Probability of getting heads in a coin toss A fair coin is tossed five times. What is the probability of getting a sequence of
three heads?
What i tried
Since the probability of getting a head and not getting a head is $0.5$. The probability of getting three heads is $0.5^3$ while there is a $0.5^2$ chance of not getting a head. But since the  three heads are in sequence out of five toss, the total probability  is a multiplication of  $ \left(  
\begin{array}{c}
   5 \\
    3 \\
  \end{array}
\right).0.5^3.0.5^2=5/16$ 
Is my working correct. Could anyone explain. Thanks
 A: Not quite right. You need a sequence of three heads, not just any three heads. You've counted HHTHT as a valid coin-toss-sequence, whereas according to the question it shouldn't be.
The only allowable sequences are HHHTT, THHHT, TTHHH, HHHHT, THHHH, HTHHH, HHHTH, HHHHH. 
A: Hint: 
${5\choose 3} =10$ is the number of sequences where you have 3 heads and 2 tails. The heads have not to occur consecutive. The number of sequences with exact 3 heads consecutive are  
$hhhtt, \ thhht, tthhh$
Therefore you have only 3 arrangements. Each arrangement has a probability of $0.5^5$
Remark:
If you can have more then 3 consecutive heads, then you have 6 arrangements.
A: If the probability of getting a head is $p$, and the probability of not getting a head is $q=1-p$, then the probability of getting 3 or more consecutive heads is 
$$\begin{align}
\underbrace{3 p^3 q^2}_{P(HHH)}+\underbrace{2p^4q^1}_{P(HHHH)}+\underbrace{p^5q^0}_{P(HHHHH)}&=p^3(3q^2+2pq+p^2)\\
&=6p^5\qquad\text{(as }q=p\text{)}\\
&=\frac 3{16}\qquad\text{(}p=\frac 12\text{)}
\end{align}$$
This method can also be used if $p\neq\frac12$.
A: You know that there are exactly $2^5 = 32$ unique outcomes in the game. Now in your "win" case, you must have exactly a sequence of three heads, so let's represent this $HHH$ sequence with a variable $W$. Now, you can simplify your question to: how many unique arrangements of $WTT$ are there? We must have $3! \over 1!2!$, or 3 unique arrangements. 
Each unique arrangement has probability $P(H)^{N(H)}P(T)^{N(T)}$, where $N$ is the number of objects in a group and $P$ is the probability function. Since every one of our three arrangements must have three heads and two tails, and since the probability of getting heads/tails is 0.5, then each arrangement has probability $0.5^30.5^2 = {1 \over 32}$. So the solution is $3({1 \over 32}) = {3 \over 32}$ .
Note: the number of unique arrangements of a group with common elements is
$$N(total)! \over N(a_1)!N(a_2)!N(a_3)!\cdots$$
Where each $a_i$ is a unique element. In our case, we have only 2 unique elements's: $W$, and $T$, so $N(W) = 1$, $N(T) = 2$, and $N(total) = 3$. Of course, this strategy fails when two elements $a_i$ and $a_j$ have common elements, so watch out for that when using this strategy.
A: This is just me, but I'd take a direct approach and (since you only have 5 events to consider) draw out a table or chart of all the possible outcomes for first only 2 or 3 events, see if you can scale up, then check your calculation against that. 
I find when people aren't inwardly certain in a solid way about the algorithm, a bit of visual intuition can really help. 
