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| Random number sequence (Posted on 2011-04-20) |
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You have N bags. Bag 1 has a black ball, Bag 2 has a black ball and a white ball, Bag 3 has a black ball and two white balls, and so on. Bag N has a black ball and N-1 white balls. You pick a ball from each bag at random and record the numbers of the bags that you picked a black ball from. For example, if you had 100 bags, then your sequence might be 1, 2, 3, 10, 14, 37. Call the last number in your sequence X. Prove that X is a random number from 1 to N with a uniform distribution.
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Submitted by Math Man
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Solution:
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Since X is the last number in the sequence, you picked a black ball from Bag X and a white ball from Bags X+1 to N. The probability of picking a black ball from Bag X is 1/X. The probability of picking a white ball from any bag Y is (Y-1)/Y. Therefore, the probability of picking a black ball from X and a white ball from every bag from X+1 to N is (1/X)*(X/(X+1))*...*((N-1)/N)=1/N. That means that the chance of X being the last number in the sequence is 1/N. Therefore, it is random.
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