N Heads in a Row
Given a coin that lands heads with probability 0 < p < 1, what is the expected number of flips needed to get N heads in a row?
Solve with p = 0.73 and N = 20. Then can someone give a general method / formula based only on p and N? Thanks 
It's the negative binomial distribution (with parameters N and p).

Pardon my ignorance but ...
maybe I'm not understanding this right, but it looks like negative binomial distribution is used to determine, for instance, the expected number of tails before a predetermined number of heads, but not necessarily in order 
[QUOTE=petrw1;484296]maybe I'm not understanding this right[/QUOTE]
I think I misunderstood you! In that case, you can work it out from first principles. Fix a positive integer \(k\geq N\). How many ways are there of throwing \(kN\) times without getting \(N\) heads in a row? What is the probability of each of those possibilities followed by \(N\) consecutive heads? 
[QUOTE=Nick;484307]How many ways are there of throwing \(kN\) times without getting \(N\) heads in a row? What is the probability of each of those possibilities followed by \(N\) consecutive heads?[/QUOTE]
Not quite right, either. The \(kN\) might have ended with one or more heads. I usually grab the density using the transition matrix for the Markov Chain of #heads in a row with an absorbing state at \(N\), but this question asks for the mean. I'd try an inductive approach on the number of heads. To get one head in a row, the first toss is a head with probability p or a tail with probability (1p). If heads, you are done. If tails, you need the to start over. So X1 = p*1 + (1p)*(X1+1) X1 = 1/p For k in a row, you need k1 in a row, then one more toss either finishes it or starts it over. Xk = p*(X(k1)+1) + (1p)(Xk+X(k1)+1) Xk = X(k1)/p + 1/p The logic of the last equation is conditioned on the results of the flip after first achieving k1 heads. Check that arithmetic  I fixed several errors before posting and might still have some left  but the logic is solid. 
[QUOTE=petrw1;484289]Given a coin that lands heads with probability 0 < p < 1, what is the expected number of flips needed to get N heads in a row?
Solve with p = 0.73 and N = 20. Then can someone give a general method / formula based only on p and N? Thanks[/QUOTE] Take 1(1p)ⁿ to be the probability of getting heads n times ? Factor in how far you've advanced and find where it hits 50% ? 
Oops  this is Homework Help. I had fun working out the inductive equations and their general solution, but I suppose I shouldn't post all that here. So I stand by the previous suggestion that N+1 heads in a row takes N heads in a row plus one more, and then you may be done or you may need to start over. To check your work, I get 2001.6 for p = 0.73 and N = 20.

All times are UTC. The time now is 22:38. 
Powered by vBulletin® Version 3.8.11
Copyright ©2000  2021, Jelsoft Enterprises Ltd.