91Ó°ÊÓ

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 $i=0,1,2,…$, 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:

$P\left(A\right)=\underset{k=0}{\overset{∞}{∑}}{\left[P\left(W_A^C\right)P\left(W_B^C\right)\right]}^{k}P\left(W_A\right)$

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

$\underset{k=0}{\overset{∞}{∑}}{q}^k=\frac{1}{1-q}$

Thus,

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

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:

localid="1647467224059" $P\left(A\right)=\frac{{1^1}^2}{1-\left(1-P_1^2\right)\left(1-P_2^2\right)}$

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 $BAA,ABA,AA$

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

$\frac{{\mathrm{P}}_1}{1-\left(1-{\mathrm{P}}_1\right)\left(1-{\mathrm{â„™}}_1\right)}$

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,

localid="1646610756857" $P\left("BA{A}^{\text{'}\text{'}}\right)=\left[1-\frac{P_1}{1-\left(1-P_1\right)\left(1-P_2\right)}\right]×\left(1-P_2\right)×{\left[1-\frac{P_1}{1-\left(1-P_1\right)\left(1-P_2\right)}\right]}^2$

By applying the same sense

localid="1647467319367" $$\begin{gathered} P\left("AB{A}^{\text{'}\text{'}}\right)=\left[1-\frac{P_1}{1-\left(1-P_1\right)\left(1-P_2\right)}\right]×\left[1-\frac{P_1}{1-\left(1-P_1\right)×1-P_2}\right]{×}^{}\left(1-P_2\right)×\left[1-\frac{{\displaystyle P_1}}{{\displaystyle 1-\left(1-P_1\right)\left(1-P_2\right)}}\right] \\ P\left(A\right)={\left[1-\frac{{\displaystyle P_1}}{{\displaystyle 1-\left(1-P_1\right)\left(1-P_2\right)}}\right]}^2×\left(\left(1-2(1-P_2\right)\left[1-\frac{{\displaystyle P_1}}{{\displaystyle 1-\left(1-P_1\right)\left(1-P_2\right)}}\right]\right) \end{gathered}$$

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 $P\left(A\right)=\frac{1}{1-\left[P\right(wF\left)P\right(WF)âˆ\pounds }P\left(W_A\right)$, the affair is

localid="1647467347425" $P\left(A\right)=\left[1-\frac{{P_1}^3}{1-\left(1-{P_1}^3\left)\right(1-{P_2}^3\right)}\right]=\frac{{P_1}^3}{P_1{}^3+P_2{}^3-P_1{{}^{3}P_1}^3}$

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

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

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

localid="1647467442963" $$\begin{gathered} P\left(Awins\right)={\left[\frac{P_1}{1-\left(1-P_1\right)\left(1-P_2\right)}\right]}^3×3(1-P_2)\left(1-\frac{{\displaystyle P_1}}{{\displaystyle 1-\left(1-P_1\right)\left(1-P_2\right)}}\right) \\ \left[1+(1-P_2)\left(1-\frac{{\displaystyle P_1}}{{\displaystyle 1-\left(1-P_1\right)\left(1-P_2\right)}}\right)+\left(\frac{{\displaystyle P_2}}{{\displaystyle 1-\left(1-P_1\right)\left(1-P_2\right)}}\right)\right] \end{gathered}$$