For (c) you must mean "find a way to choose the team fairly, even if the coin is unfair". There are infinitely many ways to do that, but is there a solution which is optimal in some precise sense?

No you are wrong Fleeting guess, Let me clear the meaning of problem with a example: Let the coint has the side $+,-$. and we have $22$ players. and two teams with the names of $A , B$(In the exam the names where SAYPA&FOLAD,I think the designer(Master ALISHAHI) is one of the fans of these teams ). For each player ,we drop the coint and $+$ will give the player to $A$ and $-$ will give the player to $B$. Find a way to claim that coin is fair or not?

Since I don't know Farsi and don't have the exam, let me just discuss the mathematical part: If the question is "how to test whether a given coin has $p = 1/2$", I think that there is no procedure guaranteed to do that (with probability 1) after some finite number of steps, let alone a simple procedure that can be done at the Takhti Stadium (probably the locations reflect the mathematical sophistication of the solution). If the question is "how to simulate a fair coin using an unfair coin with unknown constant $p \in (0,1)$", there is a simple procedure that with probability 1 will terminate in a finite number of steps. It is a nice problem whether or not it was part (c) of this olympiad.

First the designer was not Mr. Alishahi the n he is not the fan of these 2 teams. First the question was for the 2 teams Real Madrid And Barcelona

Come on man,why are you saying barcelona & real madrid,There were Iranian teams(I have misstyped Saba battery with saypa ), But As you wish ,Mr.hatami ,you are one of the coordinators but who on the earth can design such a problems beside of Mr.alishahi ?? he designed the problem ,doesnt him? Fleeting guess:I completely agree with you,but This is beauty of the problem ,It seems impossible but how can you prove your Quote: think that there is no procedure guaranteed to do that (with probability 1) after some finite number of steps, let alone a simple procedure that can be done at the Takhti Stadium

First I didn't say that it was these 2 teams in the exam. I told that they changed these 2 teams to 2 Iranian teams for the exam. And also this problem was designed by all of the members of the National Mathematical olympiad Committee. Not by a person, However some problems were given by one person.

re: impossiblity of a finite-with-probability-1 test for $p = 1/2$: Consider any testing procedure that stops (with probability 1) after a finite number of coin tosses and declares YES or NO to the question "is $p = 1/2$?". For any sequence of tosses where the procedure stops, those tosses are consistent with any $p \in (0,1)$; they do not determine $p$, so any declaration by the testing procedure is not necessarily correct. What can be guaranteed is to answer the question "is $|p - 1/2| < \delta$" with a probability greater than $1 - \epsilon$ that the answer is correct. To do this (for a given $\delta, \epsilon > 0$) one can just flip the coin some large number of times and answer YES/NO according to whether the proportion of Heads/Tails is close to 1/2.

reply of (a) : let the persons be $A,B,C$ . consider this sets : {$A,B$},{$B,C$},{$C,A$}. in the first step, we ballot by the coin, in each set , and in each set one person can win . then if one person didn't win in any sets (second step) we delete him and ballot between two other. if each of persons won in one of his sets , we ballot in each of three sets again . we know finaly have the above manner(the above manner : one of persons don't win in any sets) ( we have that manner in finitely ballots )so we do the second step and finally one lucky person wins the coin fairly.

(a) i think there is no finite procedure. Caz the probability for each person to win is a polynomial of 1/2^k. Let T denotes tail and H denotes head. At each step toss the coin twice, if it's TT, then the first person win, TH for the second one and HT for the third. If it's HH, redo the procedure. (b) similar to (a), just write a in base 2, then it's a polynomial of 1/2^k. Define A to win at each step according to each k.

cm wrote: (a) i think there is no finite procedure. Caz the probability for each person to win is a polynomial of 1/2^k. Let T denotes tail and H denotes head. At each step toss the coin twice, if it's TT, then the first person win, TH for the second one and HT for the third. If it's HH, redo the procedure. (b) similar to (a), just write a in base 2, then it's a polynomial of 1/2^k. Define A to win at each step according to each k. if you read the above solution , you will understand that you were rong .

cm wrote: (a) i think there is no finite procedure. Caz the probability for each person to win is a polynomial of 1/2^k. This shows that there is no procedure that stops within $k$ steps, where $k$ is given in advance. It is possible that, with probability 1, the procedure will stop after some finite (but not limitable in advance) number of steps. For example, this is true of the procedure "flip coin until a Heads appears".

Can one give a solution for part (b)?

Question (b) is actually Putnam 1989 A4. See http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1868731525&t=75385 or http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1868731525&t=106683