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What is the expected number of consecutive Heads from 3 coin tosses?

I see the solution available here but am unsure if it is applicable to the question above, as that solution refers to a particular number of consecutive heads whereas this one could have 2 or 3 consecutive heads.

Thank You

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    $\begingroup$ There are only 8 possible outcomes for three coin tosses. Count the number of consecutive heads in each possible outcome and average them. $\endgroup$
    – MJD
    Oct 16, 2015 at 12:52
  • $\begingroup$ @MJD So is it (2*(1/8) + 2*(1/8) + 3*(1/8))/3 = 7/24? $\endgroup$
    – Jojo
    Oct 16, 2015 at 13:16
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    $\begingroup$ What are you counting? How many "consecutive heads" are there in HTH? $\endgroup$
    – Graham Kemp
    Oct 16, 2015 at 13:31
  • $\begingroup$ @GrahamKemp I counted HHH, HHT and THH which have 3 consecutive Heads and 2,2 each. Am I misunderstanding the question? $\endgroup$
    – Jojo
    Oct 16, 2015 at 14:05
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    $\begingroup$ you have $HHH$ happening with prob $\frac18$ and $\{HHT, THH\}$ happening with prob $\frac28$. Hence the expected value is $1\times \frac18+2\times\frac28=\frac58$. $\endgroup$
    – Math-fun
    Oct 16, 2015 at 14:17

3 Answers 3

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Honestly, this is a question I found in a probability book that I am practicing and doesnt have answers, so I dont know what the correct definition is. But I feel that HHT has 1 and HHH has 2.

Well, then if the "number of consecutive heads" is counting the number of "heads which follow immediately after another head", then the expected number of consecutive heads will be : $$0\cdot \tfrac 5 8 +1\cdot\tfrac 28+ 2\cdot\tfrac 18 =\tfrac 48$$

Because of the eight equiprobable outcomes, five have no consecutive heads, two have one consecutive heads, and one has two consecutive heads.

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HHT THH :: 2/8 consecutive 2 heads

HHH :: 1/8 consecutive 3 heads

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The possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT. Of those, HHH, HHT, and THH have "consecutive heads". Now, do you mean "the number of times you get consecutive heads" or "the number of heads in flips that have consecutive heads"? In the first case, there are 3 cases with consecutive heads so the expected number of "consecutive heads" is 3/8. In the second case, there are a total of 3+ 2+ 2= 7 heads so the expected number is 7/8.

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    $\begingroup$ HTT has one consecutive head, not zero. $\endgroup$
    – MJD
    Oct 16, 2015 at 14:00

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