3.82 - A First Course in Probability

Question

A and B flip coins. A starts and continues flipping

until a tail occurs, at which point B starts flipping and continues

until there is a tail. Then A takes over, and so on.

Let P1 be the probability of the coin landing on heads

when A flips and P2 when B flips. The winner of the game

is the first one to get

(a) 2 heads in a row;

(b) a total of 2 heads;

(c) 3 heads in a row;

(d) a total of 3 heads.

In each case, find the probability that A wins

Step-by-Step Solution

Answer

(a). $P\left(A\right)=\frac{{1^1}^2}{1-\left(1-P_1^2\right)\left(1-P_2^2\right)}$

(b).$P\left(A\right)={\left[1-\frac{{\displaystyle P_1}}{{\displaystyle 1-\left(1-P_1\right)\left(1-P_2\right)}}\right]}^2\times \left(\left(1-2(1-P_2\right)\left[1-\frac{{\displaystyle P_1}}{{\displaystyle 1-\left(1-P_1\right)\left(1-P_2\right)}}\right]\right)$

(c). $P\left(A\right)=\frac{{P_1}^3}{P_1{}^3+P_2{}^3-P_1{{}^{3}P_1}^3}$

(d). 

$$\begin{gathered} P\left(A wins\right)={\left[\frac{P_1}{1-\left(1-P_1\right)\left(1-P_2\right)}\right]}^3\times 3(1-P_2)\left(1-\frac{{\displaystyle P_1}}{{\displaystyle 1-\left(1-P_1\right)\left(1-P_2\right)}}\right) \\ \left[1+(1-P_2)\left(1-\frac{{\displaystyle P_1}}{{\displaystyle 1-\left(1-P_1\right)\left(1-P_2\right)}}\right)+\left(\frac{{\displaystyle P_2}}{{\displaystyle 1-\left(1-P_1\right)\left(1-P_2\right)}}\right)\right] \end{gathered}$$

Step 1 :Calculation (Part a)

Events:

$H_A$ - Toss by A results in Heads

$H_B$- Toss by B results in Heads

$W_A$ - A wins in the fixed turn

$W_B$ - B wins in the fixed turn

Chances : 

               $$\begin{gathered} P\left(H_A\right)=P_1 \\ P\left(H_B\right)=P_2 \end{gathered}$$

It is given that the winning condition is when the outgrowth  is 2 heads in a row. Thereby, $A$ will win in the below cases:

When the both  rivals fail i times, where $i=0,1,2,\dots$, and  achieving winning condition.

Probability $={\left[P\left(W_A^C\right)P\left(W_B^C\right)\right]}^{k}P\left(W_A\right)$

It is known that that the events ${\mathrm{W}}_{\mathrm{A}}$and ${\mathrm{W}}_{\mathrm{B}}$ are independent and they could be mutually exclusive if all i-s are different, thus, the probability of their union event $A$ is equal to sum of their individual probsbilities:

$P\left(A\right)=\sum _{k=0}^{\infty }{\left[P\left(W_A^C\right)P\left(W_B^C\right)\right]}^{k}P\left(W_A\right)$

thus, the needed sum is the sum of the geometric progression with the formuls

$\sum _{k=0}^{\infty }{q}^k=\frac{1}{1-q}$

Thus,

$P\left(A\right)=\frac{1}{1-\left[P\left(W_A^C\right)P\left(W_a^C\right)\mid \right.}P\left(W_A\right)\text{ Referred as (1) }$

The winning condition of two heads in a row can be written as

          $$\begin{gathered} P\left(W_A\right)=P\left(H_A\right)P\left(H_A\right) \\ P\left(W_B\right)=P\left(H_B\right)P\left(H_B\right) \end{gathered}$$

Thus,

While substituting it into equation (1), it can be calculated:

$P\left(A\right)=\frac{{1^1}^2}{1-\left(1-P_1^2\right)\left(1-P_2^2\right)}$

Step 2 : Calculation (Part b)

Consider the case when the coin has been flipped three time, also there can be only one winner. $A$ palm in the script when the $B$ flipped only original head or the alternate   head or when the $A$ gets the first two heads. These  scripts  can be written as $B A A, A B A, A A$

It can be determined that these situations are mutually exclusive so:

$P\left(A\right)=P\left("BA{A}^{\text{'}\text{'}}\right)+P\left("AB{A}^{\text{'}\text{'}}\right)+P("AA")$

Case "BAA"

When this event occurs, $B$ will be the first one to flip a head which is analogous  to A losing if the winning game is turning a head, i.e. the formula from the morning  is

$\frac{{\mathrm{P}}_1}{1-\left(1-{\mathrm{P}}_1\right)\left(1-{\mathrm{\mathbb{P}}}_1\right)}$

The second and third flip must be tails by$B$ with probability equal to I -  $P_2$

In the alternate  chances, Aflips head with the same probability I -  $P_2$ as of winning a game of flipping head one time. After that $A$ has to flip head again also with the same probability.

Thus,

$P\left("BA{A}^{\text{'}\text{'}}\right)=\left[1-\frac{P_1}{1-\left(1-P_1\right)\left(1-P_2\right)}\right]\times \left(1-P_2\right)\times {\left[1-\frac{P_1}{1-\left(1-P_1\right)\left(1-P_2\right)}\right]}^2$

By applying the same sense 

$$\begin{gathered} P\left("AB{A}^{\text{'}\text{'}}\right)=\left[1-\frac{P_1}{1-\left(1-P_1\right)\left(1-P_2\right)}\right]\times \left[1-\frac{P_1}{1-\left(1-P_1\right)\times 1-P_2}\right]{\times }^{}\left(1-P_2\right)\times \left[1-\frac{{\displaystyle P_1}}{{\displaystyle 1-\left(1-P_1\right)\left(1-P_2\right)}}\right] \\ P\left(A\right)={\left[1-\frac{{\displaystyle P_1}}{{\displaystyle 1-\left(1-P_1\right)\left(1-P_2\right)}}\right]}^2\times \left(\left(1-2(1-P_2\right)\left[1-\frac{{\displaystyle P_1}}{{\displaystyle 1-\left(1-P_1\right)\left(1-P_2\right)}}\right]\right) \end{gathered}$$

Step 3 : Calculation(Part c)

The winning condition of three heads in a row has probability equal to

     $$\begin{gathered} P\left(W_A\right)=P\left(H_A\right)P\left(H_A\right)P\left(H_A\right)={P_1}^3 \\ P\left(W_B\right)=P\left(H_B\right)P\left(H_B\right)P\left(H_B\right)={P_2}^3 \end{gathered}$$

By substituting into formuls $P\left(A\right)=\frac{1}{1-\left[P\right(wF\left)P\right(WF)\mid }P\left(W_A\right)$, the  affair  is

$P\left(A\right)=\left[1-\frac{{P_1}^3}{1-\left(1-{P_1}^3\left)\right(1-{P_2}^3\right)}\right]=\frac{{P_1}^3}{P_1{}^3+P_2{}^3-P_1{{}^{3}P_1}^3}$

Step 4 : Calculation (Part d)

The probability of the outgrowth of a aggregate  of 3 heads can be written as:
$$\begin{gathered} P\left(A\right)=P(" AAA ")+P(" AABA")+P(" ABAA") \\ +P("BAAA")+P("BBAAA")+P("BABAA") \\ +P("BAABA")+P("ABBAA")+P("ABABA") \\ +P("AABBA") \end{gathered}$$

By using the sense  defined in part b), The $2^{\text{nd }},3^{\text{rd }},4^{\text{th }}$ term on right hand side of above equation have same probability equal to

${\left[1-\frac{P_1}{1-\left(1-P_1\right)\left(1-P_2\right)}\right]}^3\times \left[1-\frac{P_1}{1-\left(1-P_1\right)\left(1-P_2\right)}\right]\times \left(1-P_2\right)$

 Then it's veritably  important to understand the sequence in which player tosses head. This is why, if $B$ tosses only 1 head in a row, also the coming toss by $B$should be tails and the probability of it is $\left(1-P_2\right)$ time more.

Thus,

$P\left("ABAB{A}^{\text{'}\text{'}}\right)+P("BABAA")+P("BAABA")=3P("ABABA")$

Also,

$P("BRAAA)+P("ABBAA")+P("AABBA")=3P("BBAAA")$

By adding all these values

$$\begin{gathered} P\left(A wins\right)={\left[\frac{P_1}{1-\left(1-P_1\right)\left(1-P_2\right)}\right]}^3\times 3(1-P_2)\left(1-\frac{{\displaystyle P_1}}{{\displaystyle 1-\left(1-P_1\right)\left(1-P_2\right)}}\right) \\ \left[1+(1-P_2)\left(1-\frac{{\displaystyle P_1}}{{\displaystyle 1-\left(1-P_1\right)\left(1-P_2\right)}}\right)+\left(\frac{{\displaystyle P_2}}{{\displaystyle 1-\left(1-P_1\right)\left(1-P_2\right)}}\right)\right] \end{gathered}$$