Two guys are asked to think of a number between 0 and 100. If guy A chooses number greater than B than A wins 10 paid by the house and B gets nothing. If B chooses number greater than A then B gets 10 paid by the house and A gets nothing. If both choose same number then both end up paying 10 each to the house. What is the best strategy to choose a number. (guys can not communicate with each-other).
Best strategy to maximize winning between 2 competitors asked to choose number
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Partial answer:
Let's assume players after the game receive only the information about the result - pay (equal), get (higher), or nothing (lower) and the goal is to get paid more (not to win the opponent by higher score). Maximum summary for players is 10 per game. Since the game is symmetric, the maximum benefit for a player is 5 per game. It is achievable with alternate strategy where players wins every 2nd game. e.g. using strategy 100-99-100-99... for player A and 99-100-99-100-99... for player B for all games after N-th game for some finite number N.
Strategy also works with 0 (means "I really want you to win in this game") instead of 99 ("I want you to win only, if choose 100"), but if another player really always wants to loose it is a winning strategy to exploit such behaviour.
Now, what is missing here? If the other player does not or does not want to understand the optimal strategy or is a "100 always robot" the best strategy for player A is 99 always to get/lose zero. So it becomes the 10€ per game winning strategy for the other player.
What happens if B chooses greater than A? If nothing happens, then the best strategy is for A to choose 100, and B to choose 0.
A is motivated to do so because of the three possible outcomes, he either:
gets $10 - if he chooses higher
gets $0 - if he chooses lower
loses $10 - if he chooses the same
B is motivated to do so because of the three possible outcomes, he either:
gets $0 - if he chooses higher
gets $0 - if he chooses lower
loses $10 - if he chooses the same
B will be indifferent between options 1 and 2. So, to avoid the chance of losing $10 (option B-3), he will choose as low a number as possible.
A will obviously prefer option 1. So, to avoid the chance of losing $10 (Option A-3), he will choose as high as possible.
Hence A=100, B=0