Building on your interpretation, let us define the rules as follows: A and B alternately throw a single die. If the sum of the last two throws is 5, the one who just threw wins. Note that as stated, an initial 5 could win, or a series of throws of a different length than 2. The analysis would change, but follow the same route. The below requires precisely two throws to add to 5.
Let p be the probability that the first player wins. Let q be the probability that the next player to throw wins given that he has received a chance of winning, that is that the last throw is less than 5. Then if you receive a throw of 5 or 6 your chance of winning is p. So q=1/6 (that you win on this throw) + (1-p)/3 (that you throw 5 or 6 and then win) +(1-q)/2 (that you throw <5, don't win this throw, but finally win. p=(1-p)/3 (that you throw 5 or 6 and win)+2(1-q)/3 (that you throw <5 and win).
If I have the algebra right, p=15/32 and q=9/16, so the first player should withdraw $2/17.