Players and coins Three players A,B and C flip simultaneously a coin. The coin of A (B,C) gives head with probability $a$ ($b,c$), with $0<a,b,c,<1$. If two of three coins give the same result, the player who flip the third coin is tossed out of play; if the coins are all equal, players flip again the coins.

*

*What is the probability that A is the first player tossed out?


*What is the value of the probability of 1) if $a=b=c$? Could you answer without calculations?


*If $a=b=c$, what is the mean number of games to finish the game?

I'm stuck. Could you give me any hints? Thanks in advance.
 A: Letting $t$ be the probability that a given round ends in a tie, we get
$$
t=abc+(1-a)(1-b)(1-c)=1-a-b-c+ab+bc+ca
$$
Explanation:

*

*$abc$ is the probability that in a given round, all players get heads.$\\[4pt]$

*$(1-a)(1-b)(1-c)$ is the probability that in a given round, all players get tails.

Letting $p$ be the probability that the game ends with $A$ tossed out, we get
$$
p=a(1-b)(1-c)+(1-a)bc+tp
$$
Explanation:

*

*$a(1-b)(1-c)$ is the probability that in the first round, $A$ gets heads and $B,C$ both get tails.$\\[4pt]$

*$(1-a)bc$ is the probability that in the first round, $A$ gets tails and $B,C$ both get heads.$\\[4pt]$

*$tp$ is the probability that

*

*$\;$The first round ends in a tie.$\\[5pt]$

*$\;A$ gets tossed out in the game starting with round $2$ (effectively a new game).


Solving for $p$, we get
$$
p=\frac{a(1-b)(1-c)+(1-a)bc}{1-t}
$$
which simplifies to
$$
p=\frac
{a-ab+bc-ca}
{a+b+c-ab-bc-ca}
$$
By symmetry, if $a=b=c$, we should get $p={\large{\frac{1}{3}}}$, which can be checked against the above formula.

Letting $e$ be the expected number of rounds for a complete game, we get
$$
e=(1-t)(1)+t(1+e)
$$
Explanation:

*

*$(1-t)(1)$ is the probability that game ends in the first round, multiplied by $1$, since the number of rounds is $1$.$\\[4pt]$

*$t(1+e)$ is the probability the first round ends in a tie, multiplied by $1+e$, since one round has already expired, and $e$ is the expected number of rounds to follow (in what is effectively a new game).  

Solving for $e$, we get
$$
e
=
\frac{1}{1-t}
=
\frac{1}{a+b+c-ab-bc-ca}
$$
and for the case $a=b=c$, the above formula yields
$$
e=\frac{1}{3a(1-a)}
$$
