I know how to solve this problem using Markov Chain and I get 8.However, I'm trying to solve it using expected values, and I can't get it...If I were to look for HHH, I would do it this wayLet X=expected number of tosses to get HHHX=(1/8)*3+1/8*(3+X)+1/2*(1+X)+1/4*(2+X) And, I would get 14....Now, in the 'THH' case, the process doesn't always goes to the origin, so it much take fewer tosses... anybody know how I can adjust the expected value equation?

<wrong>

Last edited by Peniel on April 22nd, 2013, 10:00 pm, edited 1 time in total.

QuoteOriginally posted by: RoniNYCNow, in the 'THH' case, the process doesn't always goes to the origin, so it much take fewer tosses... anybody know how I can adjust the expected value equation?When conditioning on the first outcome you need to solve for the intermediate quantity E X_T - the expected outstanding number of tosses after having thrown T in the first one. That's easy: x = 1 + (1/2) E X_T + (1/2) xHence E X_T = x -2. Here x is the expectation you are looking for. Now condition on the second and the third toss using the above result (sinse wrong outcome gives the first element to your sequence). This yields x = 8.

For completeness, with E=E X_T:E = (1/4)*2 + (2/4)*(E+2) + (1/4)*((1/2)*3 + (1/2)*(E+3)) -> E=6 -> x=8

GZIP: On