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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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  • $\begingroup$ So it would be ( 8 choose 4) since there are 8 pairs of slots where you can place the HT pair. Is this correct? $\endgroup$ – lord12 Sep 24 '11 at 0:09
  • $\begingroup$ Not obvious, and there are more. But shape of your formula turns out to be right. $\endgroup$ – André Nicolas Sep 24 '11 at 0:27
  • $\begingroup$ @André I thought Henning was calculating it correctly... Can you suggest some sequence he is missing out? $\endgroup$ – Srivatsan Sep 24 '11 at 0:30
  • $\begingroup$ @Srivatsan Narayanan: Of course he was. However, the interpretation by lord12, immediately above my comment, is not correct. But kind of close in spirit. $\endgroup$ – André Nicolas Sep 24 '11 at 0:43
  • $\begingroup$ What am I missing? $\endgroup$ – lord12 Sep 24 '11 at 0:48

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