Flipping coin(s) to decide who pays the bill I'm not from a mathematical background. I found this video on YouTube rather confusing. I know basics of probability from school.
My question: If a single coin is flipped among 3 people say A, B and C with following combination
Say between A & B first

Between A & B

and then

Between A(or)B and C //depending on whether A or B loses in the first toss

how would that not be fair.
In the first instance, when coin is tossed between A & B - the probability of A or B paying the bill is 50% each.
Likewise, in the second instance when the coin is tossed between the loser of first instance (which could be A or B) and C - the probability is again 50% each.
Hence, a fair deal.
I do not understand why the speaker in the beginning of the video says it would not be fair and requires the person to be selected using "binary randomness"
Feel free to edit the question, if it has not been worded well
 A: Suppose we agree that we will always toss the coin twice, and then follow the schedule you described:


*

*Between $A$ and $B$: $A$ loses on heads, $B$ loses on tails

*Between $C$ and the loser of toss 1: $C$ loses on heads, the round 1 loser loses on tails


There are four ways the coin could come up, all equally likely:
$$\begin{array}{cc|c}
\text{first toss} & \text{second toss} & \text{loser} \\ \hline
H & H  & C \\
H & T & A \\
T & H & C \\
T & T & B
\end{array}
$$
We see from the chart that $C$ pays half the time, and one of $A$ or $B$ pays the other half the time, which is only $\frac14$ each.  Then $C$ is paying twice as often as $A$ or $B$ is.
It should be clear that $C$ has a disadvantage: to get stuck with the check, $A$ must lose two coin tosses, but $C$ is stuck with the check if she loses one coin toss, which is easier than losing two.
In fact there is no perfectly fair way to chose among three people with a finite number of coin tosses.  But the following method works well:
$$\begin{array}{cc|c}
\text{first toss} & \text{second toss} & \text{loser} \\ \hline
H & H  & A \\
H & T & B \\
T & H & C \\
T & T & \text{flip again}
\end{array}
$$
Several other methods are described in Is it possible to 'split' coin flipping 3 ways?
