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A fair coin is flipped repeatedly until it turns up tails five times. What is the expected number of heads before that happens?

Based on the link given in the comment, I have found it can be solved using the recursion $E(n)=\frac{1}{2}(E(n)+1)+\frac{1}{2}(E(n-1))$ which is equivalent to $E(n)=E(n-1)+1$. Is it correct?

StubbornAtom
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BitanjS
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    @StinkingBishop I'll take a look at that, thanks. I double-clicked the post the question button, I think it's a bug. I already deleted it. – BitanjS Oct 02 '20 at 10:40
  • I don't know, I have not looked at the original post - I've just linked yours to that one as a duplicate. You may want to comment on that original post. –  Oct 02 '20 at 10:59
  • Should clarify: did you mean "five $\textit {consecutive}$ tails" or any five tails? The supposed duplicate is for consecutive. If you just meant any five tails, then note that you expect to get $5$ tails in $10$ tosses, so the answer is $5$. – lulu Oct 02 '20 at 11:22
  • @lulu I've seen this question before with the same exact wording. It is not five consecutive. Don't know why it was flag for duplicate though.. – Adola Oct 02 '20 at 12:45
  • @Adola The non-consecutive case is much easier than the consecutive case, and the OP hasn't corrected the impression that "consecutive" was intended. So who knows? – lulu Oct 02 '20 at 13:18
  • @lulu https://321da88a-a-62cb3a1a-s-sites.googlegroups.com/site/imocanada/2009-summer-camp/2009SummerCamp-DavidArthur-ProbabilisticMethod.pdf?attachauth=ANoY7cpyl9eVhEnUUYzDipqxfZbwrK--Y6D6NMX06XWdaAKRlzATVW1bhE8l2p5TOCaIYPXpZwqPz5y-G2F5bA3WX_WVheXUAaraikbiqUeINE2qoavOI3LdHUwvUG7FNRlggAvvPsp-7sRjo44ycmQMeOdADlrxWNlJwKwRBsh5cTR_k4POkEAml9QbL8p0CDQJYh4rlwild4mEG58zU5Uh6L6FkHkKzexnGzzcrha89swMBRt2bs8CP-COrEmLHehN2jSaMK7FW6OZjJRwL25Ngb1njebkWA%3D%3D&attredirects=1 at the bottom of page 3 – Adola Oct 02 '20 at 15:52

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