0

Let’s say, we are tossing a fair coin.

How many times, on average, should we toss it so as to have N more heads than tails or tails than heads?

In other words, what is the mathematical expectation of the number of tosses till |number(heads) - number(tails)| == N?

Nestor
  • 101
  • What have you tried so far since this sounds like a homework question? – Gregory Dec 17 '19 at 20:58
  • I’ve managed to solve the Markov System for the simplified problem “how long does it take to have N more heads?”. Yet don’t see how to get from there to here. The experiments show that it will take somewhere around twice as long (with the ratio depending on N) to reach N more heads than N more heads or tails. – Nestor Dec 17 '19 at 21:12
  • The random variable given the absolute difference of number of heads and number of tails is also Markov. Did you try using the same technique on that? – sudeep5221 Dec 17 '19 at 21:21
  • 2
    This is the hitting time of ${-N,N}$ in simple random walk and should be a well-studied problem in the literature. You could probably find a solution in many textbooks. – Math1000 Dec 18 '19 at 01:20

0 Answers0