Let $ a_n $ denote the act of ways to flip $ newton $ 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 last few coin flips in such a sequence of $ normality $ mint flips must be one of $ H, HT, HTT, HTTT, HTTTT, HTTTTT $, or $ HTTTTTT $. After deleting this last moment, 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 $ $

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

with initial conditions $ a_k = 2^k, 0 \le kilobyte \le 6 $. Using a calculator it would not be very hard to compute $ a_ { 150 } $ from here, particularly if you use the matrix method that David Speyer suggests .

In any case, let ‘s see what we can say approximately. The asymptotic growth of $ a_n $ is controlled by the largest plus 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 – adam = \frac { 1 } { x^7 } $, so to a first approximation the largest root is $ roentgen \approx 2 – \frac { 1 } { 128 } $. This means that $ a_n $ is approximately $ \lambda \left ( 2 – \frac { 1 } { 128 } \right ) ^n $ for some constant $ \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 lease ‘s approximate $ \lambda $. I claim that the generating function for $ a_n $ is

Read more: How to Make Money as a Coin Collector

$ $ 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 controversy in the second paragraph, we can decompose any valid sequence of mint flips into a sequence of one of seven blocks $ H, HT, … $ uniquely, except that the initial segment does not inevitably start with $ H $. To simplify the above expression, write $ A ( x ) = \frac { 1 – ten } { 1 – 2x + x^8 } $. now, the partial fraction 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 principle ; we find that $ \lambda $ is adequate to

$ $ \lim_ { x \to \frac { 1 } { gas constant } } \frac { r ( 1 – rx ) ( 1 – ten ) } { 1 – 2x + x^8 } = \lim_ { x \to \frac { 1 } { roentgen } } \frac { -r ( r+1 ) + 2r^2x } { -2 + 8x^7 } = \frac { r^2-r } { 2 – \frac { 8 } { r^7 } } \approx 1. $ $

So my official guess at the actual value of the answer is $ 1 – 0.56 = 0.44 $. Anyone care to validate it ?

Sequences like $ a_n $ count the issue of words in objects called regular languages, whose enumerative behavior is described by linear recurrences and which can besides be analyzed using finite department of state machines. Those are all good keywords to look up if you are matter to in generalizations of this method acting. I discuss some of these issues in my notes on generating functions, but you can find a more exhaustive introduction in the relevant department of Stanley ‘s Enumerative Combinatorics .

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