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Expected number of games in a series that terminates

  1. Oct 31, 2011 #1
    The Question:
    2 people A, B play a series of independent games. We have the following probabilities:

    P(A wins a game) = p
    P(B wins a game) = q = 1 - p

    Both players begin with X units of money, and in each game the winner takes 1 unit from the
    other player. The series terminates when either A or B loses all their money. The assumption is that p > q.

    Derive the expected number of games in a series.

    Attempt at a solution:

    Z = number of games, then we are after E[Z]:

    So if the number of games is X and the series terminated, that means that either that A won all the games from the start or B won all the games from the start. e.g. For E[Z]=N

    p[itex]^{X}[/itex]q[itex]^{0}[/itex] or p[itex]^{0}[/itex]q[itex]^{X}[/itex]

    In order for the series to terminate, the number of wins either for A or B has to be X greater than the number of wins for the other player. e.g. If B wins 1 game, then A needs to win X+1 games in order for the series to terminate. I'm not sure whether my attempt actually helps in obtaining a solution, can anyone advise? Thanks.
  2. jcsd
  3. Nov 2, 2011 #2
    I can confirm that I've solved this problem.
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