2in 3

50% Having already took out a white marble, says that we are using bag A or bag C (bag B contains 2 black ones). If it's bag A, the second one will be white, if it's bag C it will be black: one possibility on two.

2/3. Bayes theorem.

2/3. There are 3 white balls if you play the game 3 times, picking a different white ball each time, you’ll win twice.

it is 0.5 based entirely on the phrasing of the question. if the question asked "what is the probability that if you picked a white marble, the second marble drawn would also be white" then that would be 0.66666. but that is not what the question is asking. as asked, it is 0.5. as asked, this is not bayes theorem.

Everyone saying 2 in 3, it's Hayes Theorem, needs to re read the last line of the question. Asking about a white marble as the second draw makes this a 1/2 question, the third bag with only black marbles is irrelevant to the problem.

Everyone saying it’s 50-50 and not 2/3… if you think of the dilemma as this: you pick a random bag and take out one marble, which is white. Which bag did you probably pick? It’s fairly intuitive to see that you’ve probably picked the all-white bag. Thus, the probability of the remaining ball being white is greater than 50%. It’s 67%.