The passing ball game

Its very clear that Tom got the ball 1 time. Adam starts the game. Also Adam passes ball equal time as Robert gets it. As Adam and Tom did not get the last pass, so Tom got the ball once. As Robert passes the ball 3 times and he could have got the ball the last time, so he could possibly get it 3 or 4 times. Combined with Tom, it makes 4 or 5 times. As Robert did not get the ball most, so Adam or Brian should get it most. And we have already seen that Adam gets one less than Robert, so Brian gets max times. That leaves a possibility of 3 or 4 for Brian. But by the previous statement, Brian gets 4 times.

Let us name the players R,A,B,T. Let the times they got the ball be Rg,Ag,Bg,Tg and times they passed be Rp,Ap,Bp,Tp respectively.

It is clear that Tg=Tp as T neither started the game nor ended with the ball .Thus Tg=1-------(1) .

Since A started with the ball and didn't end up with it, Ag=Ap-1 . As given Rg=Ap, hence Rg=Ag+1-------(2) .

Now since the passing is 10 times, Ag+Rg+Tg+Bg=10, i.e,from eqns (1) and (2), Ag+Ag+1+1+Bg=10; 2Ag+Bg=8;-------(3)

Since Rp=3 and R did not start with the ball, Rg=3 or 4 . Thus from above equation (2), Ag=Rg-1=2 or 3.

If Ag=3,from eqn (3),Bg=2. But it is given Bg not less than Ag.Hence Ag=2 and Bg=4. Hence answer is 4 .

It is easy to see that Tom gets the ball only once. Coupled with the fact that Robert did not get the ball the most, this means that Adam and Brian got the ball the most. Since we know that Robert passes the ball thrice, he must have gotten the ball at least thrice. Therefore, our answer must be greater than 3. Hence it is $4$ . How convenient.