This problem can be solved by exchange. The decision of equipment selection is to decide α and β, and the two sides of a coin are A and B. Let the probability of Party A appearing be Pa and Party B appearing be Pb, then Pa+Pb= 1.

Step 1: Use the A side of the coin to determine α, and use the B side of the coin to determine β, tossing for n times.

Then it is determined that the expected number of times that the surface A appears is N*Pa and the expected number of times that the surface B appears is N*Pb.

Step 2: Use the B side of the coin to determine α, and use the A side of the coin to determine β, tossing N times.

Then the expected times of B-plane appearance obtained by determining α are N*Pb, and the expected times of A-plane appearance obtained by β are N * Pa.

Because in the above throwing, * * * threw 2N times in total, then:

The expected number of faces obtained by determining α * * is N*(Pa+Pb)=N times, and the probability of occurrence is n/2n =1/2; The expected number of faces obtained by determining β * * is N*(Pb+Pa)=N times, and the probability of occurrence is N/2N= 1/2.

So this method is fair.