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A fair coin is flipped 7 times. What is the probability that the coin will come up heads five times in a row?

When I first did this problem, I determined that there would be 128 ($2^7$) different combinations. Determining how many cases, however, I decided that since you need 5 coins out of 7, there were $\binom75$ ways. So the probability I determined was $\frac{21}{128}$. This is not the answer as soon as I realized that my solution only determined that there were at least 5 coins chosen as head in no particular order; however, order does matter in this case.

How would one go about solving this problem? Is it possible to proceed as I did and some how reduce the number of possibilities from the one I determined?

Also any good probability books? (Unless you don't think this small question is appropriate, please answer this in the comments. It's just that I have trouble with probability)

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    $\begingroup$ You can list out all the favourable cases. There isn't many. $\endgroup$ – Kenny Lau Aug 4 '16 at 4:41
  • $\begingroup$ To recommend a probability book, we would need to know your mathematics background. $\endgroup$ – N. F. Taussig Aug 5 '16 at 9:42
  • $\begingroup$ I'm still in high school and know up to integral calculus (I will like to mention my ability in calculus doesn't consist of a full "analysis") $\endgroup$ – Ian Limarta Aug 5 '16 at 17:37
  • $\begingroup$ Samuel Goldberg's Probability: An Introduction is a nice introduction to discrete probability. $\endgroup$ – N. F. Taussig Aug 5 '16 at 18:17
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Consider cases.

Case 1: Exactly five consecutive heads.

The string of five consecutive heads can begin on the first, second, or third throw.

If the first five tosses are heads, the sixth toss must be a tail. The seventh toss can be a head or a tail. Two possibilities.

If the middle five tosses are heads, the first and last tosses must be tails. One possibility.

If the last five tosses are heads, the second toss must be a tail. The first toss can be a head or a tail. Two possibilities.

Total: $2 + 1 + 2 = 5$

Case 2: Exactly six consecutive heads.

The string of six consecutive heads can begin on the first or second throw.

If the first six tosses are heads, the last toss must be a tail. One possibility.

If the last six tosses are heads, the first toss must be a tail. One possibility.

Total: $1 + 1 = 2$.

Case 3: Seven consecutive heads.

All the tosses must be heads. One possibility.

Total: $1$

Total: Since the three cases are disjoint, the number of sequences with at least five consecutive heads is $5 + 2 + 1 = 8$.

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  • $\begingroup$ It appears the solution is 9/128 but your approach seems completely valid. Wonder why... $\endgroup$ – Ian Limarta Aug 5 '16 at 17:40
  • $\begingroup$ @IanLimarta I suspect that there is a typographical error in the answer key. $\endgroup$ – N. F. Taussig Aug 6 '16 at 0:03
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First of all, the total number of cases are $128$, because each coin flips head or tail, and seven such flips are happening, so $2^7=128$.

Now, suppose five consecutive heads do appear. Let us number the tosses from $1$ to $7$. the following cases arise, and we will eliminate overlaps when they happen:

1) Tosses $1$ to $5$ are heads, and tosses $6$ and $7$ are whatever. Then we have four total events: HHHHHHH,HHHHHTT,HHHHHTH,HHHHHHT.

2) Tosses $2$ to $6$ are heads, and tosses $1$ and $7$ are whatever. Then we have the following cases: HHHHHHH,THHHHHH,THHHHHT,HHHHHHT. Two of these cases are covered above, namely HHHHHHH and HHHHHHT. So we have $4+2$ = six favourable cases.

3) Tosses $3$ to $7$ are heads, and tosses $1$ and $2$ are whatever. Then we have the following cases: HHHHHHH,TTHHHHH,THHHHHH,HTHHHHH. Of these, HHHHHHH and THHHHHH are the repeated cases. So we have $6+2=8$ favourable cases.

Thus the probability is $\frac{8}{128}$.

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    $\begingroup$ The case $THHHHHH$ is also repeated (second and third cases), so there are only eight favorable outcomes. $\endgroup$ – N. F. Taussig Aug 5 '16 at 9:56

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