A and B are playing a game, simultaneously exposing
one or two fingers.
If the total number of fingers is odd, then A pays B that number of dollars.
If it’s even, then B pays A accordingly.
Is it a fair game?
a. Assume random decision by both players.
or
b. Both chose the optimal strategy.
(In reply to
A's strategy by Math Man)
However, if A knows that B is not sophisticated enough to know the correct strategy and would blindly use the 1/2 and 1/2 randomization as in part a, he could exploit that by choosing 2 fingers more than half the time. Theoretically the more the better, but it really shouldn't be enough to tip off B of what he's doing.
This is unlike B, who can assure a win if he knows the correct strategy. For A the "best strategy" merely assures a loss. And even if B knows his own best strategy, A will fare no worse than losing 1/12 dollar anyway, as that's what B is assuring.
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Posted by Charlie
on 2015-09-17 19:29:03 |
re: A's strategy
