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I was reading: https://math.stackexchange.com/a/1839505/1015824

But still I can't understand how to calculate any of the following: $(3|), (3|H)...$

I know that $E(x)=E(E(X|Y))$ but I don't know how to calculate the above expected value...

Jose Avilez
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John
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1 Answers1

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The expected number of tosses to obtain three consecutive heads given that the first toss is a tail equals one plus the expected number of tosses to obtain three consecutive heads (starting from that point). $$\mathsf E(3H\mid T) = 1+\mathsf E(3H)$$

This is the $(1+x)$ factor in the original solution. The factor of $\tfrac 12$ is the probability for the first toss showing a tail.

So are the other terms evaluated.

Graham Kemp
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  • and $(3)$ is equal to?... We are back to the original problem as we want to calculate this... 2) Can you explain why "equals one plus the expected number"... it's the first time I see something like this (I was used to use the definition of expected value to calculate it)
    • – John Feb 03 '22 at 02:12
  • $$\small\begin{align}\mathsf E(3H)&=\mathsf P(T)\mathsf E(3H\mid T)+\mathsf P(HT)\mathsf E(3H\mid HT)+\mathsf P(HHT)\mathsf E(3H\mid HHT)+\mathsf P(HHH)\mathsf E(3H\mid HHH)\&=\tfrac 12(1+\mathsf E(3H))+\tfrac 14(2+\mathsf E(3H))+\tfrac 18(3+\mathsf E(3H))+\tfrac18\cdot3\end{align}$$ Then solve for $\mathsf E(3H)$
    • – Graham Kemp Feb 03 '22 at 02:25
  • The expected count of tosses to terminate given the first toss is a tail equals that first toss plus the expected count of the rest of the tosses.
    • – Graham Kemp Feb 03 '22 at 02:28