Talk:2023 AMC 12B Problems/Problem 19

Concerning solutions on 2023 AMC 12B #19

Stars and Bars does not provide an exact probability. However, it does provide a good estimate for the approximate answer as on average, the number of arrangements will be almost the same when each container has an odd # of balls or when each container has an even # of balls. (similar binomial distributions)

The above clarification is a good explanation of why the stars and bars argument still yields an expression sufficiently close to . However, I think the content of this clarification should be clearly explained WITHIN any stars and bars solution. Even with this clarification, this page is currently a redundant mess of poorly explained fakesolves:

Solutions 1 and 2 both use stars and bars to arrive at an incorrect probability and fail to explain why their method is a good approximation of the actual probability.

The explanation in solution 4 does not make sense at all.

The nature of this problem makes it so that pretty much any approach will arrive at an answer close to . However, I'm concerned that the current state of this page will make most readers come away with a worsened understanding of the structure of the problem, especially since the first two solutions are both fakesolves. This page should be fixed ASAP.

Please don't delete this message without addressing any of my concerns. If you have any questions, feel free to PM me. ~ CT17

(Solution 5 by Dissmo does a very good job at explaining Solution 4.) -Multpi12

Is the reason why the answers aren't exact because the bins are not distinct? Is it because of the situations where some bins have the same number of balls? ~ Bread10

The reason why solutions using stars and bars are not exact is because they make incorrect assumptions about the probability space. In the stars and bars solution, the sample space is the set of all orderings such that are nonnegative integers satisfying . However, not every element in this sample space is equally likely, so one cannot calculate probability by simply dividing the # of favorable outcomes over the # of total outcomes (# of elements in the sample space). ~tsun26

Yeah Bread10 here's an example: Say we were counting the number of ways to put 6 objects into 3 bins. Stars and Bars tells us there are ways for this to happen, which is perfectly fine and accurate for counting. However, for probabilities, it treats all arrangements as being equally likely to happen. So using Stars and Bars for probabilities would mean having 2 objects per bin is just as likely as having 3 objects in one bin, 2 in another, and 1 in the leftover bin, when in reality, the case is more likely to happen because there are different ways to permute the number of objects in the bins (3-2-1, 3-1-2, 1-2-3, 1-3-2, 2-1-3, 2-3-1) versus only one way for the objects per bin case (2-2-2).

~grogg007