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What is the best strategy that maximizes the probability of winning?

38 viewed last edited 11 months ago
Mahesh Godavarti
0
A and B are to play a two game chess. If they tied after two games, they continue to the first win. A is the weaker of the two. With a daring game A has a probability of 45% of winning and 55% of losing; with a conservative game A's probability of a draw is 90% and of losing 10%. What is A's strategy that could lead to their winning the match? And with what probability?
Mahesh Godavarti
0

Let's consider some alternate problems.


  1. Let's consider the case where A and B play a sequence of games and the first person to win a game, wins the match-up. In such a case, the best strategy for A to win is to play aggressively in the first game itself which gives A the probability of winning as 45%. Playing conservatively does not help A's cause as A HAS to play aggressively to win.
  2. Interestingly, if the problem was stated as what is A's strategy to engage B in as many games as possible then the A's strategy to play conservatively all the time.


Essentially, if A has to win, A has to play as few games as possible.


Now, to our problem.


There are three possibilities after two games:

  1. A loses
  2. A wins
  3. A ties with B - In this case, the only option A has is to play aggressively in the third game


Essentially, A has to adopt a strategy where the probability of A winning or tying the game is maximum.


A has the following strategies for the first game to minimize loss in the first two games.

  1. Play aggressively - if it results in a loss, play aggressively in the second game to stay alive. If it results in a win, play conservatively to play for the win. Probability of winning = 45% X 90%, Probability of a tie = 45% X 10% + 55% X 45%, Probability of losing = 55% X 55%
  2. Play conservatively - if it results in a loss, play aggressively in the second game to stay alive. If it results in a draw, play aggressively for the win. Probability of winning = 90% X 45%, Probability of a tie = 10% X 45%, Probability of losing = 10% X 55% + 90% X 55% = 55%


Based on the probabilities above, we see that A's strategy is to:

  1. Play aggressively in the first game
  2. Play aggressively in the second game, if the first results in a loss and play conservatively if it results in a win
  3. Play aggressively in the third game, if the first two games result in a tie.


Therefore, total probability of winning = 45% X 90% + 45% X 65% X 45% = 0.536625

Probability of losing = 55% X 55% + 45% X 65% X 55% = 0.463375 = 1 - Probability of winning.


Note to make sure that all probabilities add up to 1.