Counting Technique: Probability of only $3$ hearts in a row when $5$ cards are drawn from a standard deck 
Problem: Draw $5$ cards from a pack of $52$ cards. Probability of only $3$ hearts in a row ?
  I am now struggling with this problem. I think that if we draw $5$ cards we will have $3$ cases which contain $3$ hearts is a row:
  \begin{align}
HHHAA
\end{align}
\begin{align}
AHHHA
\end{align}
\begin{align}
AAHHH
\end{align}
\begin{align}
H - \text{heart}, A-\text{any}
\end{align} 
  So I think the probability of $3$ hearts in a row can be calculated:
  \begin{align}
P(3 \text{ hearts in a row})=P(HHHAA)+P(AHHHA)+P(AAHHH)
\end{align} 
  However, I have no idea how to calculate the probability of each case. At first I think that the probability of the case in which first three draws are heart can be calculated:
  \begin{align}
P(HHHAA)=\frac{13*12*11*13*12}{52*51*50*49*48}
\end{align} 
  But I realized that this actually the probability of getting $3$ hearts after drawing $5$ cards. After thinking a lot i still don't figure out the solution for the problem, I have thought that I can list all the events that can appear on each case but here the number of events is too large for listing and counting.
  Thanks a lot for reading and helping me! I would appreciate a lot if you guys can show me some tips or some topic to read on consecutive problems

 A: If the question requires that there are only three hearts in a row (but it's not allowed to have four or five of them in a row), then your calculation is incorrect.
If "A" stands for anything, i.e. any card from the deck, then "HHHAA" is not what you want to count — because those two "A" may be hearts as well. For example, drawing all five hearts is also included in your count here (first three hearts, then another any card which just happens to be hearts, and then another one like that), even though we don't want to include this case.
So the problem is actually with your designation of "A" as being "any card". You want to clearly distinguish between hearts and non-hearts, in order to be able to count correctly. How about you use "A" to represent "any card that is not hearts"? Then you can state all the cases that you need to account for, without including undesired possibilities and without overlaps that lead to overcounting. There will be five cases total: the three that you listed and two more (such as "HHHAH"). Then you can count their probabilities by the same multiplication rule, and then add then together.
A: Your $A$ should not be any, they should be any non-heart.  Then the chance of drawing $HHHAA$ is $\frac {13\cdot 12 \cdot 11 \cdot 39 \cdot  38}{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48}$ because at each stage we have multiplied by the number of acceptable cards and divided by the number of available cards.  Similarly, the chance of $AHHHA$ is $\frac {39\cdot 13 \cdot 12 \cdot 11 \cdot  38}{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48}$, which you can see is equal.  The chance of $AAHHH$ is again the same, so the total chance is $3\cdot\frac {13\cdot 12 \cdot 11 \cdot 39 \cdot  38}{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48}$ as the possibilities are disjoint.
