A Bob's friend removed one ball from the box without telling Bob what was its color. Now what's the probability for Bob picking up one green ball from the box. But I would like to get one total probability, not two conditions like these. Then let's go back to the original problem. Assuming the friend picked at random, what is the chance that he picked a white ball and what is the chance the he picked the green one. I think I got what you said. The problem is that if we sum we get colored balls quantum physics than 100%. For instance, if Friend picked the white ball, Bob has 100% chance of picking up a green ball. How do you normalize it to 1. I think I got what you said. The problem is that if we sum we get more than 100%. For instance, if Friend picked the white ball, Bob has 100% chance of picking up a green ball. How do you normalize it to 1. If you don't want an instant solution using Bayes' Theorem, you can get to the solution using a more intuitive counting method. Imagine doing the experiment 100 times, with random choices each time. The experiments fall into two groups: Group I 90 experiments : player 1 picks green. In the 90 group-I experiments, player 2 faces an urn having 8 green and 1 white ball, so in how many of these 90 cases will he pick green. How many times in 100 would player 2 pick green. But why did you choose doing exactly 90 experiments of the group 1 and exactly 10 experiments of the group 2. Of course, that means that in any actual group of 100 experiments we will not always get 90 in group I, but averaged out over many different sets of 100 we will get 90. And, of course, the same can be said about how many greens are chosen by player 2 in the group I experiments. colored balls quantum physics You would get a more accurate accounting if you did 100 billion experiments, and said that group I had 90 billion members, and player 2 picks a green ball 80 billion times withing group I. Or, if that is not accurate enough for you, do 100 trillion experiments, etc. The point is that the argument is supposed to be intuitive, not rigorous. For beginners who may have trouble relating to Bayes' rule, that type of counting argument can be very helpful in clarifying the basic concepts. To be sure, it can all be done exactly in about 2 lines using Bayes' rule; see, eg. Oh, I finally figuered it. Let's assume player 1 goes to the urn and picks a ball at random. He can pick one white ball or one green ball. Oh, I finally figuered it. Let's assume player 1 goes to the urn and picks a ball at random. He can pick one white ball or one green ball. Would it surprise you to find out that the probability of any last-picked ball is 0. If player 1 takes 4 balls without showing them to player 2the probability that player 2 picks green is still exactly 0. Even if player 1 takes 9 balls types unknown to player 2 and so player 2 faces just one single ball, the chance that it is green is still 0. I did not want to bring this up at first, when you were still struggling with the issues, but we can talk some more about it if you want. Would it surprise you to find out that the probability of any last-picked ball is 0. Is it because we know for sure that player 1 picked a certain amount of balls. I mean, let's take the most interesting case, that player 1 removes 9 balls. Oh I see This is very interesting. Is it because we know for sure that player 1 picked a certain amount of balls. I mean, let's take the most interesting case, that player 1 removes 9 balls. For the more general problem more than 2 balls drawn the easiest way is to avoid using conditional probabilities and Bayes' rule, but instead to go back to sample-space fundamentals. Say player 1 draws k-1 balls without showing the results to player 2 and player 2 draws ball k. You can view this as a permutation problem: number the balls from 1-10; balls 1-9 are green and ball 10 is white. Now any drawing of balls corresponds to a permutation of the numbers from 1-10. Basically, we lay out the entire permutation of which there are 10. So, when drawing just 1 ball we lay out all 10. In the original version of your problem we are really asking for the probability that the ball in position 2 is green. It works the same way for any position k. Bob's colored balls quantum physics removed one ball from the boxWe are left to guess the basis on which the friend chose the ball. If it was eyes closed then, yes, the answer to the question is 0. But put colored balls quantum physics in the friend's position. Eyes open, would you pick any ball, equally likely, or any colour, equally likely. I would look at it like this. There are 10 balls in the bag. Bob is equally likely to get any ball. The probability it is white is, therefore, 10%. It doesn't matter how many balls his friend removes at random before he chooses. One must be careful about relying on that intuition. It's often safer to beat these problems to death with a more tedious calculation of the different possibilities. One must be careful about relying on that intuition. It's often safer to beat these problems to death with a more tedious calculation of the different possibilities. Right, and colored balls quantum physics argument in post 12 is perfectly safe. It would have been faster and easier to look at the probability of player 2 drawing white on draw k. The number of permutations with W in position k is 9. We are left to guess the basis on which the friend chose the ball. If it was eyes closed then, yes, the answer to the question is 0. But put yourself in the friend's position. Eyes open, would you pick any ball, equally likely, or any colour, equally likely. Friend's is supposed to see what ball he is picking up. It's just that he doesn't tell Bob what ball it was. But as pointed out even that Friend's doesn't see what ball he is picking up Bob still has the same chances. Colored balls quantum physics that, I guess, is because what matters is whether Friend's tells Bob what ball it was or not, i. This seems to be the same thing as it happens in the Monty Hall problem mentioned by later on. Friend's is supposed to see what ball he is picking up. It's just that he doesn't tell Bob what ball it was. But as pointed out even that Friend's doesn't see what ball he is picking up Bob still has the same chances. And that, I guess, is because what matters is whether Friend's tells Bob what ball it was or not, i. This seems to be the same thing as it happens in the Monty Hall problem mentioned by later on. It's not the same as the Monty Hall problem. The key feature of the Monty Hall problem is that Monte will never open a door with the prize. That has a filtering effect which changes the probabilities. Unless Bob's friend does something intentional like that we have no new information upon which to adjust the probabilities. Whether his friend tells Bob or not does not matter. It doesn't really matter if the friend intentionally did anything. What he drew is what he drew regardless of any intention. If the friend tells Bob what he drew, that definitely changes the probabilities for Bob, given what the friend drew. So our calculated probability would change given that information. But the probability without specifying what the friend drew is not changed from the original. In the Monty Hall problem, the result of one door changes the probabilities of the other doors. If colored balls quantum physics was not intentionally avoiding the prize, the probabilities of both other doors would remain equal. Because of his intentional filtering it changes one door probability more than your door which is unchanged. In this problem, the result of the friend's draw if known colored balls quantum physics the result of Bob's draw in a known way if the friens's draw is known. If the friend's draw is not known, there is no new information and the probabilities are unchanged. One must be careful about relying on that intuition. It's often safer to beat these problems to death with a more tedious calculation of the different possibilities. Each player has picked a single ball at random. Each is equally likely to have the white ball. More generally, if they each get five picks, then they each have an equal chance of getting the white, regardless of the order they get picked. To calculate that with your method you'd need 5 pages of conditional probabilities: If Bob gets the first, second, fourth, sixth and ninth picks, then that is one laborious calculation. And you'd need a different calculation for every combination of picks for each player. Monty Hall is a different problem, because Monty's pick is not random. He knows where the car is. Monty Hall is a different problem, because Monty's pick is not random. He knows where the car is. I realize that, but the fact remains that it is the same intuition on the part of often very well educated people which leads to the wrong answers in the Monty Hall problem. It's only when the conditional probabilities are specifically considered that the true answer is obvious to all. That's why I call it treacherous.
