Question

A player \[X\] has a biased coin whose probability of showing heads is \[p\]and a player \[Y\] has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If \[X\] starts the game, and the probability of winning the game by both the players is equal, then the value of ‘\[p\]’ is
A. \[\dfrac{1}{3}\]
B. \[\dfrac{1}{5}\]
C. \[\dfrac{1}{4}\]
D. \[\dfrac{2}{5}\]

Answer 1

Hint: Here a biased coin has a higher probability of heads or tails. A fair coin is a mythical gadget which has probability exactly ½ of showing heads and 1/2of showing tails. First write the set of outcomes so that \[X\]wins, then find out the probability. Similarly, do the same to \[Y\]. Then equate the probabilities of\[X\]wins and \[Y\]wins. So, use this concept to reach the solution of the problem.

Complete step-by-step answer:

\[X\] wins, when the outcome is one of the following set of outcomes:
\[H,TTH,TTTTH,...........\]
Since subsequent tosses are independent, the probability that \[X\] wins is
\[ \Rightarrow P\left( H \right) + P\left( {TTH} \right) + P\left( {TTTTH} \right) + ..................\]
As the probability of showing heads is \[p\], we have
\[
   \Rightarrow p + \left( {1 - p} \right)\dfrac{1}{2}p + {\left( {1 - p} \right)^2}\dfrac{1}{{{2^2}}}p + ...................... \\
   \Rightarrow p\left[ {1 + \left( {\dfrac{{1 - p}}{2}} \right) + {{\left( {\dfrac{{1 - p}}{2}} \right)}^2} + .............} \right] \\
\]
As the terms are in infinity G.P the sum of the terms are given by \[{S_\infty } = \dfrac{a}{{1 - r}}\] where \[a\] is the first term and \[r\]is the common ratio of the infinity series.
\[
   \Rightarrow p\left[ {\dfrac{1}{{1 - \left( {\dfrac{{1 - p}}{2}} \right)}}} \right] \\
   \Rightarrow \dfrac{p}{{1 - \dfrac{{1 - p}}{2}}} = \dfrac{{2p}}{{1 + p}} \\
\]
So, the probability of \[X\]wins are \[\dfrac{{2p}}{{1 + p}}....................................\left( 1 \right)\]
Similarly, \[Y\] wins if the outcome is one of the following:
\[TH,TTTH,TTTTTH,....................\]
We know that the sum of probabilities of showing head and showing tail is equal to 1.
So, the probability of showing tile is \[1 - p\], we have
\[
   \Rightarrow P\left( H \right) + P\left( {TTH} \right) + P\left( {TTTTTH} \right) + ......................... \\
   \Rightarrow \dfrac{{1 - p}}{2} + \dfrac{{{{\left( {1 - p} \right)}^2}}}{4} + \dfrac{{{{\left( {1 - p} \right)}^3}}}{8} + ............................ \\
   \Rightarrow \left( {\dfrac{{1 - p}}{2}} \right)\left[ {1 + \dfrac{{1 - p}}{2} + {{\left( {\dfrac{{1 - p}}{2}} \right)}^2} + .........................} \right] \\
\]
As the terms are in infinity G.P the sum of the terms are given by \[{S_\infty } = \dfrac{a}{{1 - r}}\] where \[a\] is the first term and \[r\]is the common ratio of the infinity series.
\[
   \Rightarrow \left( {\dfrac{{1 - p}}{2}} \right)\left[ {\dfrac{1}{{1 - \left( {\dfrac{{1 - p}}{2}} \right)}}} \right] \\
   \Rightarrow \dfrac{{\dfrac{{1 - p}}{2}}}{{1 - \left( {\dfrac{{1 - p}}{2}} \right)}} = \dfrac{{1 - p}}{{1 + p}} \\
\]
So, the probability of \[Y\]wins are \[\dfrac{{1 - p}}{{1 + p}}.....................................\left( 2 \right)\]
We know that
Probability of \[X\]wins = Probability of \[Y\]wins
\[
   \Rightarrow \dfrac{{2p}}{{1 + p}} = \dfrac{{1 - p}}{{1 + p}} \\
   \Rightarrow 2p = 1 - p \\
   \Rightarrow 2p + p = 1 \\
   \Rightarrow 3p = 1 \\
  \therefore p = \dfrac{1}{3} \\
\]
Thus, the value of \[p\] is \[\dfrac{1}{3}\].

Note: The terms which are in infinity G.P the sum of the terms are given by \[{S_\infty } = \dfrac{a}{{1 - r}}\] where \[a\] is the first term and \[r\]is the common ratio of the infinity series. The sum of probabilities of showing head and showing tail is equal to one.