import itertools
#初始化
allsuits = ['1', '2', '3', '4']
ts = list(input("请输入target拥有的手牌, 0表示不可见：")) #target手牌
psuits = list(input("请输入player拥有的花色：")) #player总花色
pseen = [] #player可见花色
tseen = [] #target可见牌
allseen = int(input("player的牌是否全部可见？(1/0)：")) #player牌是否全部可见
unseen = 0 #target不可见牌数量
nohave = [] #player没有的花色
for i in ts:
if i != '0': tseen.append(i)
else: unseen += 1
for i in psuits:
if allseen or int(input(f"player拥有的'{i}'花色是否可见(1/0)：")) != 0: pseen.append(i)
if allseen: nohave = [x for x in allsuits if x not in psuits]
elif len(psuits) != 4:
nohave = list(input("请输入target已知player没有的花色(0则没有)："))
if nohave[0] == '0': nohave = []
#穷举计算成功率
allcases = list(itertools.product(allsuits, repeat=unseen)) #为target未知手牌生成所有可能的情况
#计算target尽可能选择player有的花色的成功率
odds1 = 0
for tcards in allcases: #穷举所有情况，下同
fz = 0 #此条件下成功情况数
fm = 0 #此条件下总情况数
temp = None
tcards = list(tcards) + tseen #加上可见手牌
for i in tcards:
if i in pseen: #player肯定有此花色
temp = 1
break
if i not in nohave: #player可能有此花色
fm += 1
if i in psuits: fz += 1
if temp == 1: odds1 += 1/len(allcases) #这种情况必然成功
elif fm != 0: odds1 += fz/fm/len(allcases) #此条件下成功率/总条件数
#计算target尽可能选择player没有的花色的成功率
odds2 = 0
for tcards in allcases:
fz = 0
fm = 0
temp = None
tcards = list(tcards) + tseen
for i in tcards:
if i in nohave: #player肯定无此花色
temp = 0
break
if i not in pseen: #player可能无此花色
fm += 1
if i in psuits: fz += 1
if temp == 0: continue
elif fm == 0: odds2 += 1/len(allcases)
else: odds2 += fz/fm/len(allcases)
print(f"target尽可能选择player有的花色的情况下player的成功率为{odds1}，尽可能选择player没有的花色的情况下为{odds2}")

代码方便起见直接用'1', '2', '3', '4'代表四种花色，前面是初始化数据，player用手牌是否全部可见来让target推测player有无没有的花色。此外我发现似乎除了(1)情况双方有同花色明牌和(2)情况target明牌里有player明确没有的花色的牌外，同一事件下两个情况概率之和好像都是1