Probability of HHTH possibly switching between normal and weighted coin after each flip I'm having trouble solving this question: Your friend has two coins: one is a normal coin which lands heads $50\%$ of the time, the other is a weighted coin which lands heads $75\%$ of the time.
When your friend starts flipping his coins, he’s equally likely to start with either coin, and switches coins with $30\%$ probability.
Find $P(HHTH)$
My solution:
$P(H)P(H)P(T)P(H)=(0.5(0.5+.75))(0.3(0.5+0.75))(0.3(0.5+.25))(0.3(0.5+.75))=0.0198$
At first it is equally likely to start with either coin, so I am using $0.5$.
But, then we switch coins with $30\%$ probability, so I am using $0.3$.
I think, if we don't switch it will be $70\%$ probability, but I am not sure how to use this info in formula. Please help.
 A: If we start with the balanced coin, then the probability of getting a head at any subsequent stage is 
$$
P(H)=0.7\cdot 0.5+0.3\cdot 0.75=0.575
$$
and we can model you situation with a single weighted coin with $P(H)=0.575$ and $P(T)=0.425$. Thus
$$
P(HHTH)=0.5(0.575)^2(0.425)\approx 0.0703.
$$
If we start with the unbalanced coin, then the probability of getting a head at any subsequent stage is 
$$
P(H)=0.7\cdot 0.75+0.3\cdot 0.5=0.675
$$
and we can model you situation with a single weighted coin with $P(H)=0.675$ and $P(T)=0.325$. Thus
$$
P(HHTH)=0.75(0.675)^2(0.325)\approx 0.1111.
$$
Finally since we are equally likely to start with either coin, the total probability is
$$
P(HHTH)\approx 0.5\cdot 0.0703 + 0.5\cdot 0.1111 = 0.0907.
$$
A: Let $\{h_i\}_{i=1}^4$ be the probability of heads in the 4 tosses. The probability of switching coins between tosses is $s = \dfrac{1}{3}$.
$$ \mathbb{P}(\textrm{HHTH}) 
= \sum_{h_1,h_2,h_3,h_4 \in \{\frac{1}{2},\frac{3}{4}\}}\mathbb{P}(\textrm{HHTH}|h_1,h_2,h_3,h_4)\; \mathbb{P}(h_1,h_2,h_3,h_4) $$
The coin chosen for a certain toss is statistically independent of the all the tosses expect the preceding one. This means we can simplify $\mathbb{P}(h_1,h_2,h_3,h_4)  \equiv \mathbb{P}(h_4|h_3,h_2,h_1) \mathbb{P}(h_3|h_2,h_1) \mathbb{P}(h_2|h_1) \mathbb{P}(h_1) $ to  $\mathbb{P}(h_4|h_3) \mathbb{P}(h_3|h_2) \mathbb{P}(h_2|h_1) \mathbb{P}(h_1)$. If the probability of switching is $s$, then $\mathbb{P}(h_{k+1}|h_k) = s $ if there was a switching, i.e., $h_{k+1} \neq h_{k}$, and $ = 1-s $ if there was no switching, i.e., $h_{k+1} = h_{k}$. In other words, $\mathbb{P}(h_{k+1}|h_k)  = s \mathbb{1}_{h_{k+1} \neq h_{k}} + (1-s) \mathbb{1}_{h_{k+1} = h_{k}} = s (1-\mathbb{1}_{h_{k+1} = h_{k}}) + (1-s) \mathbb{1}_{h_{k+1} = h_{k}} = s + (1-2s)\mathbb{1}_{h_{k+1} = h_{k}}$.
\begin{align}
\mathbb{P}(h_1,h_2,h_3,h_4) 
&= \mathbb{P}(h_4|h_3)\;\mathbb{P}(h_3|h_2)\;\mathbb{P}(h_2|h_1)\;\mathbb{P}(h_1) \\
&= [s+ (1-2s)\mathbb{1}_{h_4=h_3}][s+ (1-2s)\mathbb{1}_{h_3=h_2}][s+ (1-2s)\mathbb{1}_{h_2=h_1}]\frac{1}{2}\\
&= \frac{1}{54}[1+ \mathbb{1}_{h_4=h_3}][1+ \mathbb{1}_{h_3=h_2}][1+ \mathbb{1}_{h_2=h_1}]\\
\mathbb{P}(\textrm{HHTH}) 
&= \frac{1}{54} \sum_{h_1,h_2,h_3,h_4 \in \{\frac{1}{2},\frac{3}{4}\}} h_1h_2(1-h_3)h_4\;[1+ \mathbb{1}_{h_4=h_3}][1+ \mathbb{1}_{h_3=h_2}][1+ \mathbb{1}_{h_2=h_1}]\\
&=\frac{35}{432} \approx 0.08 
\end{align}
