Hi All,

Well, it has been years since I've posted here, but it has also been years since I faced a problem like this. This was always my go-to place for getting mathematical solutions to problems I could only solve via simulation. The problem I have here is one such problem. Here is the set-up:

You have 4 balls: A, B, C, and D. You are going to draw them from a hat with replacement. Your goal is to draw 4 balls of A, 4 balls of C, and 4 balls of D, while drawing 0 balls of B. So, you'd like to end up with 4, 0, 4, 4 respectively.

Now, there are two catches to this problem. The first is that once you have drawn 4 of a ball, you can NO LONGER draw that ball. The ball is removed from the hat. The second is that you may, at any time, remove: (a) Ball D, (b) Balls C and D, or (c) Balls B, C, and D from the hat and continue drawing (i.e., you cannot remove Ball B without removing C and D also). For example, say you were to draw your first 8 balls and be at 0, 0, 4, 4. Now you could remove ball B from the bag (C and D being already removed for being at their maximum) and only draw Ball A, yielding the desired 4, 0, 4, 4 result.

Now, I have two questions:
1. What is the probability of getting 4, 0, 4, 4?
2. Does the probability change if you start by removing balls B, C, and D to first draw all 4 of Ball A (effectively starting at 4, 0, 0, 0 with Ball A removed).

My simulation suggests that the probability of getting 4, 0, 4, 4 is about 2.1%, regardless of whether you start at 0, 0, 0, 0 or at 4, 0, 0, 0. But I'd like to see how this could be solved mathematically, rather than via simulation.

Cheers,

Sherman

After a bit more thought, I think I was able to confirm that my simulation is definitely right. I still don't know the exact solution to this specific problem, but I created a simpler version that makes it more obvious that both strategies have equally probable outcomes.

Instead of 4 balls drawing to 4, imagine only 3 balls (A, B, C) drawing to 1. My goal is to get 101 (1A, 0B, 1C). I'm permitted, if I wish, to remove balls B and C from the hat and guarantee myself to draw A. I can also remove just ball C from the hat to draw between A and B, but that doesn't seem like a logical thing to do, so I'll omit that one. And, once I have one ball of each, I must remove it from the hat.

So, let's say I decide to remove balls B and C from the hat. I draw Ball A obviously. On turn two, it is now 50/50 that I get my desired result. So, the first strategy clearly yields a .50 probability of success.

The second strategy (leaving all three balls in and drawing twice) has six sequences of outcomes:

AB = 1/3 * 1/2 = 1/6 (Bad Result)
AC = 1/3 * 1/2 = 1/6 (Good Result)
BA = 1/3 * 1/2 = 1/6 (Bad Result)
BC = 1/3 * 1/2 = 1/6 = (Bad Result)
CA = 1/3 * 1/1 = 1/3 = (Good Result; remember after drawing C, I can remove ball A)
CB = 1/3 * 0 = 0 = (Impossible result; once I draw C, I'll certainly remove B)

So, we have 1/6 + 1/6 + 16 = 3/6 bad results and 1/3 + 1/6 = 3/6 good results. In other words, it's 50-50.

I'm still interested in a possible combinatorics solution to the larger problem. But this simpler version demonstrates without a doubt that the strategies result in the same probability of success.

Hi Sherman, glad to see you back!

No, and this part doesn't require calculation. The following is key:

Ball A might as well not exist. It's similar to asking, "When repeatedly rolling a die, what's the chance of rolling a 5 before a 6?" The existence of the numbers 1-4 is irrelevant because you'll keep re-rolling until a 5 or 6 comes. All that matters is P(5 | 5 or 6)

This reduces to, "What is the probability of getting 4C and 4D before any B?"

Given that a draw isn't Ball A, each other ball is 1/3 until a ball is removed, at which point it becomes 1/2. The change to 1/2 can happen on any draw after the 4th. The possibilities are as follows:

- Ball C gets removed in 4 draws and then Ball D gets picked 4 times = (1/3)^4 * 1/2^4

- Ball C gets removed on the 5th draw and then Ball D gets picked 3 times. For Ball C to be removed on the 5th draw, we need Ball D to get picked exactly once out of the first 4 draws, the probability of which is 4(1/3)^4. Then the 5th draw needs to be C, so multiply by another 1/3.

- And so on with the later draws. Altogether we have [(1/3)^4 / 16] + [4(1/3)^5 / 8] + [C(5,2)*(1/3)^6 / 4] + [C(6,3)*(1/3)^7 / 2]

Multiply that by 2 because Ball D can be the first one to 4 instead of C.

Answer = 2.16620942 %