- #1
Tazz01
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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:
If:
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^{X}q^{0} or p^{0}q^{X}
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 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:
If:
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^{X}q^{0} or p^{0}q^{X}
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.