# What is the probability of a coin landing tails 7 times in a row in a series of 150 coin flips?

$\begingroup$ hera are some details ; I will lone work out the sheath where you want $7$ tails in a quarrel, and the general sheath is like. I am interpreting your doubt to mean “ what is the probability that, at least once, you flip at least 7 tails in a row ? ”
Let $a_n$ denote the number of ways to flip $north$ coins such that at no point do you flip more than $6$ consecutive tails. then the number you want to compute is $1 – \frac { a_ { 150 } } { 2^ { 150 } }$. The death few coin flips in such a sequence of $nitrogen$ coin flips must be one of $H, HT, HTT, HTTT, HTTTT, HTTTTT$, or $HTTTTTT$. After deleting this last bite, what remains is another sequence of mint flips with no more than $6$ consecutive tails. So it follows that
 a_ { n+7 } = a_ { n+6 } + a_ { n+5 } + a_ { n+4 } + a_ { n+3 } + a_ { n+2 } + a_ { n+1 } + a_n 

with initial conditions $a_k = 2^k, 0 \le k \le 6$. Using a computer it would not be very hard to compute $a_ { 150 }$ from here, specially if you use the matrix method acting that David Speyer suggests .
In any event, let ‘s see what we can say approximately. The asymptotic growth of $a_n$ is controlled by the largest positive root of the characteristic polynomial $x^7 = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$, which is a little less than $2$. Rearranging this identity gives $2 – x = \frac { 1 } { x^7 }$, so to a first estimate the largest rout is $radius \approx 2 – \frac { 1 } { 128 }$. This means that $a_n$ is approximately $\lambda \left ( 2 – \frac { 1 } { 128 } \right ) ^n$ for some ceaseless $\lambda$, which means that $\frac { a_ { 150 } } { 2^ { 150 } }$ is roughly
 \lambda \left ( 1 – \frac { 1 } { 256 } \right ) ^ { 150 } \approx \lambda e^ { – \frac { 150 } { 256 } } \approx 0.56 \lambda 
although $\lambda$ still needs to be determined .
Edit: So let ‘s approximate $\lambda$. I claim that the generating function for $a_n$ is

 A ( x ) = 1 + \sum_ { n \ge 1 } a_ { n-1 } x^n = \frac { 1 } { 1 – adam – x^2 – x^3 – x^4 – x^5 – x^6 – x^7 }. 
This is because, by iterating the argument in the moment paragraph, we can decompose any valid sequence of coin flips into a sequence of one of seven blocks $H, HT, …$ uniquely, except that the initial segment does not necessarily start with $H$. To simplify the above formula, write $A ( x ) = \frac { 1 – ten } { 1 – 2x + x^8 }$. nowadays, the overtone divide decomposition of $A ( x )$ has the shape
 A ( x ) = \frac { \lambda } { r ( 1 – rx ) } + \text { other terms } 
where $\lambda, r$ are as above, and it is this first term which determines the asymptotic behavior of $a_n$ as above. To compute $\lambda$ we can use l’Hopital ‘s convention ; we find that $\lambda$ is equal to

 \lim_ { x \to \frac { 1 } { radius } } \frac { r ( 1 – rx ) ( 1 – adam ) } { 1 – 2x + x^8 } = \lim_ { x \to \frac { 1 } { gas constant } } \frac { -r ( r+1 ) + 2r^2x } { -2 + 8x^7 } = \frac { r^2-r } { 2 – \frac { 8 } { r^7 } } \approx 1. 
So my official estimate at the actual value of the answer is $1 – 0.56 = 0.44$. Anyone concern to validate it ?
Sequences like $a_n$ count the phone number of words in objects called regular languages, whose enumerative behavior is described by analogue recurrences and which can besides be analyzed using finite state machines. Those are all good keywords to look up if you are interest in generalizations of this method. I discuss some of these issues in my notes on generating functions, but you can find a more thorough initiation in the relevant section of Stanley ‘s Enumerative Combinatorics .

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