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November 11th, 2015, 09:03 AM   #1
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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.
  • 1a) Each drawn element is redrawable again even after being drawn (i.e. each draw is independent)
  • 1b) Each drawn element is undrawable after being drawn once (i.e. each draw is dependent on previous draws)
2) We are asked to repeatedly apply P(k), where k is an integer such that k=10^1000 on 1st draw and then k = some random integer from 2nd draw onward.

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/(k-1) 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
Aladdin is offline  
 
November 15th, 2015, 12:52 PM   #2
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Case 1a) why do you say the worst case is k? The question says the draws can be repeated
davidmoore63 is offline  
November 16th, 2015, 05:23 AM   #3
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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/(k-1) * the probability you didn't get a 1 on the first draw

see if you can carry on this argument...
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