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.