How would you divide 50 black and 50 white marbles into two piles so that the probability of picking a white marble as follows is maximized: we first pick one of the piles uniformly at random, then we pick a marble in that pile uniformly at random?
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4 comments:
Solution :
1.> The probability of choosing a pile is 1/2
2.> Next we need to maximize the probability of choosing a white marble. What is the max probability - '1'.
When can the probability be '1' - when the even is certain. This is possible IF AND ONLY IF there is one one marble to choose from and that marble is white.
Conclusion :
Divide the marbles in two piles :
First pile : 50 black + 49 white marbles
Second pile : 1 white marble
The probability of choosing a white marble is maximum.
There is a mathematical proof here, but I could not understand that approach. If someone could decipher it and help that would be great!
Hey madhur,
how are you doing?
I just went through the mathematical proof. Want I found is, some minor (obvious) changes are required in the proof (just my opinion).
1."If the two piles are equi-sized, then no matter how we distribute the balls, the probability of drawing a white is \frac{w}{w+b}"
This is not w/(w+b), coz the probability of choosing any of the piles is 1/2 and then the probaility of drawing the balls is from the selected pile (assuming all the balls are in the same pile) is w/(w+b) .. so the total probability is 1/2 * w/(w+b) ....
OOPs i might take a lot of space commenting here.. will put thru a mail ;-)
@Ranganath
Here, w and b are total no. of white and black balls respectively. So, if the two piles are of equal size (lets say total no. of balls in each pile is z). Now, assume pile 1 has x white balls. Then pile 2 will have w-x white balls. So, total probability of drawing a white ball is -
1/2(x/z + (w-x)/z) = 1/2(w/z) = w/(w+b)
Do the calculations, to understand.
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