Question

Three players simultaneously toss coins. The coin tossed by $A\left(B\right)\left[C\right]$turns up heads with probability $P_1\left(P_2\right)\left[P_3\right]$. If one person gets an outcome different from those of the other two, then he is the odd man out. If there is no odd man out, the players flip again and continue to do so until they get an odd man out. What is the probability that $A$ will be the odd man ?

Short answer

The probability of$P(A\text{is the odd man out}\mid \text{someone is the odd man out})=\frac{P_1\left(1-P_2\right)\left(1-P_3\right)+\left(1-P_1\right)P_2{P}_3}{P_1{P}_2{P}_3+\left(1-P_1\right)\left(1-P_2\right)\left(1-P_3\right)}$.

Step-by-step solution

Given Information

Three players toss coins with probabilities $P_1,P_2$and $P_3$respectively for $A,B$and $C$ to get heads.

Explanation 

Event in question is:

$P\left(A\text{gets different result from}B\text{and}C\right)=P_1\left(1-P_2\right)\left(1-P_3\right)+\left(1-P_1\right)P_2{P}_3$

And that is the probability that $A$is excluded in a single set of tosses. But we know that one of the players will be the odd man out, so this probability is the one we are looking for:

$P(A\text{is the odd man out}\mid \text{someone is the odd man out})$

$=\frac{P\left(A\text{is the odd man out, and someone is the odd man out}\right)}{P\left(\text{someone is the odd man out}\right)}$

Explanation

Nominator:

$P\left(A\text{is the odd man out, and someone is the odd man out}\right)$

$=P\left(A\text{is the odd man out}\right)$

$=P\left(A\text{gets different result from}B\text{and}C\right)$

$=P_1\left(1-P_2\right)\left(1-P_3\right)+\left(1-P_1\right)P_2{P}_3$

$=P_1-P_1{P}_3-P_1{P}_2+P_2{P}_3$

Explanation

Denominator:

$P\left(\text{someone is the odd man out}\right)=1-P\left(\text{no one is the odd man out}\right)$

$=1-P\left(\text{all get the same result}\right)$

$=P_1{P}_2{P}_3+\left(1-P_1\right)\left(1-P_2\right)\left(1-P_3\right)$

$P(A\text{is the odd man out}\mid \text{someone is the odd man out})$

$=\frac{P_1\left(1-P_2\right)\left(1-P_3\right)+\left(1-P_1\right)P_2{P}_3}{P_1{P}_2{P}_3+\left(1-P_1\right)\left(1-P_2\right)\left(1-P_3\right)}$

Final Answer

The probability of $P(A\text{is the odd man out}\mid \text{someone is the odd man out})$

$=\frac{P_1\left(1-P_2\right)\left(1-P_3\right)+\left(1-P_1\right)P_2{P}_3}{P_1{P}_2{P}_3+\left(1-P_1\right)\left(1-P_2\right)\left(1-P_3\right)}$.