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I toss a fair coin. All tosses are independent. Let $T_{HH}$ denote the waiting time to get heads two times in a row and likewise for $T_{TT}$, $T_{HT}$

Find the probability when the sequence of $HH, TT, HT$ occur at respectively toss 2 and 3 toss

$P(T_{HH}=2)=\frac{1}{2}*\frac{1}{2}=\frac{1}{4}$

$P(T_{HT}=2)=\frac{1}{2}*\frac{1}{2}=\frac{1}{4}$

First two are straight forward, as they are independent I simply multiple the probabilities of $\frac{1}{2}$

$P(T_{HT}=3)=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}$

The sequence can start with both $H$ and $T$. Therefore we have the union of $T_{HHT}$ and $T_{THT}$. As $T$ and $H$ cannot appear at the same toss, they are disjoint, therefore we add the probabilities together. The probability is calculated below as 1 over 8.

$P(T_{HH}=3)=\frac{1}{2}^3=\frac{1}{8}$

This is equivalent to getting $P(T_{THH})$. So we simple multiple each probability.

Is this correctly understood or am I missing some information as why it is possible, besides the tosses being independent and fair?

Sirmimer
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    Why is $P(T_{HT}=3)=\frac{1}{4}$? Did you take into account the sequences of length 3 $HTH$ and $HTT$? They both have $HT$ at position 2. So maybe it's less than $\frac{1}{4}$ – Joao Noch Sep 14 '17 at 08:24
  • Maybe of interest https://math.stackexchange.com/questions/112726/coin-tossed-until-two-consecutive-heads-or-tails-appear?rq=1 – cgiovanardi Sep 15 '17 at 00:27
  • Thanks for the responses. I'll update this post ones I've figured it out – Sirmimer Sep 18 '17 at 19:33
  • I found the correct solution. So i'll see if i can figure out how to mark this as solved. Thanks for the help! – Sirmimer Sep 19 '17 at 12:40
  • In order to mark it as solved, you can answer your own question. – spaceisdarkgreen Sep 19 '17 at 13:01

1 Answers1

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solved (see comments and post)

Sirmimer
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