No answer so far. This is the way I think the game goes as the bag is passed from beggar to beggar.
All five beggars start with the same number of pennies, say N. Lets call the five A, B, C, D, and E, knowing that A=1, B=2 and so on, because it will get confusing using numbers for the beggars and for the pennies they put into and take out of the bag.
So A puts N pennies into the bag and passes it to C.
C takes out 2 pennies and gives to B who now has N+2 and takes out 4 pennies and gives to D who now has N+4. C then puts 1/2N pennies into the bag and passes it to E; C now has N-1/2N = 1/2N pennies and the bag has N-6 +1/2N = 3/2N -6 pennies.
E takes out 1 penny and gives to A, who now has 1 penny. E takes out 4 pennies and gives to D who now has N+8 pennies. E puts into the bag 1/2(3/2N -6) and passes the bag to B. E now has N- 1/2(3/2N -6) = 1/4N +3 pennies and the bag has 3/2N -11 +(3/4N -3) = 9/4N -14 pennies.
B takes out 1 penny and gives it to A, who now has 1+1=2 pennies and takes out 3 pennies and gives to C, who now has 1/2N +3 pennies. B now puts 1/2(9/4N -14) pennies into the bag and passes it to D. B now has N+2 - (9/8N -7) = -1/8N +9 pennies and the bag has 9/4N -18 + (9/8N -7) =27/8N -25 pennies.
If N =72 then at this point B will have 0 pennies and the bag will have 243. If N is less than 72 then B will have had to borrow extra pennies from C, D, or E. We also know that after D completes his turn he will pass the bag to A, and A will have only 2 pennies, which could be the end of the game. What seems confusing to me (and I may already be confused by keeping track of what is in the bag after each turn) is that when D receives the bag he takes out pennies and gives them to C and takes out 5 pennies and gives to E. So even if B had to borrow pennies from C, D, and E, C and D will still have some pennies, which contradicts the story that A leaves with the bag plus his 2 pennies and all the rest have none.