Re: [閒聊] 開黃泉後盤面五屬珠都>=3顆的機率
看板ToS (神魔之塔(Tower of Saviors))作者znmkhxrw (QQ)時間8年前 (2017/09/10 14:58)推噓51(53推 2噓 9→)留言64則, 58人參與討論串3/3 (看更多)
※ 引述《znmkhxrw (QQ)》之銘言:
: 大家好(‧^ω^‧)
: 剛剛狂日大大問了個 TOS-MATH 問題如題
: 正常關卡(非禁珠)下開黃泉洗珠,每一屬(不含心)都≧3顆的機率是?
答案:0.544634557955
---------------------------------------
原本認為手算就算捏爛LP也算不出來
不過在今天早上先捏一下下後,發現其實很多項可以couple在一起(‧^ω^‧)
接著再藉由計算網頁就得到許多網友跑程式的結果了
【問題】
假設全版洗盤後版面是平均分布六色珠(含自消盤面),那五屬性各至少3顆以上機率是?
【答案】
分母 = 全部盤面 = 6^30 = 221073919720733357899776
分子 = 五屬性各至少3顆
= 全部盤面 - 至少有一屬≦2顆
= 221073919720733357899776 - 100669423178347649484640
= 120404496542385708415136
機率 = 分子/分母 = 0.544634557955
【證明】
令 S為所有盤面所成的集合
#(any set) := 此集合的元素個素
令 A 為水珠≦2顆的盤面所成的集合
A_1為水珠=2顆的盤面所成的集合
A_2為水珠=1顆的盤面所成的集合
A_3為水珠=0顆的盤面所成的集合
則 A=A_1∪A_2∪A_3, 且三者互斥(兩兩交集為空集合)
同樣的令出 B(火), C(木), D(光), E(暗)
因此我們所求的"五屬珠至少有一屬≦2顆"
其實就是 #(A∪B∪C∪D∪E)
之後根據排容原理
#(A∪B∪C∪D∪E)
= Σ#(單一集合) - Σ#(兩兩交集) + Σ#(三三交集) - Σ#(四四交集) + Σ#(全交集)
接著觀察到A,B,C,D,E是對稱的
因此每個Σ內只要算一次,之後乘上交集可能方法數即可,具體如下:
(1) Σ#(單一集合) = C(5,1) * #(A)
(2) Σ#(兩兩交集) = C(5,2) * #(A∩B)
(3) Σ#(三三交集) = C(5,3) * #(A∩B∩C)
(4) Σ#(四四交集) = C(5,4) * #(A∩B∩C∩D)
(5) Σ#(全交集) = C(5,5) * #(A∩B∩C∩D∩E)
最後,就是分別計算以下項目:(" ** "代表次方" ^ ")
(a) #(A) = #(A_1) + #(A_2) + #(A_3)
= c(30,2)*5**28+c(30,1)*5**29+5**30
(b) #(A∩B) = #(A_1∩B_1) + #(A_1∩B_2) + #(A_1∩B_3)
+ #(A_2∩B_1) + #(A_2∩B_2) + #(A_2∩B_3)
+ #(A_3∩B_1) + #(A_3∩B_2) + #(A_3∩B_3)
= c(30,4)*c(4,2)*4**26+c(30,2)*c(2,1)*4**28+4**30
+ 2*[c(30,3)*c(3,2)*4**27+c(30,2)*4**28+c(30,1)*4**29]
【紅色部分也是簡化關鍵,因為淺藍那兩個項一樣、綠色一樣、紫色一樣】
(c) #(A∩B∩C)
= c(30,6)*c(6,2)*c(4,2)*3**24+c(30,3)*c(3,1)*c(2,1)*3**27+3**30
+ 3*[ c(30,5)*c(5,2)*c(3,2)*3**25+c(30,4)*c(4,2)*3**26+c(30,2)*3**28
+c(30,2)*c(2,1)*3**28+c(30,1)*3**29]
+ 6*[c(30,3)*c(3,2)*3**27]
(d) #(A∩B∩C∩D)
= c(30,8)*c(8,2)*c(6,2)*c(4,2)*2**22+c(30,4)*c(4,1)*c(3,1)*c(2,1)*2**26+2**30
+ 4*[ c(30,7)*c(7,2)*c(5,2)*c(3,2)*2**23+c(30,6)*c(6,2)*c(4,2)*2**24
+c(30,5)*c(5,2)*c(3,1)*c(2,1)*2**25+c(30,2)*2**28
+c(30,3)*c(3,1)*c(2,1)*2**27+c(30,1)*2**29]
+ 6*[ c(30,6)*c(6,2)*c(4,2)*c(2,1)*2**24+c(30,4)*c(4,2)*2**26
+c(30,2)*c(2,1)*2**28]
+ 12*[ c(30,5)*c(5,2)*c(3,2)*2**25+c(30,4)*c(4,2)*c(2,1)*2**26
+c(30,3)*c(3,2)*2**27]
(e) #(A∩B∩C∩D∩E)
= c(30,10)*c(10,2)*c(8,2)*c(6,2)*c(4,2)+c(30,5)*c(5,1)*c(4,1)*c(3,1)*c(2,1)+1
+ 5*[ c(30,9)*c(9,2)*c(7,2)*c(5,2)*c(3,2)+c(30,8)*c(8,2)*c(6,2)*c(4,2)
+c(30,6)*c(6,2)*c(4,1)*c(3,1)*c(2,1)+c(30,2)+c(30,4)*c(4,1)*c(3,1)*c(2,1)
+c(30,1)]
+ 10*[ c(30,8)*c(8,2)*c(6,2)*c(4,2)*c(2,1)+c(30,6)*c(6,2)*c(4,2)
+c(30,7)*c(7,2)*c(5,2)*c(3,1)*c(2,1)+c(30,4)*c(4,2)
+c(30,3)*c(3,1)*c(2,1)+c(30,2)*c(2,1)]
+ 20*[c(30,7)*c(7,2)*c(5,2)*c(3,2)+c(30,5)*c(5,2)*c(3,1)*c(2,1)+c(30,3)*c(3,2)]
+ 30*[c(30,6)*c(6,2)*c(4,2)*c(2,1)+c(30,5)*c(5,2)*c(3,2)+c(30,4)*c(4,2)*c(2,1)]
===========================================================================
最後一步了!藉由心算或是電腦計算即可得
#(A∪B∪C∪D∪E) = 100669423178347649484640
接著答案就是 [6^30-#(A∪B∪C∪D∪E)]/6^30
掐指一算 = 0.544634557955 (‧^ω^‧)
-------------------------------------------
說真的...按run的那瞬間好怕答案跟版友的程式解不一樣
我沒有勇氣回頭再check一次(╯°□°)╯ ~ /(_□_,,)\
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 111.255.238.56
※ 文章網址: https://www.ptt.cc/bbs/ToS/M.1505026716.A.04D.html
→
09/10 15:00, , 1F
09/10 15:00, 1F
真沒擋頭
推
09/10 15:00, , 2F
09/10 15:00, 2F
→
09/10 15:00, , 3F
09/10 15:00, 3F
推
09/10 15:01, , 4F
09/10 15:01, 4F
→
09/10 15:01, , 5F
09/10 15:01, 5F
推
09/10 15:01, , 6F
09/10 15:01, 6F
→
09/10 15:01, , 7F
09/10 15:01, 7F
→
09/10 15:04, , 8F
09/10 15:04, 8F
推
09/10 15:06, , 9F
09/10 15:06, 9F
推
09/10 15:07, , 10F
09/10 15:07, 10F
推
09/10 15:08, , 11F
09/10 15:08, 11F
推
09/10 15:08, , 12F
09/10 15:08, 12F
推
09/10 15:08, , 13F
09/10 15:08, 13F
推
09/10 15:09, , 14F
09/10 15:09, 14F
推
09/10 15:10, , 15F
09/10 15:10, 15F
→
09/10 15:14, , 16F
09/10 15:14, 16F
推
09/10 15:21, , 17F
09/10 15:21, 17F
噓
09/10 15:22, , 18F
09/10 15:22, 18F
推
09/10 15:22, , 19F
09/10 15:22, 19F
推
09/10 15:24, , 20F
09/10 15:24, 20F
推
09/10 15:24, , 21F
09/10 15:24, 21F
推
09/10 15:27, , 22F
09/10 15:27, 22F
→
09/10 15:28, , 23F
09/10 15:28, 23F
推
09/10 15:30, , 24F
09/10 15:30, 24F
你才老濕
這是高中排組阿
推
09/10 15:31, , 25F
09/10 15:31, 25F
推
09/10 15:34, , 26F
09/10 15:34, 26F
推
09/10 15:39, , 27F
09/10 15:39, 27F
推
09/10 15:55, , 28F
09/10 15:55, 28F
※ 編輯: znmkhxrw (111.255.238.56), 09/10/2017 16:26:02
推
09/10 16:30, , 29F
09/10 16:30, 29F
推
09/10 16:34, , 30F
09/10 16:34, 30F
推
09/10 16:39, , 31F
09/10 16:39, 31F
推
09/10 16:43, , 32F
09/10 16:43, 32F
→
09/10 16:43, , 33F
09/10 16:43, 33F
推
09/10 16:46, , 34F
09/10 16:46, 34F
推
09/10 17:01, , 35F
09/10 17:01, 35F
推
09/10 17:08, , 36F
09/10 17:08, 36F
推
09/10 17:10, , 37F
09/10 17:10, 37F
推
09/10 17:11, , 38F
09/10 17:11, 38F
推
09/10 17:13, , 39F
09/10 17:13, 39F
推
09/10 17:25, , 40F
09/10 17:25, 40F
推
09/10 17:26, , 41F
09/10 17:26, 41F
推
09/10 17:37, , 42F
09/10 17:37, 42F
推
09/10 17:45, , 43F
09/10 17:45, 43F
→
09/10 17:45, , 44F
09/10 17:45, 44F
推
09/10 17:46, , 45F
09/10 17:46, 45F
推
09/10 18:17, , 46F
09/10 18:17, 46F
推
09/10 18:23, , 47F
09/10 18:23, 47F
推
09/10 19:11, , 48F
09/10 19:11, 48F
推
09/10 19:12, , 49F
09/10 19:12, 49F
推
09/10 19:17, , 50F
09/10 19:17, 50F
推
09/10 19:32, , 51F
09/10 19:32, 51F
推
09/10 19:35, , 52F
09/10 19:35, 52F
老實說我算到#(A∩B)後 還沒觀察到有些項相等時 這個有9項
但是如果繼續算下去
#(A∩B∩C) 有27項
#(A∩B∩C∩D) 有81項
#(A∩B∩C∩D∩E) 有243項 乾你老師
還好藉由對稱性發現其實這243項有可以分成1個 5個 10個 20個 30個
※ 編輯: znmkhxrw (111.255.238.56), 09/10/2017 19:44:00
推
09/10 21:28, , 53F
09/10 21:28, 53F
推
09/10 23:40, , 54F
09/10 23:40, 54F
推
09/11 00:00, , 55F
09/11 00:00, 55F
推
09/11 00:43, , 56F
09/11 00:43, 56F
推
09/11 01:12, , 57F
09/11 01:12, 57F
推
09/11 01:33, , 58F
09/11 01:33, 58F
推
09/11 03:24, , 59F
09/11 03:24, 59F
噓
09/11 04:51, , 60F
09/11 04:51, 60F
推
09/11 10:34, , 61F
09/11 10:34, 61F
推
09/11 13:15, , 62F
09/11 13:15, 62F
推
09/11 13:55, , 63F
09/11 13:55, 63F
推
09/15 17:28, , 64F
09/15 17:28, 64F
討論串 (同標題文章)
ToS 近期熱門文章
PTT遊戲區 即時熱門文章
18
23
13
20
44
70
18
28