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５．game2の部分を抜粋する
URLﾘﾝｸ(www.ma.huji.ac.il) PUZZLES ”Choice Games”Sergiu Hart November 4, 2013
(抜粋)
A similar result, but now without using the Axiom of Choice.^2 Consider the following two-person game game2:
・Player 1 chooses a rational number in the interval [0,1] and writes down its infinite decimal expansion^3 0.x1x2...xn..., with all xn ∈ {0,1,...,9}.
・Player 2 asks (in some order) what are the digits xn except one, say xi; then he writes down a digit ξ ∈ {0,1,...,9}.
・If xi = ξ then Player 2 wins, and if xi 6= ξ then Player 1 wins.
By choosing i arbitrarily and ξ uniformly in {0,1,...,9}, Player 2 can guarantee a win with probability 1/10. However, we have:
Theorem 2 For every ε > 0 Player 2 has a mixed strategy in game2 guaranteeing him a win with probability at least 1 ? ε.
Proof.
The proof is the same as for Theorem 1, except that here we do not use the Axiom of Choice.
Because there are only countably many sequences x ∈ {0,...,9}N that Player 1 may choose (namely, those x that become eventually periodic),
we can order them － say x(1),x(2),...,x(m),...－ and then choose in each equivalence class the element with minimal index (thus F(x) = x(m) iff m is the minimal natural number such that^4 x ～ x(m)).
Remark. When the number of boxes is finite Player 1 can guarantee a win with probability 1 in game1, and with probability 9/10 in game2, by choosing the xi independently and uniformly on [0,1] and {0,1,...,9}, respectively.
Note:
^2 Due to Phil Reny.
^3 When there is more than one expansion, e.g., 0.1000000... = 0.0999999..., Player 1 chooses which expansion to use.
^4 Explicit strategies σj may also be constructed, based on Rj being the index where the sequence yj becomes periodic.
(引用終り)