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If $10$ coins are tossed simultaneously. Find Probability of getting six or more consecutive heads.

My Method:

(i) Six Consecutive heads

(a) HHHHHHTTTT

(b) THHHHHHTTT

(c) TTHHHHHHTT

(d) TTTHHHHHHT

(e) TTTTHHHHHH

(ii) Seven consecutive heads

(a) HHHHHHHTTT

(b) THHHHHHHTT

(c) TTHHHHHHHT

(d) TTTHHHHHHH

(iii) Eight consecutive heads

(a) HHHHHHHHTT

(b) THHHHHHHHT

(c) TTHHHHHHHH

(iii) Nine consecutive heads

(a) HHHHHHHHHT

(b) THHHHHHHHH

(iv) Ten consecutive Heads

(a) HHHHHHHHHH

So my answer is $(5+4+3+2+1)\dfrac{1}{2^{10}}$

But Given Answer is $\dfrac{3}{2^6}$

I can't figure out what mistake I am making.

Can anyone here help me out?

Related: Fair coin tosses: Probability of $\geq 4$ consecutive heads

mathophile
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    Far easier to count the binary strings without a block like $H^6$. – lulu Oct 03 '23 at 15:50
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    Your method omits a lot of strings. For instance $H^6TH^3$ isn't on the list. Nor is $HTH^6TH$, and so on. But, really, count the complement. So much easier. I confirm the official solution. – lulu Oct 03 '23 at 15:57
  • @trueblueanil Actually I need help in my method only. I know other method which bring me to result of $\dfrac{3}{2^6}$ – mathophile Oct 03 '23 at 16:00
  • @lulu I'll try once again – mathophile Oct 03 '23 at 16:01
  • I seriously wouldn't. There are so many cases. That's why the complementary method works better...there is only one way to fail to have $H^6$, while there are many ways to have that string. – lulu Oct 03 '23 at 16:02
  • @lulu Can you provide your method as answer? – mathophile Oct 03 '23 at 16:03
  • I assume it's the same as the "other method" you said you knew. What method was that? – lulu Oct 03 '23 at 16:04

3 Answers3

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To summarize (and elaborate on) the discussion in the comments:

The indicated method is fine in principle, but the actual enumeration here assumed that all the $H's$ appeared in one block, which of course was not part of the assumption. Thus strings like $HTH^6TH$ were omitted.

To proceed via the complementary method, let $s_n$ be the number of binary strings (in $H,T$) without a block of the form $H^6$. Then:

$$n≤5\implies s_n=2^n\quad \quad \&\quad \quad s_6=2^6-1$$

If $n>6$ then any good word of length $n$ must have the form $H^iTW$ where $0≤i≤5$ and $W$ is a good word of length $n-i-1$. It follows that we have the recursion $$n>6\implies s_n=\sum_{i=1}^6 s_{n-i}$$

It's easy to confirm the official solution from there.

lulu
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I tried to do this in between brute force (listing all explicitly) and a pure solution that is more elegant.

Here the observation is that you have to work with a fixed block (for case of 6) $TH^6T$ with freedom of $T$ and $H$ on either side.

This gives for example $TH^6T \_\space\_\space\_ $ which is 8 successes and by moving the block around and using symmetry you get 28 of these. Proceeding similarly for more than $6$ heads yields 48 total and then $48/2^{10}$ is $3/2^6$.

Not the most efficient but I found it intuitive.

AlgTop1854
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You're making the mistaken assumption that all of the heads are in one consecutive block, which isn't necessary true.

For example, the case you have written as HHHHHHTTTT should really be HHHHHHT???, where each ? can independently be H or T. This counts as $2^3 = 8$ outcomes, not just one.

A full enumeration of “six or more consecutive heads” gives 48 (not 15) total outcomes:

  • HHHHHHHHHH
  • HHHHHHHHHT
  • HHHHHHHHTH
  • HHHHHHHHTT
  • HHHHHHHTHH
  • HHHHHHHTHT
  • HHHHHHHTTH
  • HHHHHHHTTT
  • HHHHHHTHHH
  • HHHHHHTHHT
  • HHHHHHTHTH
  • HHHHHHTHTT
  • HHHHHHTTHH
  • HHHHHHTTHT
  • HHHHHHTTTH
  • HHHHHHTTTT
  • HHHTHHHHHH
  • HHTHHHHHHH
  • HHTHHHHHHT
  • HHTTHHHHHH
  • HTHHHHHHHH
  • HTHHHHHHHT
  • HTHHHHHHTH
  • HTHHHHHHTT
  • HTHTHHHHHH
  • HTTHHHHHHH
  • HTTHHHHHHT
  • HTTTHHHHHH
  • THHHHHHHHH
  • THHHHHHHHT
  • THHHHHHHTH
  • THHHHHHHTT
  • THHHHHHTHH
  • THHHHHHTHT
  • THHHHHHTTH
  • THHHHHHTTT
  • THHTHHHHHH
  • THTHHHHHHH
  • THTHHHHHHT
  • THTTHHHHHH
  • TTHHHHHHHH
  • TTHHHHHHHT
  • TTHHHHHHTH
  • TTHHHHHHTT
  • TTHTHHHHHH
  • TTTHHHHHHH
  • TTTHHHHHHT
  • TTTTHHHHHH

The probability is thus $\frac{48}{1024} = \frac{3}{64} = 0.046875$.

Dan
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