季後賽種子序可能性 - NBA
By Victoria
at 2020-08-12T12:26
at 2020-08-12T12:26
Table of Contents
今天打完之後
剩餘可能變動的排名區如下:
1.東區4和5或5和6
熱火 44-27
溜馬 43-28
76人 42-29
各剩2場比賽且同戰績比序也是熱火>溜馬>76人
今年無主客場所以4和5意思是一樣的
除非76人兩場(對暴龍、火箭)皆贏且溜馬兩場(對火箭、熱火)皆輸
不然東區對戰組合將如下
1.公鹿 2.暴龍
8.魔術 7.籃網
4.熱火 3.賽提
5.溜馬 6.76人
2.西區2~6名
快艇 47-23
金塊 46-25
火箭 44-26
雷霆 43-27
爵士 43-28
除了爵士剩一場比賽,其餘隊伍皆剩兩場比賽
快艇最單純,只要不要二連敗且金塊二連勝就第2種子
金塊只要不要二連敗且火箭二連勝也就有第3種子以上
至於箭雷爵三隊
雷2:1爵
雷2:1箭
箭2:1爵
因此爵士要爬到第五必須雷霆全敗且自己贏球
箭和雷誰4誰5就不討論了
3.西區8~11名
拓荒 34-39
灰熊 33-39
太陽 33-39
馬刺 32-38
四隊都是周五打最後一戰
所以排序將會是看這一戰
贏的排前面輸的排後面
都贏或都輸的隊伍按照拓>熊>陽>刺的順序排列
因此如無意外西區季後賽對戰組合的雛型也差不多出來了
1.湖人 2.快艇
8.拓/熊/陽/刺 7.小牛
4.火箭 3.金塊
5.雷霆 6.爵士
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