至少有两张A的概率为
1-C(48,4)/C(52,4)-C(48,3)*C(4,1)/C(52,4)
=1-48*47*46*45/(52*51*50*49)-48*47*46*4*4/(52*51*50*49)
≈1-0.25555-0.71874
≈0.0257