# Thread: sequential picking 1080 balls from 21 barrels

1. ## sequential picking 2700 balls from 100 barrels

hello. i'm trying to optimize job performance in a warehouse. the problem is somewhat simplified here, but it gives all the necessary into.

i have 100 bins + 1 giant barrel. the 100 bins each contain 24 red balls + 3 black balls.

of the total 300 black balls, 16 have green dots, 3 have yellow dots, and 1 has a white dot. As a rule, no bin necessarily has a dotted black ball, however no bin will contain more than 1 dotted ball -- in other words each bin will contain 2 black balls with no dots and each bin will contain a 3rd black ball either with or without a dot.

in the giant barrel there are an even distribution of 2700 white balls, each numbered 1-100 (there are 27 balls with the number 1, 27 balls with the number 2, etc).

My task is to randomly select (and discard) a numbered ball from the giant barrel, then choose a random ball from whichever bin is marked on the numbered ball. the random ball from the bin is also discarded.

If i select a red ball, i try again. if i select the black ball with the white dot, i win.

Problem 1:
what is the probability at any given time that i will draw the winning ball?

Problem 2:
as i progress, i can retain information about which type of balls i have drawn from the various bins. So if i could choose the numbered balls from the giant bin, how should i calculate it?

Problem 3:
there is $1 million at stake. It costs me$1000 for each ball i take from the giant bin. I will be awarded $500 thousand for drawing the ball with the white dot,$100 thousand for each ball with a yellow dot, and $12,500 for each ball with a black dot. what is the optimum number of balls to draw (I must prepay the fee before drawing balls)? Any ideas would be greatly appreciated. - Jack in TO 2. If I read this correctly: 1) Every time you draw a white ball, you draw a ball from that bin, so the number of white balls with a certain bin number on it is always equal to the number of balls in that bin. (You start out with 27 of both). So, the two steps together are equivalent (probability-wise) to just drawing a ball at random from all the balls left in the bins. The probability of drawing the winning ball, given that this is your n-th try, is 1/(2700+1-n) 2) When you draw a black ball, you're done. Therefore, all the other dotted balls except the winning ball are, so far, equivalent to any other black ball. The chance of the winning ball being in a certain bin has no correlation to the number of red balls you have already pulled out of it, as far as I can see - however the chance of drawing it, if it should happen to be there, obviously increases as the number of red balls shrinks. Imagine if there were only two bins, one untouched and one with only black balls left. The probabilities for each bin having the winner ball is 1/2. For the first bin, you have a probability of 1/2*1/100=1/200 of winning. For the other, it's 1/2*1/3=1/6. So, which should you pick? If you pick from the almost-empty bin, your chance of winning is 1/6 but your chance of the game being over is also 5/6. If you pick from the other, the chance of winning is 1/200, but the chance of the game ending is also a lot less - 1/2*2/100 + 1/2*3/100 = 5/200. Notice that the ratio of probabilities of winning to "game over" is a constant 1:5 (because there are six black balls and one has a white dot). I'm not sure about this, but my intuition is that the red balls are just "stalling" and won't matter in the end - the probability of reaching a decision increases, but the chances of winning don't, so you might as well just pick one at random. 3: I'm not sure if I get this question, but if you assume that you're back to random drawing, plus the game is over as soon as you draw any black ball (dots or no dots): Eventually if you go on long enough, you will draw a black ball. When this happens, your average return on your investment will be (1,000,000/300)$, approx. 33,333$. (Most of the time it will be 0, but if you do it enough times, this is what the average will tend to). The probability of drawing a black ball increases as you go along. For n tries, the probability of drawing a black ball at some time during those is 1-(the probability of drawing n red balls in a row) = 1 - (2600/2700)*(2599/2699)...(2600-n+1)/(2500-n+1)= 1 - 2600!(2500-n)!/(2600-n)!2500!. In terms of binomial coefficient, that's (2600 pick n)/(2500 pick n). My guess is that you're looking for the number of tries where the ratio of (this number times the average prize) to n*1000$ is the best.