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Making a Multiple of Four (Posted on 2014-04-30) Difficulty: 3 of 5
I want to make a multiple of four in a curious way: First I choose a random integer 1 to 4, each equally likely. If I get the '4', I stop immediately and that is my number. As long as I do not have a multiple of four, I continue to choose more random integers 1 to 4, keeping track of the running total. When the running total is a multiple of four, then I stop.

What is the expected number of random integers I need to get my multiple of four and what is the expected value of the running total?


Variation: The first random integer is the same, but the second and subsequent integers are equal to, one less, or one more than the previous integer. (If the previous random integer was 3, then I would choose one of 2, 3, or 4.)

In this case what is the expected number of integers needed and what is the expected total?

No Solution Yet Submitted by Brian Smith    
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Comments: ( Back to comment list | You must be logged in to post comments.)
re: Or maybe I'm simple (part I spoiler) | Comment 4 of 6 |
(In reply to Or maybe I'm simple (part I spoiler) by Steve Herman)

The intuitive way of thinking about the total of the random number selections is that the randomizer doesn't care who's invoking it. Over the long run, the average roll will be 2.5; then, since each person has the same expected total, that must be 10.
  Posted by Charlie on 2014-04-30 12:42:26

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