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 June 29th, 2010, 06:20 PM #1 Member   Joined: Jun 2010 Posts: 64 Thanks: 0 toss coin until the first run of r consecutive tails occurs if a single coin, with probability q = 1 - p of obtaining tails on each toss, is tossed repeatedly until the first run of r consecutive tails occurs, How is the expected number of tosses????.....and how is the Probability tha we toss the coin n-times(n>=r) ?????????? THANK YOU VERY MUCH.
 June 30th, 2010, 12:09 AM #2 Member   Joined: Jun 2010 Posts: 80 Thanks: 0 Re: toss coin until the first run of r consecutive tails occurs Hi, nice to meet you. I'm new to this stuff. I think I can't help you much, I could only find the following : let A be the event : thrown r times tail consecutively, prob. of A knowing n tosses were thrown were : p(A|n)={ 0, if n
 June 30th, 2010, 06:56 AM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: toss coin until the first run of r consecutive tails occurs After the first r-1 tosses, you have roughly a q^r chance per flip of getting r tails in a row. There was a (long) thread about this on the Mersenne forums if you need more detail.
 June 30th, 2010, 06:47 PM #4 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: toss coin until the first run of r consecutive tails occurs I tried solving this problem using Markov chains. Consider the first r throws as leading to one of the r+1 initial states {"...H", "...HT", "...HTT", "...HTTT", "...HTTTT", ... "T...TTTT"} and create a substochastic matrix with r rows: $T=\begin{bmatrix}p&q=&...=&0\\p=&q=&...=&0\\p=&q=&...=&0\\\vdots=&\vdots=&\vdots=&\vdots=&\vdots=$ Then the expected number of throws is $r+p\begin{bmatrix}1&q&q^2&q^3&...&q^{r-1}\end{bmatrix}(I-T)^{-1}\begin{bmatrix}1&1&...&1&1\end{bmatrix}^{\top}$
 June 30th, 2010, 08:38 PM #5 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: toss coin until the first run of r consecutive tails occurs If p=q=½, the answer is $2(2^r-1)$.
 July 1st, 2010, 06:54 AM #6 Member   Joined: Jun 2010 Posts: 80 Thanks: 0 Re: toss coin until the first run of r consecutive tails occurs Hi, do you know how to link this result to the probabilities in post 2 ? this gave : =r+p(r+1)+p^2(2r+1)+..p^k(kr+1) but I couldn't achieve to sum this up. (r tail+ 1head + (any r-1)+ 1head+...i thought maybe the develop. of (1-t)^{-1}=1+t+t^2+t^3, could help ? ) thanks. Bye.
 July 1st, 2010, 11:49 PM #7 Member   Joined: Jun 2010 Posts: 80 Thanks: 0 Re: toss coin until the first run of r consecutive tails occurs >Erratum : sorry : r+p(r+1+r+2+...r+r)+...p^m(mr+mr+1+...mr+r)+...=$r+\sum_{m=1}^\infty p^m(mr+\frac{r(r+1)}{2})$ ?

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