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?

My solution to the first part:

On any given turn, the probability of drawing an integer that will bring the running total to 0 mod 4 is 1/4.  Thus the expected number of turns until you stop is 4. 

The expected value of any individual turn is 2.5, so the expected value of the running total when you stop is 4 x 2.5 = 10.

A question about the variation:

Do the integers cycle?  For instance, if I draw a 1, on my next turn are my choices limited to just 1 and 2, or could I draw a 1, 2, or 4?

 

 

Posted by tomarken on 2014-04-30 10:33:57