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A fair coin is tossed $20$ times. Find the probability that exactly $6$ heads were tossed one of which was tossed on the first toss and one of which was tossed on the last toss, and that no consecutive tosses were heads.

Typically, the probability would be ${20 \choose 6}(1/2)^{6}(1/2)^{14}$.

However, we are restricted in that we cannot include all combinations of successes and failures.

Would the adjusted answer be:

${10 \choose 4}(0.5)^4 (1.5)^{14}$?

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Just count how many of the $2^{20}$ possible sequences satisfy the conditions. Every legal sequence must have the form HT...H, such that the "..." consists of 4 "HT" combinations and 9 "T"s that are not part of such a combination.

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    So it would be ( 8 choose 4) since there are 8 pairs of slots where you can place the HT pair. Is this correct?2011-09-24
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    Not obvious, and there are more. But shape of your formula turns out to be right.2011-09-24
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    @André I thought Henning was calculating it correctly... Can you suggest some sequence he is missing out?2011-09-24
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    @Srivatsan Narayanan: Of course he was. However, the interpretation by lord12, immediately above my comment, is not correct. But kind of close in spirit.2011-09-24
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    What am I missing?2011-09-24
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    We have HT at beginning, TH at end, $16$ spaces left in between. You probably thought of them as divided into $8$ groups of two, of which you would pick $4$ for $HT$. But this leaves out, for example, for the middle $16$, THTHTHTHTTTTTTTT. Point is that the H's need not occupy "odd" positions. Your $\binom{8}{4}$ only counts those.2011-09-24
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    Would it be 2(8 choose 4) then?2011-09-24
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    Now you're just guessing. Again, there are 4 "HT"s and 9 "T"s that you have to give in some order. What is four plus nine?2011-09-24
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    Actually, since you have HT....TH you have to arrange 4 H's and 8 T's so the answer would be (12 choose 4)/2^202011-09-27
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    @lord12, no. I assume your idea is to place all the T's there an put a H next to 4 of them. Which is fine as far as it goes, but you're counting the possible position for H's wrong. Both sequences starting with HTH... and sequences ending with ...HTH. So, starting with HT$^{14}$H, there are 13 (not 12) places you can put a H, and choosing 4 of them gives $\binom{13}{4}$ possibilities.2011-09-27