一場比賽， 雙方各派出5支隊伍，五局三勝制，同一個隊伍的遇到不同的對手勝率都不一樣，如何求總勝率？

大家好 我這邊有一道玩手游時遇到的概率問題 想請教一下大家（不是為了遊戲，純粹是好奇心在作祟）現在我要與一名對手進行一場比賽， 雙方各派出5支隊伍，五局三勝。我有隊伍12345，對方則是隊伍ABCDE。現有數據如下，1_A=100%（即1隊遇到A隊100%取勝），1_B=0, 1_C=50%, 1_D=0, 1_E=100%, 2_A=100%, 2_B=80%, 2_C=0, 2_D=0, 2_E=0, 3_A=100%, 3_B=0, 3_C=0, 3_D=0, 3_E=100%, 4_A=100%,4_B=100%, 4_C=20%, 4_D=50%, 4_E=0, 5_A=100%, 5_B=0, 5_C=20%, 5_D=0, 5_E=100%.

請問

1.我的總勝率是多少？求過程或公式/解釋

2. 條件一定的情況下，如何改進隊伍更有效率？是著重提高三/四個隊的勝率，放棄另外一/兩個，還是盡量平均提高5個隊伍質量？

謝謝

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根據題面描述，題主的總勝率大約是39.48%。計算過程沒什麼可解釋的，寫個程序窮舉就行了。

ABCDE: 90% (1vA=100%, 2vB= 80%, 3vC= 0%, 4vD= 50%, 5vE=100%)
ABCED: 0% (1vA=100%, 2vB= 80%, 3vC= 0%, 4vE= 0%, 5vD= 0%)
ABDCE: 84% (1vA=100%, 2vB= 80%, 3vD= 0%, 4vC= 20%, 5vE=100%)
ABDEC: 16% (1vA=100%, 2vB= 80%, 3vD= 0%, 4vE= 0%, 5vC= 20%)
ABECD: 84% (1vA=100%, 2vB= 80%, 3vE=100%, 4vC= 20%, 5vD= 0%)
ABEDC: 92% (1vA=100%, 2vB= 80%, 3vE=100%, 4vD= 50%, 5vC= 20%)
ACBDE: 50% (1vA=100%, 2vC= 0%, 3vB= 0%, 4vD= 50%, 5vE=100%)
ACBED: 0% (1vA=100%, 2vC= 0%, 3vB= 0%, 4vE= 0%, 5vD= 0%)
ACDBE: 100% (1vA=100%, 2vC= 0%, 3vD= 0%, 4vB=100%, 5vE=100%)
ACDEB: 0% (1vA=100%, 2vC= 0%, 3vD= 0%, 4vE= 0%, 5vB= 0%)
ACEBD: 100% (1vA=100%, 2vC= 0%, 3vE=100%, 4vB=100%, 5vD= 0%)
ACEDB: 50% (1vA=100%, 2vC= 0%, 3vE=100%, 4vD= 50%, 5vB= 0%)
ADBCE: 20% (1vA=100%, 2vD= 0%, 3vB= 0%, 4vC= 20%, 5vE=100%)
ADBEC: 0% (1vA=100%, 2vD= 0%, 3vB= 0%, 4vE= 0%, 5vC= 20%)
ADCBE: 100% (1vA=100%, 2vD= 0%, 3vC= 0%, 4vB=100%, 5vE=100%)
ADCEB: 0% (1vA=100%, 2vD= 0%, 3vC= 0%, 4vE= 0%, 5vB= 0%)
ADEBC: 100% (1vA=100%, 2vD= 0%, 3vE=100%, 4vB=100%, 5vC= 20%)
ADECB: 20% (1vA=100%, 2vD= 0%, 3vE=100%, 4vC= 20%, 5vB= 0%)
AEBCD: 0% (1vA=100%, 2vE= 0%, 3vB= 0%, 4vC= 20%, 5vD= 0%)
AEBDC: 10% (1vA=100%, 2vE= 0%, 3vB= 0%, 4vD= 50%, 5vC= 20%)
AECBD: 0% (1vA=100%, 2vE= 0%, 3vC= 0%, 4vB=100%, 5vD= 0%)
AECDB: 0% (1vA=100%, 2vE= 0%, 3vC= 0%, 4vD= 50%, 5vB= 0%)
AEDBC: 20% (1vA=100%, 2vE= 0%, 3vD= 0%, 4vB=100%, 5vC= 20%)
AEDCB: 0% (1vA=100%, 2vE= 0%, 3vD= 0%, 4vC= 20%, 5vB= 0%)
BACDE: 50% (1vB= 0%, 2vA=100%, 3vC= 0%, 4vD= 50%, 5vE=100%)
BACED: 0% (1vB= 0%, 2vA=100%, 3vC= 0%, 4vE= 0%, 5vD= 0%)
BADCE: 20% (1vB= 0%, 2vA=100%, 3vD= 0%, 4vC= 20%, 5vE=100%)
BADEC: 0% (1vB= 0%, 2vA=100%, 3vD= 0%, 4vE= 0%, 5vC= 20%)
BAECD: 20% (1vB= 0%, 2vA=100%, 3vE=100%, 4vC= 20%, 5vD= 0%)
BAEDC: 60% (1vB= 0%, 2vA=100%, 3vE=100%, 4vD= 50%, 5vC= 20%)
BCADE: 50% (1vB= 0%, 2vC= 0%, 3vA=100%, 4vD= 50%, 5vE=100%)
BCAED: 0% (1vB= 0%, 2vC= 0%, 3vA=100%, 4vE= 0%, 5vD= 0%)
BCDAE: 0% (1vB= 0%, 2vC= 0%, 3vD= 0%, 4vA=100%, 5vE=100%)
BCDEA: 0% (1vB= 0%, 2vC= 0%, 3vD= 0%, 4vE= 0%, 5vA=100%)
BCEAD: 0% (1vB= 0%, 2vC= 0%, 3vE=100%, 4vA=100%, 5vD= 0%)
BCEDA: 50% (1vB= 0%, 2vC= 0%, 3vE=100%, 4vD= 50%, 5vA=100%)
BDACE: 20% (1vB= 0%, 2vD= 0%, 3vA=100%, 4vC= 20%, 5vE=100%)
BDAEC: 0% (1vB= 0%, 2vD= 0%, 3vA=100%, 4vE= 0%, 5vC= 20%)
BDCAE: 0% (1vB= 0%, 2vD= 0%, 3vC= 0%, 4vA=100%, 5vE=100%)
BDCEA: 0% (1vB= 0%, 2vD= 0%, 3vC= 0%, 4vE= 0%, 5vA=100%)
BDEAC: 20% (1vB= 0%, 2vD= 0%, 3vE=100%, 4vA=100%, 5vC= 20%)
BDECA: 20% (1vB= 0%, 2vD= 0%, 3vE=100%, 4vC= 20%, 5vA=100%)
BEACD: 0% (1vB= 0%, 2vE= 0%, 3vA=100%, 4vC= 20%, 5vD= 0%)
BEADC: 10% (1vB= 0%, 2vE= 0%, 3vA=100%, 4vD= 50%, 5vC= 20%)
BECAD: 0% (1vB= 0%, 2vE= 0%, 3vC= 0%, 4vA=100%, 5vD= 0%)
BECDA: 0% (1vB= 0%, 2vE= 0%, 3vC= 0%, 4vD= 50%, 5vA=100%)
BEDAC: 0% (1vB= 0%, 2vE= 0%, 3vD= 0%, 4vA=100%, 5vC= 20%)
BEDCA: 0% (1vB= 0%, 2vE= 0%, 3vD= 0%, 4vC= 20%, 5vA=100%)
CABDE: 75% (1vC= 50%, 2vA=100%, 3vB= 0%, 4vD= 50%, 5vE=100%)
CABED: 0% (1vC= 50%, 2vA=100%, 3vB= 0%, 4vE= 0%, 5vD= 0%)
CADBE: 100% (1vC= 50%, 2vA=100%, 3vD= 0%, 4vB=100%, 5vE=100%)
CADEB: 0% (1vC= 50%, 2vA=100%, 3vD= 0%, 4vE= 0%, 5vB= 0%)
CAEBD: 100% (1vC= 50%, 2vA=100%, 3vE=100%, 4vB=100%, 5vD= 0%)
CAEDB: 75% (1vC= 50%, 2vA=100%, 3vE=100%, 4vD= 50%, 5vB= 0%)
CBADE: 95% (1vC= 50%, 2vB= 80%, 3vA=100%, 4vD= 50%, 5vE=100%)
CBAED: 40% (1vC= 50%, 2vB= 80%, 3vA=100%, 4vE= 0%, 5vD= 0%)
CBDAE: 90% (1vC= 50%, 2vB= 80%, 3vD= 0%, 4vA=100%, 5vE=100%)
CBDEA: 40% (1vC= 50%, 2vB= 80%, 3vD= 0%, 4vE= 0%, 5vA=100%)
CBEAD: 90% (1vC= 50%, 2vB= 80%, 3vE=100%, 4vA=100%, 5vD= 0%)
CBEDA: 95% (1vC= 50%, 2vB= 80%, 3vE=100%, 4vD= 50%, 5vA=100%)
CDABE: 100% (1vC= 50%, 2vD= 0%, 3vA=100%, 4vB=100%, 5vE=100%)
CDAEB: 0% (1vC= 50%, 2vD= 0%, 3vA=100%, 4vE= 0%, 5vB= 0%)
CDBAE: 50% (1vC= 50%, 2vD= 0%, 3vB= 0%, 4vA=100%, 5vE=100%)
CDBEA: 0% (1vC= 50%, 2vD= 0%, 3vB= 0%, 4vE= 0%, 5vA=100%)
CDEAB: 50% (1vC= 50%, 2vD= 0%, 3vE=100%, 4vA=100%, 5vB= 0%)
CDEBA: 100% (1vC= 50%, 2vD= 0%, 3vE=100%, 4vB=100%, 5vA=100%)
CEABD: 50% (1vC= 50%, 2vE= 0%, 3vA=100%, 4vB=100%, 5vD= 0%)
CEADB: 25% (1vC= 50%, 2vE= 0%, 3vA=100%, 4vD= 50%, 5vB= 0%)
CEBAD: 0% (1vC= 50%, 2vE= 0%, 3vB= 0%, 4vA=100%, 5vD= 0%)
CEBDA: 25% (1vC= 50%, 2vE= 0%, 3vB= 0%, 4vD= 50%, 5vA=100%)
CEDAB: 0% (1vC= 50%, 2vE= 0%, 3vD= 0%, 4vA=100%, 5vB= 0%)
CEDBA: 50% (1vC= 50%, 2vE= 0%, 3vD= 0%, 4vB=100%, 5vA=100%)
DABCE: 20% (1vD= 0%, 2vA=100%, 3vB= 0%, 4vC= 20%, 5vE=100%)
DABEC: 0% (1vD= 0%, 2vA=100%, 3vB= 0%, 4vE= 0%, 5vC= 20%)
DACBE: 100% (1vD= 0%, 2vA=100%, 3vC= 0%, 4vB=100%, 5vE=100%)
DACEB: 0% (1vD= 0%, 2vA=100%, 3vC= 0%, 4vE= 0%, 5vB= 0%)
DAEBC: 100% (1vD= 0%, 2vA=100%, 3vE=100%, 4vB=100%, 5vC= 20%)
DAECB: 20% (1vD= 0%, 2vA=100%, 3vE=100%, 4vC= 20%, 5vB= 0%)
DBACE: 84% (1vD= 0%, 2vB= 80%, 3vA=100%, 4vC= 20%, 5vE=100%)
DBAEC: 16% (1vD= 0%, 2vB= 80%, 3vA=100%, 4vE= 0%, 5vC= 20%)
DBCAE: 80% (1vD= 0%, 2vB= 80%, 3vC= 0%, 4vA=100%, 5vE=100%)
DBCEA: 0% (1vD= 0%, 2vB= 80%, 3vC= 0%, 4vE= 0%, 5vA=100%)
DBEAC: 84% (1vD= 0%, 2vB= 80%, 3vE=100%, 4vA=100%, 5vC= 20%)
DBECA: 84% (1vD= 0%, 2vB= 80%, 3vE=100%, 4vC= 20%, 5vA=100%)
DCABE: 100% (1vD= 0%, 2vC= 0%, 3vA=100%, 4vB=100%, 5vE=100%)
DCAEB: 0% (1vD= 0%, 2vC= 0%, 3vA=100%, 4vE= 0%, 5vB= 0%)
DCBAE: 0% (1vD= 0%, 2vC= 0%, 3vB= 0%, 4vA=100%, 5vE=100%)
DCBEA: 0% (1vD= 0%, 2vC= 0%, 3vB= 0%, 4vE= 0%, 5vA=100%)
DCEAB: 0% (1vD= 0%, 2vC= 0%, 3vE=100%, 4vA=100%, 5vB= 0%)
DCEBA: 100% (1vD= 0%, 2vC= 0%, 3vE=100%, 4vB=100%, 5vA=100%)
DEABC: 20% (1vD= 0%, 2vE= 0%, 3vA=100%, 4vB=100%, 5vC= 20%)
DEACB: 0% (1vD= 0%, 2vE= 0%, 3vA=100%, 4vC= 20%, 5vB= 0%)
DEBAC: 0% (1vD= 0%, 2vE= 0%, 3vB= 0%, 4vA=100%, 5vC= 20%)
DEBCA: 0% (1vD= 0%, 2vE= 0%, 3vB= 0%, 4vC= 20%, 5vA=100%)
DECAB: 0% (1vD= 0%, 2vE= 0%, 3vC= 0%, 4vA=100%, 5vB= 0%)
DECBA: 0% (1vD= 0%, 2vE= 0%, 3vC= 0%, 4vB=100%, 5vA=100%)
EABCD: 20% (1vE=100%, 2vA=100%, 3vB= 0%, 4vC= 20%, 5vD= 0%)
EABDC: 60% (1vE=100%, 2vA=100%, 3vB= 0%, 4vD= 50%, 5vC= 20%)
EACBD: 100% (1vE=100%, 2vA=100%, 3vC= 0%, 4vB=100%, 5vD= 0%)
EACDB: 50% (1vE=100%, 2vA=100%, 3vC= 0%, 4vD= 50%, 5vB= 0%)
EADBC: 100% (1vE=100%, 2vA=100%, 3vD= 0%, 4vB=100%, 5vC= 20%)
EADCB: 20% (1vE=100%, 2vA=100%, 3vD= 0%, 4vC= 20%, 5vB= 0%)
EBACD: 84% (1vE=100%, 2vB= 80%, 3vA=100%, 4vC= 20%, 5vD= 0%)
EBADC: 92% (1vE=100%, 2vB= 80%, 3vA=100%, 4vD= 50%, 5vC= 20%)
EBCAD: 80% (1vE=100%, 2vB= 80%, 3vC= 0%, 4vA=100%, 5vD= 0%)
EBCDA: 90% (1vE=100%, 2vB= 80%, 3vC= 0%, 4vD= 50%, 5vA=100%)
EBDAC: 84% (1vE=100%, 2vB= 80%, 3vD= 0%, 4vA=100%, 5vC= 20%)
EBDCA: 84% (1vE=100%, 2vB= 80%, 3vD= 0%, 4vC= 20%, 5vA=100%)
ECABD: 100% (1vE=100%, 2vC= 0%, 3vA=100%, 4vB=100%, 5vD= 0%)
ECADB: 50% (1vE=100%, 2vC= 0%, 3vA=100%, 4vD= 50%, 5vB= 0%)
ECBAD: 0% (1vE=100%, 2vC= 0%, 3vB= 0%, 4vA=100%, 5vD= 0%)
ECBDA: 50% (1vE=100%, 2vC= 0%, 3vB= 0%, 4vD= 50%, 5vA=100%)
ECDAB: 0% (1vE=100%, 2vC= 0%, 3vD= 0%, 4vA=100%, 5vB= 0%)
ECDBA: 100% (1vE=100%, 2vC= 0%, 3vD= 0%, 4vB=100%, 5vA=100%)
EDABC: 100% (1vE=100%, 2vD= 0%, 3vA=100%, 4vB=100%, 5vC= 20%)
EDACB: 20% (1vE=100%, 2vD= 0%, 3vA=100%, 4vC= 20%, 5vB= 0%)
EDBAC: 20% (1vE=100%, 2vD= 0%, 3vB= 0%, 4vA=100%, 5vC= 20%)
EDBCA: 20% (1vE=100%, 2vD= 0%, 3vB= 0%, 4vC= 20%, 5vA=100%)
EDCAB: 0% (1vE=100%, 2vD= 0%, 3vC= 0%, 4vA=100%, 5vB= 0%)
EDCBA: 100% (1vE=100%, 2vD= 0%, 3vC= 0%, 4vB=100%, 5vA=100%)
對手出戰順序：120種排列。題主綜合期望勝率：39.483333%。
總120場中，
題主有18場吊打對手（三必勝）；
有44場被對手吊打（三必敗）；
有58場掙扎局，掙扎局的勝率為50.6551724137931%。


拚命練3馬、捨棄另2馬的戰術不是不可以，代價是你只能靠這3馬贏，它們當中任何一個輸了你就廢了。
舉個例子：總戰鬥力上限30000，你的分配是10000+10000+10000+0+0，碰到6000+6000+6000+6000+6000這種平均流對手你可以穩贏，但是如果遇到個13000+13000+1000+1000+1000的奇葩，除非對方的兩優馬撞到你的兩劣馬，否則你穩輸。

當然，上例中的這個奇葩遇到平均流就跪了。靠譜的遊戲一般都能弄出點循環相剋，到底怎麼配更有效率，還是得看具體環境。

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信息是否是對稱的？如果對方也知道勝率並且兩方選擇出戰隊伍是同時的，那麼對方很可能會捨棄A隊，從而將輸贏壓縮在前四局以內，如果四局你贏了兩局就可以獲勝了。

這種情況下我的建議是放棄2隊，如果有決勝局再上2隊去贏對面A隊。如果你的4隊能抓到B隊，那你就能贏下比賽。

另外這並不是一道簡單的求概率問題，隨著比賽進行，雙方能派出的隊伍變少，都會根據形勢改變策略，必輸的策略誰也不會使用，因此窮舉法的結果未必準確。當然，對方也有可能故意在前四場用A隊輸一場來打亂你的策略，讓遊戲變的更為複雜。個人覺得這是一個達不到均衡的博弈，不過具體分析我實在是才疏學淺，希望有研究方向是game theory的大牛關注這個問題給你詳細的解答。

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