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November 11th, 2015, 09:03 AM  #1 
Newbie Joined: Nov 2015 From: Paris Posts: 1 Thanks: 0  Tricky question
Here's the problem: Let P(k) be a random draw of integers between 1 and k (inclusive). Assume you repeatedly apply P, starting at 10^1000. What's the expected number of repeated applications until you get 1? Why? What is your understanding of the following statement? Assume you repeatedly apply P, starting at 10^1000. 1) We are asked to repeatedly apply P(k), where k is given and equal to 10^1000.
Case 1a) Since best case is that it takes just one draw and worst case is that it takes k draws, I'd guess the expected number of repeated applications until you get 1 is k/2 but I can't really explain why. Case 1b) My understanding is that the probability of getting 1 is 1/k on 1st draw, 1/(k1) on 2nd draw, ..., all the way up to 1 on kth draw. But how does this help me get to the expected number of repeated applications until you get 1? Case 2) Am I overcomplicating it? Thank you for your hints 
November 15th, 2015, 12:52 PM  #2 
Newbie Joined: Sep 2015 From: New york Posts: 17 Thanks: 4 
Case 1a) why do you say the worst case is k? The question says the draws can be repeated

November 16th, 2015, 05:23 AM  #3 
Newbie Joined: Sep 2015 From: New york Posts: 17 Thanks: 4 
Case 1b) your argument is almost correct prob (getting a 1 on first draw)= 1/k but prob (getting a 1 on second draw) = 1/(k1) * the probability you didn't get a 1 on the first draw see if you can carry on this argument... 

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