Question

A player tosses a coin and is to score one point for every head turned up and two for every tail. He is to play on until his score reaches or passes $n$. If $p_n$ is the chance for attaining exactly $n$, show that $p_n=\frac{1}{2}(p_{n-1}+p_{n-2})$ and hence find the value of $p_n$

• A. $p_n=\frac{1}{3}\times \{2+\frac{1}{2^n}\}$
• B. $p_n=\frac{1}{3}\{2+{(-1)}^n\frac{1}{2^n}\}$
• C. $p_n=\frac{2}{3}\{2+{(-1)}^n\frac{1}{2^n}\}$
• D. $p_n=\frac{2}{3}\times \{2+\frac{1}{2^n}\}$

Answer 1

The correct option is B $p_n=\frac{1}{3}\{2+{(-1)}^n\frac{1}{2^n}\}$

N points can be scored in the following ways.
Either with n number of heads
$\to \frac{1}{2^n}$
Or
1 tails and (n-2) number of heads
That is
$T,H,H...(n-2)times$ ...total n-1 tosses
Internally this could be permuted in $\frac{n-1!}{(n-2)!.1!}$
$={}^{n-1}{C}_1$ ways.
Required probability
$={}^{n-1}{C}_1.\frac{1}{2^{n-1}}$
Or
2 tails and (n-4) number of heads
That is
$T,T,H,H...(n-4)times$ ...total n-2 tosses
Internally this could be permuted in $\frac{n-2!}{(n-4)!.2!}$
$={}^{n-2}{C}_2$ ways.
Required probability
$={}^{n-2}{C}_2.\frac{1}{2^{n-2}}$ and so on
Hence
$P_n$
$=\frac{1}{2^n}+{}^{n-1}{C}_1\frac{1}{2^{n-1}}+{}^{n-2}{C}_2\frac{1}{2^{n-2}}+....$
$=\frac{1}{2^n}[1+{}^{n-1}{C}_1.2+{}^{n-2}{C}_2{2}^2+{}^{n-3}{C}_3{2}^3+...]$

Replacing $n$ by $n-1$, we get
$P_{n-1}$
$=\frac{1}{2^{n-1}}[1+{}^{n-2}{C}_1.2+{}^{n-3}{C}_2{2}^2+{}^{n-4}{C}_3{2}^3+...]$

$=\frac{1}{2^n}[2+{}^{n-2}{C}_1{.2}^2+{}^{n-3}{C}_2{2}^3+{}^{n-4}{C}_3{2}^4+...]$ ...(a)
Replacing $n$ by $n-2$ we get
$P_{n-2}$
$=\frac{1}{2^{n-2}}[1+{}^{n-3}{C}_1.2+{}^{n-4}{C}_2{2}^2+{}^{n-5}{C}_3{2}^3+...]$
$=\frac{1}{2^n}[4+{}^{n-3}{C}_1{.2}^3+{}^{n-4}{C}_2{2}^4+{}^{n-5}{C}_3{2}^5+...]$ ...(b)

adding a and b, we get
$P_{n-1}+P_{n-2}$
$=\frac{1}{2^n}[6+{}^{n-2}{C}_1{2}^2+2^3[{}^{n-3}{C}_1+{}^{n-3}{C}_2]+2^4[{}^{n-4}{C}_2+{}^{n-4}{C}_3]+...]$
$\frac{1}{2^n}[6+{}^{n-2}{C}_1{2}^2+2^3({}^{n-2}{C}_2)+2^4({}^{n-3}{C}_3)+...]$
$=\frac{2}{2^n}[1+(2+{}^{n-2}{C}_1{2}^1)+2^2({}^{n-2}{C}_2)+2^3({}^{n-3}{C}_3)+...]$
$=\frac{2}{2^n}[1+2({}^{n-2}{C}_0+{}^{n-2}{C}_1)+2^2({}^{n-2}{C}_2)+2^3({}^{n-3}{C}_3)+...]$
$=\frac{2}{2^n}[1+2({}^{n-1}{C}_1)+2^2({}^{n-2}{C}_2)+2^3({}^{n-3}{C}_3)+...]$
$=2P_n$
Hence
$P_{n-1}+P_{n-2}=2\left(P_n\right)$
Now
$P_n=\frac{1}{2^n}+{}^{n-1}{C}_1\frac{1}{2^{n-1}}+{}^{n-2}{C}_2\frac{1}{2^{n-2}}+....$

Substituting n=1, we get $P_1=\frac{1}{2}$
For n=2 $P_2=\frac{3}{4}$

For n=3 $P_3=\frac{5}{8}$

for n=4 $P_4=\frac{11}{16}$
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Hence $P_n=\frac{1}{3}[2+(-1{)}^n\left(\frac{1}{2^n}\right)]$