Not able to understand the solution of this puzzle

DonAntonio

The 3rd possibility is: we have two black marbles in the urn but this time the first one drawn wasn't N1 but N2...these are DIFFERENT, identifiable

black marbles for this problem's sake.

DonAntonio

jedishrfu

Good answer except she gets brought up on assault charges and beheaded.

The best solution was to pick a stone and drop it quickly then say well you can what stone I picked by whats left and of course the king wont say he cheated cause it isn't kingly and so she gets to leave peacefully.

moonman239

This.

As I recall, there was a story about a king who had taken some people captive. One of their wise men faced the king, who gave him challenges and told him he and his people could go if he completed the challenges. In the last game, the king said he would write or have written "Stay" on one piece of paper and "Go" on the other. If the guy picked the one that said "Go," he and his people could leave.

The wise man, however, was not fooled. He knew there'd be no paper that said "Go." Both of them would say "Stay." So he took a piece of paper and ate it. He turned over the remaining one and said, "This one says 'Stay,' so the one I ate must have said 'Go.'"

Anyways, OP, maybe running a Monte Carlo simulation would help.

D H

You do not know that there is 50% chance that it is black initially. That 50% probability your naive best guess, based on a complete lack of information. After performing the experiment you have evidence that lets you alter that initial naive guess. I'll look at two variations on the problem to illustrate this point.

Suppose instead of drawing a black marble, the marble pulled from the urn had been white. Now the answer to question what is the probability of drawing a black marble from the urn now? is 1. There is no doubt that the remaining marble is black. You have to alter your initial guess based on this overwhelming evidence. The evidence is not quite so overwhelming when the drawn marble is black, but it is still useful information, and it still does let one improve upon that initial naive guess.

Suppose you repeated the test many times: a black marble is added to the urn and a marble is randomly drawn from the urn. You do this ten times, and each time the marble drawn from the urn is black. This accumulation of evidence is quite overwhelming. The answer to the question what is the probability of drawing a black marble from the urn now? is 1024/1025.

viraltux

Hi DH,

Yes he knows, that is not part of his answer but part of the problem.

If you don't know the probability of the first marble being black then you can never know the probability to draw a black marble; if this is the case you can only give an estimation within a certain confidence.

SW VandeCarr

Assuming one ball is has a 0.5 probability of being black vs white and you've drawn a ball, there is only one ball left in the urn. The possibilities are the original black ball you put in, the other ball that happens to be black, the other ball that happens to be white. Therefore there are three possibilities, two of which are a back ball. So the probability you would have drawn a black ball is 2/3..

D H

No, he doesn't know this. Once the initial ball is placed in the urn its color is fixed. It is either white or black with a probability of one. The key problem here is that the initial state (initial color) is unknown. It is a hidden variable. That 50% is just a guess. This is an estimation problem, something that Bayesian statistics are quite good at addressing. Another problem is that drawing a black marble is imperfect information. (In contrast, drawing a white marble yields perfect information.)

Bayes law provides a way to deal with this imperfect information and with the unknown initial state. A simple treatment is to use Bayes law as is, assuming some reasonable prior probability regarding the hidden state. More advanced Bayesian techniques provide the ability to express the uncertainty in that initial guess. Some of these more advanced techniques even provide the ability to indicate that the initial state is completely unknown in the form of a singular a priori covariance matrix. That, however, is way beyond the scope of this thread.


I should have qualified this part of my response in post #21 as once again assuming an a priori probability of 50%. A different prior will result in a different posterior probability for this repeated experiment.

For example, suppose the initial ball was randomly selected from an urn containing 99 white balls and 1 black ball. Now the prior probability is measly 1/100 but the posterior probability after drawing ten black marbles via the experiment is 1024/1123 per Bayes law. It's only when the prior probability is very small that this accumulation of evidence no longer strongly indicates that the remaining marble is black.

viraltux

I love the smell of a flame in the morning...

Well, I'd say the initial ball's color is fixed even before is placed in the urn, unless of course it is a Michael Jackson kind of ball which are known to be able to change its color spontaneously.

That the initial state is unknown does not imply that the probability for a particular state is a guess. For instance, if I flip a perfect coin and I ask you "heads or tails?" you are not guessing that the probability for heads is 1/2, you know.

So does Frequentist / Fisherian aproaches.

I would not call the Bayesian treatments to estimate parameters simple, oh... I feel so tempted to bite the bait, but for that we should start a new Thread "To Bayes, or not to Bayes"

For Bayesians everything is a parameter which I don't think it has to be so, also Bayesian estimation methods do not allow probabilities 1 or 0, which, again, I don't think either it has to be so. But I'd say this is quite off topic and probably we're confusing the original poster musicgold.

So for musicgold's sake let's agree the book is right from a non Bayesian approach

D H

The only one who is flaming is you. Stop now.

Of course. If we ran my repeated experiment and somehow drew a white ball at one point and then later drew another white ball, both a frequentist and a Bayesianist would say that there is something fishy going on here.

You're right. Both approaches are useful. Neither approach is perfect unless the information is perfect (in which case, why use statistics at all?) In the words of Donald Rumsfeld, "Data is like a captured spy. Torture it enough and it will tell you anything."

That's nonsense. If the marble drawn from the urn was white rather than black a Bayesian approach would yield an answer of 1 in response to "what is the probability of drawing a black marble from the urn now."

The book is right from a Bayesian approach too, assuming a prior based on the principle of indifference.

Note that nowhere in the question (the blue text in the original post) is there any indication of the probability distribution for the initial ball. The principle of indifference is about all we have to go on.

viraltux

I think I will apply this principle right now.

DonAntonio

I think Viraltux has the correct approach in this case: for the contestant, the probability of the marble which is already in the

urn being black (white) is 0.5, unless there's some other piece of info available (like, say, knowing that the person "perpetrating" the game

has a huge bias for white marbles or whatever). So if a person participates for the first time in the game and heard nothing about it in

the past, then he can comfortably assume the prob. is 0.5 of the marble being black. Then, under this pretty natural (imo) assumption, the

evaluation of the prob. is, as already noted, 2/3.

DonAntonio

D H

That is exactly what Bayes' theorem says. With no additional knowledge, the best guess regarding the color of the hidden marble is that it is black with probability 1/2. Now we perform experiment of adding a black marble to the urn and randomly drawing a marble from the urn. If the drawn marble was white, we would know with 100% certainty that the remaining marble is black. That's not what happened. We drew a black marble. So what is the posterior probability that the remaining marble is black given that we drew a black marble? Bayes' theorem says that it's 2/3, the same answer as in the book.