My Math Forum Number of coin flips required to have a variance of ten?

 October 4th, 2015, 09:29 PM #1 Newbie   Joined: Oct 2015 From: Us Posts: 1 Thanks: 0 Number of coin flips required to have a variance of ten? If we play a game and flip a coin, heads I win your dollar, tails you win my dollar, every time you lose a flip, you lose a dollar. How many times would we have to flip the coin for me to be up 10 dollars? I want to make a distinction, I'm not asking the number of flips to get 10 heads in a row, instead how many flips for me to be up 10 dollars. I get it's expected value is zero, but on a long enough timeline there will be a variance of positive 10, so how would you calculate to find the number of flips needed to get there? Thanks guys, the math is a little over my head, ok a lot lol but I'd really love it if someone could explain the formula.
 October 5th, 2015, 07:00 PM #2 Global Moderator   Joined: May 2007 Posts: 6,788 Thanks: 708 The variance for a Bernoulli variable for n tries is np(1-p), where p is the probability of success. For coin flips, p=1/2, so the variance is n/4. What you are looking for is the standard deviation (square root of variance) = 10. This gives you n = 400.
 October 5th, 2015, 11:51 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,835 Thanks: 2162 That doesn't imply you can expect to be up 10 dollars after 400 flips.
 October 15th, 2015, 11:18 AM #4 Newbie   Joined: Sep 2015 From: New york Posts: 17 Thanks: 4 I think you are asking the question: given a fair coin, how many flips, on average, would it take for me to get to +10 dollars? The answer, believe it or not, is infinity. There is a theorem that the average number of steps in a symmetric random walk, where the walk stops if you get to -b dollars or + a dollars, is ab. But in your case b is infinity, hence the result.

 Tags coin, flips, number, required, ten, variance

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