I threw a coin n times, and never got three tails in a row. I calculated the odds of this event, and found out they were just about even; 50%-50%. How many times did I throw the coin?

A second question: what were the chances of having not gotten three heads in a row either?

This problem can be solved using Markov chains. For the first part, we define states H (last toss was heads), T (last toss was tails), TT (we got two tails in a row) and TTT (three tails in a row). We also define P(I,S) as the probabilitiy of being in state S after the I-th throw. By definition P(1,H)=P(1,T)=0.5, and P(1,TT)=P(1,TTT)=0. The probabilities satisfy P(i+1.H)=0.5x(P(i,H)+P(i,T)+P(i,TT)), P(i+1,T)=0.5xP(i,H), P(i+1,TT)=0.5xP(i,T), P(i+1,TTT)=P(i,TTT)+0.5xP(i,TT). Doing the numbers, for i=10 we get P(10,TTT)=50.8% so there were 10 coin tosses.

Posted by Oskar on 2004-06-12 23:04:29