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February 24th, 2016, 01:21 AM   #1
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Advanced combinatorics

There are 2 buckets and 100 balls. A man throws the balls one at a time and it always falls in one of the two buckets. At which point(how many balls thrown) the difference between the number of balls in the two buckets is statistically likely to be the greatest?

// My solution
When you throw 1 ball the difference is always 1 so - 1; For 2 - 1// 3 - 1.5// 4-1.5// //5-1.86(prox) calculating for all cases 5-0 3-2 4-1 and their probability(For example the probaility of 3- 2 is 20/32 so 20/32*1(the difference)

// 6-1.87(prox) and so on.It is a growing function but when the number is even its drops a bit , because of the cases where the balls split in two equal parts and that is why is 99 and not 100

A bit harder would be to work out the task with 3 buckets and when its most likely the the sum of 2 to be equal to the 3rd bucket.
goking is offline  
February 24th, 2016, 04:18 AM   #2
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I would say that after the first ball is thrown, the difference will probably be the greatest because when only first ball is thrown, we know that it can go to only one bucket and the other bucket will be empty. Starting with the throw of the second ball, we don't know where the ball goes.
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February 24th, 2016, 08:38 AM   #3
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If the probability of throwing in a box is the same for all boxes, after three balls, one expects a difference of 1.5.

Last edited by Hoempa; February 24th, 2016 at 08:41 AM.
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February 24th, 2016, 05:12 PM   #4
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The expected difference will always increase with the number of balls thrown, no matter the true probability of each throw landing in each bucket. Assuming iid throws. Interesting case if p_1=p_2=.5. It's a random walk.
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