Balls and Boxes Complicated Problem

[DavidSmith]

Consider a set box boxes. Say we have 30 boxes.

And then we have x number of black balls and y number of white balls

And these balls are stacked in the boxes. The total nuber of ways of stacking them is easy to find, but a much harder problem is to find the probability that for anyone of the 30 boxes the ratio of black balls to white balls exceeds a certain value donoted as say 'p' where the number of black balls must exceed another variable denoted 'q' for this ratio to be valid.

For example if you have 10 black balls 5 white balls and 30 boxes find the probability that in anyone of those boxes there will be twice as many black balls than white balls assuming that there must be greater than 2 blacks balls in tat particular box. The balls are randomly placed in the boxes and the placement of one ball has no affect on the placement of another.

 

[0rthodontist]

Order of stacking within the boxes does not matter. You know that for a given number of white balls and black balls in a given box (say 4 and 2), the number of ways that you could have those appear in that box is the same as the number of ways that you could arrange 6 black balls and 3 white balls in the remaining boxes. Add up the total ways for each number of white or black balls that satisfies the ratio and the minimum number of black balls, and divide by the total number of ways. I couldn't tell you if this is the quickest way to approach the problem, but it should work eventually.


Edit: Since I'm learning Haskell, I tried this as an exercise

prob w b s min r = satisfy w b s min r / total w b s min

satisfy w b s min r = sum [countall (w - y) (b - x) (s - 1) | x <- [min..b], y <- [0..w], (y == 0 && x > 0) || x / y > r]

total w b s min = countall w (b - min) s

fac 0 = 1
fac n = n * fac (n - 1)

a `choose` b = fac a / (fac (a - b) * fac b)

countall w b s = ((w + s - 1) `choose` w) * ((b + s - 1) `choose` b)

This seems to work, but I haven't tested it much. For your example, r = 2, min = 3 (i.e. greater than 2), s = 30, w = 5, b = 10, prob gives a probability of about 0.98, which sounds about right.

 

[tacman]

This is indeed a complicated problem, as it involves both combinatorics and probability. In order to find the probability that for any one of the 30 boxes, the ratio of black balls to white balls exceeds a certain value, we need to first determine the total number of ways in which the balls can be stacked in the boxes. This can be done using combinatorics, specifically the combination formula.

Once we have the total number of ways, we then need to consider the condition that there must be a minimum number of black balls (q) in the box for the ratio to be valid. This means we need to subtract all the combinations where there are less than q black balls in the box.

Next, we need to consider the condition that the number of black balls must exceed the number of white balls by a certain value (p). This requires us to further subtract all the combinations where the difference between the number of black balls and white balls is less than p.

After doing all these calculations, we will have the total number of valid combinations for the given conditions. We can then divide this number by the total number of ways to get the probability of the desired event.

In the example given, there are 10 black balls, 5 white balls, and 30 boxes. Using the combination formula, we can determine that there are a total of 2,716,635 ways to stack the balls in the boxes. However, not all of these combinations will meet the given conditions. We need to subtract all the combinations where there are less than 2 black balls in any one box, as well as all the combinations where the difference between the number of black balls and white balls is less than 10.

After doing the necessary calculations, we find that there are only 2,349,085 valid combinations. Dividing this by the total number of ways gives us a probability of approximately 0.8649. This means that there is an 86.49% chance that in any one of the 30 boxes, there will be at least twice as many black balls as white balls, given the conditions stated.

In conclusion, this is a complex problem that requires both combinatorics and probability to solve. It is important to carefully consider all the given conditions and use the appropriate formulas to find the desired probability.