难度:++
二人猜数。甲在1到9中选一个奇数,乙在2到8中选一个偶数。如果两数差1,乙胜,否则甲胜。
问两人的最佳策略各是什么?如果两人都采取最佳策略,获胜的概率各是什么?
This can be formulated as a max-min problem..., is there an simpler approach? |
Could you give some details about your max-min approach? |
Thanks. |
Assume 甲's strategy is to randomly pick an odder number with probablity p1,p3,...,p9 and 乙's stragetgy is to randomly pick an even number with probability q2,q4,...q8 Then 乙 pick q's so as to maximize p1*q2+p3*(q2+q4)+....p7*(q6+q8)+p9*q8 and 甲 minimize the resulting formular. |
The best strategy for A is to choose 1, 5, and 9 with probablity 1/3 each. It can be calculated that A can win at least 2/3 of the time no matter what B does. The best strategy for B is to choose 2 and 8 with probability 1/3 each, and 4/6 with a combined probability 1/3. It can be calculated that B wins at least 1/3 of the time no matter what A does. |
.....A can win at least 2/3 of the time ..... .....B wins at least 1/3 of the time ..... Are these two statements a little contradicting? |
These two statements are not contradicting to each other, since we have >= and <=. Together they guarantee that both strategies are optimal: If applied together, A wins 2/3 of the time, B 1/3, which answers part 2.
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really worth. |
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