A question about tossing coin

Finatos · November 7th, 2012, 8:16 pm

What's the average number of times for tossing a coin to get head twice(like HH). I use traditional way to calculate it and get the correct answer 6. But my teacher pointed out that we can use some knowledge of stochastic process and solve the problem quickly. I wonder how can we solve the problem in this way? I think is to calculate the stopping time for a martingale. But I don't know how to do it. Anybody can help?THANKS!

alpher · November 7th, 2012, 10:07 pm

not sure but you're right...it takes 6 on average and based on my quick calculations:it takes 14 on average to get 3 in a row. in general the whole process satisFies this:head_n = (1−p)(1+head_n )+p(1−p)(2+head_n )+p2 (1−p)(3+head_n )+. . .+p_n−1 (1−p)(n+head_n )+p_n X ni suppose you can easily prove this...// and...obviously "p" stands For probability above...

Ultraviolet · November 7th, 2012, 10:54 pm

The expected number N of coin tosses for obtaining HH is E[N] = 1/2 * E[N|T] + 1/2 *E[N|H]E[N|H] = 1/2*2 + 1/2 * (1 + E[N|T])E[N|T] = 1 + E[N]E[N] = 1/2 * (1 + E[N]) + 1/2 * (2 + 1/2 * E[N])n = 1/2 + 1/2n + 1 + 1/4n = 3/2 + 3/4n1/4n = 3/2n = 6

animeshsaxena · November 9th, 2012, 5:27 am

use induction method....u can easily get a general result of getting p continuous tosses.assume that u already have p-1 tosses...and then go to final pth toss....

Landscape · November 9th, 2012, 12:26 pm

1. What is the expected time until we hit first head? Answer: 2.2. What is the probability that you will get head again and thus have your first HH? Answer: 1/2. (If fail, we will start from zero)3. Expected time to hit HH must therefore be 3/(1/2) = 6.Or in mathematics:t_0 = first time to hit HE[t_0] = 2.t_1 = first time to hit HHE[t_1] = E[t_0] + E[t_1 | Standing in H] = E[t_0] + 1*P(H) + (1 + E[t_1])*P(T) => E[t_1] = 2*E[t_0] + 2 = 6

mpuang · November 17th, 2012, 4:37 am

Can we actually create an edge from these analysis of consecutive H or T?http://quant.stackexchange.com/question ... p?t=652687