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HH

Following the method outlined in this answer on Math.SEthis answer on Math.SE, we get the following:

Let $e$ be the expected number of tosses.

Start tossing. If we get a tail immediately (probability $\tfrac{1}{2}$) then the expected number is $e+1$. If we get a head then a tail (probability $\tfrac{1}{4}$), then the expected number is $e+2$. Finally, if our first $2$ tosses are heads (again with probability $\tfrac{1}{4}$), then the expected number is $2$. Thus $$e=\frac{1}{2}\cdot(e+1)+\frac{1}{4}\cdot(e+2)+\frac{1}{4}\cdot(2)$$ Solving this for $e$ yields $e=6$.

HT

Again, let $e$ be the expected number of tosses and start tossing.

If we get a tail immediately (probability $\tfrac{1}{2}$) then the expected number is $e+1$. If we get a head followed by another head (probability $\tfrac{1}{4}$), then the expected number is $e$, since only the first head can't be part of the sequence and were already $1$ deep into the sequence. Finally, if our first $2$ tosses are exactly heads followed by tails (again with probability $\tfrac{1}{4}$), then the expected number is $2$. Thus $$e=\frac{1}{2}\cdot(e+1)+\frac{1}{4}\cdot(e)+\frac{1}{4}\cdot(2)$$ Solving this for $e$ yields $e=4$.

HH

Following the method outlined in this answer on Math.SE, we get the following:

Let $e$ be the expected number of tosses.

Start tossing. If we get a tail immediately (probability $\tfrac{1}{2}$) then the expected number is $e+1$. If we get a head then a tail (probability $\tfrac{1}{4}$), then the expected number is $e+2$. Finally, if our first $2$ tosses are heads (again with probability $\tfrac{1}{4}$), then the expected number is $2$. Thus $$e=\frac{1}{2}\cdot(e+1)+\frac{1}{4}\cdot(e+2)+\frac{1}{4}\cdot(2)$$ Solving this for $e$ yields $e=6$.

HT

Again, let $e$ be the expected number of tosses and start tossing.

If we get a tail immediately (probability $\tfrac{1}{2}$) then the expected number is $e+1$. If we get a head followed by another head (probability $\tfrac{1}{4}$), then the expected number is $e$, since only the first head can't be part of the sequence and were already $1$ deep into the sequence. Finally, if our first $2$ tosses are exactly heads followed by tails (again with probability $\tfrac{1}{4}$), then the expected number is $2$. Thus $$e=\frac{1}{2}\cdot(e+1)+\frac{1}{4}\cdot(e)+\frac{1}{4}\cdot(2)$$ Solving this for $e$ yields $e=4$.

HH

Following the method outlined in this answer on Math.SE, we get the following:

Let $e$ be the expected number of tosses.

Start tossing. If we get a tail immediately (probability $\tfrac{1}{2}$) then the expected number is $e+1$. If we get a head then a tail (probability $\tfrac{1}{4}$), then the expected number is $e+2$. Finally, if our first $2$ tosses are heads (again with probability $\tfrac{1}{4}$), then the expected number is $2$. Thus $$e=\frac{1}{2}\cdot(e+1)+\frac{1}{4}\cdot(e+2)+\frac{1}{4}\cdot(2)$$ Solving this for $e$ yields $e=6$.

HT

Again, let $e$ be the expected number of tosses and start tossing.

If we get a tail immediately (probability $\tfrac{1}{2}$) then the expected number is $e+1$. If we get a head followed by another head (probability $\tfrac{1}{4}$), then the expected number is $e$, since only the first head can't be part of the sequence and were already $1$ deep into the sequence. Finally, if our first $2$ tosses are exactly heads followed by tails (again with probability $\tfrac{1}{4}$), then the expected number is $2$. Thus $$e=\frac{1}{2}\cdot(e+1)+\frac{1}{4}\cdot(e)+\frac{1}{4}\cdot(2)$$ Solving this for $e$ yields $e=4$.

Corrected. It is 4.
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SQB
SQB
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HH

Following the method outlined in this answer on Math.SE, we get the following:

Let $e$ be the expected number of tosses.

Start tossing. If we get a tail immediately (probability $\tfrac{1}{2}$) then the expected number is $e+1$. If we get a head then a tail (probability $\tfrac{1}{4}$), then the expected number is $e+2$. Finally, if our first $2$ tosses are heads (again with probability $\tfrac{1}{4}$), then the expected number is $2$. Thus $$e=\frac{1}{2}\cdot(e+1)+\frac{1}{4}\cdot(e+2)+\frac{1}{4}\cdot(2)$$ Solving this for $e$ yields $e=6$.

HT

Again, let $e$ be the expected number of tosses and start tossing.

If we get a tail immediately (probability $\tfrac{1}{2}$) then the expected number is $e+1$. If we get a head followed by another head (probability $\tfrac{1}{4}$), then the expected number is $e+1$$e$, since only the first head can't be part of the sequence and were already $1$ deep into the sequence. Finally, if our first $2$ tosses are exactly heads followed by tails (again with probability $\tfrac{1}{4}$), then the expected number is $2$. Thus $$e=\frac{1}{2}\cdot(e+1)+\frac{1}{4}\cdot(e+1)+\frac{1}{4}\cdot(2)$$$$e=\frac{1}{2}\cdot(e+1)+\frac{1}{4}\cdot(e)+\frac{1}{4}\cdot(2)$$ Solving this for $e$ yields $e=5$$e=4$.

HH

Following the method outlined in this answer on Math.SE, we get the following:

Let $e$ be the expected number of tosses.

Start tossing. If we get a tail immediately (probability $\tfrac{1}{2}$) then the expected number is $e+1$. If we get a head then a tail (probability $\tfrac{1}{4}$), then the expected number is $e+2$. Finally, if our first $2$ tosses are heads (again with probability $\tfrac{1}{4}$), then the expected number is $2$. Thus $$e=\frac{1}{2}\cdot(e+1)+\frac{1}{4}\cdot(e+2)+\frac{1}{4}\cdot(2)$$ Solving this for $e$ yields $e=6$.

HT

Again, let $e$ be the expected number of tosses and start tossing.

If we get a tail immediately (probability $\tfrac{1}{2}$) then the expected number is $e+1$. If we get a head followed by another head (probability $\tfrac{1}{4}$), then the expected number is $e+1$, since only the first head can't be part of the sequence. Finally, if our first $2$ tosses are exactly heads followed by tails (again with probability $\tfrac{1}{4}$), then the expected number is $2$. Thus $$e=\frac{1}{2}\cdot(e+1)+\frac{1}{4}\cdot(e+1)+\frac{1}{4}\cdot(2)$$ Solving this for $e$ yields $e=5$.

HH

Following the method outlined in this answer on Math.SE, we get the following:

Let $e$ be the expected number of tosses.

Start tossing. If we get a tail immediately (probability $\tfrac{1}{2}$) then the expected number is $e+1$. If we get a head then a tail (probability $\tfrac{1}{4}$), then the expected number is $e+2$. Finally, if our first $2$ tosses are heads (again with probability $\tfrac{1}{4}$), then the expected number is $2$. Thus $$e=\frac{1}{2}\cdot(e+1)+\frac{1}{4}\cdot(e+2)+\frac{1}{4}\cdot(2)$$ Solving this for $e$ yields $e=6$.

HT

Again, let $e$ be the expected number of tosses and start tossing.

If we get a tail immediately (probability $\tfrac{1}{2}$) then the expected number is $e+1$. If we get a head followed by another head (probability $\tfrac{1}{4}$), then the expected number is $e$, since only the first head can't be part of the sequence and were already $1$ deep into the sequence. Finally, if our first $2$ tosses are exactly heads followed by tails (again with probability $\tfrac{1}{4}$), then the expected number is $2$. Thus $$e=\frac{1}{2}\cdot(e+1)+\frac{1}{4}\cdot(e)+\frac{1}{4}\cdot(2)$$ Solving this for $e$ yields $e=4$.

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SQB
SQB
  • 5.2k
  • 32
  • 58

HH

Following the method outlined in this answer on Math.SE, we get the following:

Let $e$ be the expected number of tosses.

Start tossing. If we get a tail immediately (probability $\tfrac{1}{2}$) then the expected number is $e+1$. If we get a head then a tail (probability $\tfrac{1}{4}$), then the expected number is $e+2$. Finally, if our first $2$ tosses are heads (again with probability $\tfrac{1}{4}$), then the expected number is $2$. Thus $$e=\frac{1}{2}\cdot(e+1)+\frac{1}{4}\cdot(e+2)+\frac{1}{4}\cdot(2)$$ Solving this for $e$ yields $e=6$.

HT

Again, let $e$ be the expected number of tosses and start tossing.

If we get a tail immediately (probability $\tfrac{1}{2}$) then the expected number is $e+1$. If we get a head followed by another head (probability $\tfrac{1}{4}$), then the expected number is $e+1$, since only the first head can't be part of the sequence. Finally, if our first $2$ tosses are exactly heads followed by tails (again with probability $\tfrac{1}{4}$), then the expected number is $2$. Thus $$e=\frac{1}{2}\cdot(e+1)+\frac{1}{4}\cdot(e+1)+\frac{1}{4}\cdot(2)$$ Solving this for $e$ yields $e=5$.