is still 50/50 one or the other.

Honestly, I’m not sure of the percentage, but I would definitely put it above 50%.

Gotcha!

I’m also inclined to say that the probability is 2/3, my reasoning being that when you reach into the box (without looking) and pull out a marble, there are the following four equally likely cases (at first):

1. The first marble is black and you pull out the first marble. (Pulled out marble is black, marble in the box is white.)
2. The first marble is black and you pull out the second marble. (Pulled out marble is white, marble in the box is black.)
3. The first marble is white and you pull out the first marble. (Pulled out marble is white, marble in the box is white.)
4. The first marble is white and you pull out the second marble. (Pulled out marble is white, marble in the box is white.)

When you then look at the marble and see that it’s white that means that it can’t be 1., which leaves us with three equally likely cases, in two of which the marble in the box is white.

This seems correct to me.

Looking at the marble in your hand culls the possibility space.

If you threw away the marble you withdrew without looking at it, then you didn’t get any additional information, and the answer is still 50%. But if you DO look at it, you’re given additional information, which changes the probabilities.

This is my reasoning as well. :)

I really like anymunym’s explanation as well, and will use it as my proof for my original 66% guess. I just spent the last hour thinking about this problem rather than the conference I was attending, and was getting close to anymunym’s proof, but I wasn’t quite there. I’d like to think I might have gotten there if I hadn’t been distracted.

I hope that it was not a terribly important conference, belouski! :)

Thanks to all of you for pondering this. I agree with the 2/3 answer and for the same reasons. But one can never be too sure, where probability is concerned…

I was looking at things on wikipedia and it led to me to Bertrand’s box paradox. Your question seems very similar but phrased in such a way as to invite people to actually think through to the correct guess instead of the incorrect one per usual with BBP.

Ironically, Bertrand’s Box is one of the few my brain groks right off.

I wonder if the probability that a randomly chosen probability problem will have the answer 2/3 is also 2/3.

It’s 2/3 to infinity and beyond. The conference was luckily covering mostly preliminary data, so my brain’s absence wasn’t critical. Then again, my brain’s absence is rarely critical.

I think the main point is the difference between probabilities of a system vs. the elements of a system. The Monty Hall problem is just such a system; what are the odds that you’re in system A instead of system B or system C? Just like asking what are the odds of drawing a system of two heads or tails, not whether or not the odds of flipping heads per coin is always 50%.

A better example is maybe is to show the system being broken down. Imagine there is the box with marbles which you’ve set up, however you are not playing but are watching as an observer. The player takes a marble out which you cannot see. What is the probability that the remaining marble is white? Since you added a white marble and this information is provided, it cannot be 50% even if you don’t know the color of the marble withdrawn.

That’s a different scenario with a different probability (3/4 rather than 2/3), isn’t it?

I think Endigm’s remaining marble has a 75% chance of being white as well. Using anymunym’s previous scenario alignment, none of them have been ruled out, so you have 3/4 chances that the remaining marble is white.

Unsolvable without knowing the probability that the first marble is white or black. If it is 50/50 than 2/3rds but otherwise it can be anywhere from 0 to 100 based on the distribution of the original marble colors.

I am on to your tricks Perkins.

If you are Bayesian, then when you don’t know the probability you should assume it’s evenly distributed over all possible outcomes.