有人問到是不是吃八張效益最高,
故這邊試著把主樓的運算模式直接擴展出來。
如果不想看計算,請直接拉到最下面的效益比較。
吃六張:
000000:0.8^6=0.262144
000001:6*0.2*0.8^5=0.393216
000011:15*0.2^2*0.8^4=0.24576
000111:20*0.2^3*0.8^3=0.08192
001111:15*0.2^4*0.8^2=0.01536
011111:6*0.2^5*0.8=0.001536
111111:0.2^6=0.000064
期望值:
0.262144*1+0.393216*2+0.24576*2+0.08192*3
+0.01536*4+0.001536*5+0.000064*6=1.85536
吃七張:
0000000:0.8^7=0.2097152
0000001:7*0.2*0.8^6=0.3670016
0000011:21*0.2^2*0.8^5=0.2752512
0000111:35*0.2^3*0.8^4=0.114688
0001111:35*0.2^4*0.8^3=0.0286722
0011111:21*0.2^5*0.8^2=0.0043008
0111111:7*0.2^6*0.8=0.0003584
1111111:0.2^7=0.0000128
期望值:
0.2097152*1+0.3670016*2+0.2752512*3+0.114688*3+0.0286722*4
+0.0043008*5+0.0003584*6+0.0000128*7=2.2519688
吃八張:
00000000:0.8^8=0.16777216
00000001:8*0.2*0.8^7=0.33554432
00000011:28*0.2^2*0.8^6=0.29360128
00000111:56*0.2^3*0.8^5=0.14680064
00001111:70*0.2^4*0.8^4=0.0458752
00011111:56*0.2^5*0.8^3=0.00917504
00111111:28*0.2^6*0.8^2=0.00114688
01111111:8*0.2^7*0.8=0.00008192
11111111:0.2^8=0.00000256
期望值:
0.16777216*1+0.33554432*2+0.29360128*3+0.14680064*4+0.0458752*4
+0.00917504*5+0.00114688*6+0.00008192*7+0.00000256*8=2.5437184
吃九張:
000000000:0.8^9=0.134217728
000000001:9*0.2*0.8^8=0.301989888
000000011:36*0.2^2*0.8^7=0.301989888
000000111:84*0.2^3*0.8^6=0.176160768
000001111:126*0.2^4*0.8^5=0.066060288
000011111:126*0.2^5*0.8^4=0.016515072
000111111:84*0.2^6*0.8^3=0.002752512
001111111:36*0.2^7*0.8^2=0.000294912
011111111:9*0.2^8*0.8=0.000018432
111111111:0.2^9=0.000000512
期望值:
0.134217728*1+0.301989888*2+0.301989888*3+0.176160768*4+0.066060288*5+0.016515072*5
+0.002752512*6+0.000294912*7+0.000018432*8+0.000000512*9=2.78041856
效益比較:
6張有1.85536
7張有2.2519688
8張有2.5437184
9張有2.78041856
取6,7,8,9的最小公倍數得到504;換言之,假設有504張同技卡,
分配成1組6,1組7等等來吃,看能有多少跳技期望值。
1組6張吃:155.85024
1組7張吃:162.1417536
1組8張吃:160.2542592
1組9張吃:155.70343936
所以如果樓主的模型是對的,效益最高應該是1組7張的吃
應觀眾要求額外新增5和10的部分,並同樣做最小公倍數算法的效益比較
吃五張:
00000:0.8^5=0.32768
00001:5*0.2*0.8^4=0.4096
00011:10*0.2^2*0.8^3=0.2048
00111:10*0.2^3*0.8^2=0.0512
01111:5*0.2^4*0.8=0.0064
11111:0.2^5=0.00032
期望值:
0.32768*1+0.4096*1+0.2048*2+0.0512*3+0.0064*4+0.00032*5=1.32768
吃十張:
0000000000:0.8^10=0.1073741824
0000000001:10*0.2*0.8^9=0.268435456
0000000011:45*0.2^2*0.8^8=0.301989888
0000000111:120*0.2^3*0.8^7=0.201326592
0000001111:210*0.2^4*0.8^6=0.088080384
0000011111:252*0.2^5*0.8^5=0.0264241152
0000111111:210*0.2^6*0.8^4=0.005505024
0001111111:120*0.2^7*0.8^3=0.000786432
0011111111:45*0.2^8*0.8^2=0.000073728
0111111111:10*0.2^9*0.8=0.000004096
1111111111:0.2^10=1.024*10^(-7)
期望值:
0.1073741824*2+0.268435456*2+0.301989888*3+0.201326592*4+0.088080384*5
+0.0264241152*6+0.005505024*6+0.000786432*7+0.000073728*8+0.000004096*9
+1.024*10^(-7)*10=3.1010048
取5,6,7,8,9,10的最小公倍數得到2520
換言之,假設有2520張同技卡,
分成1組5、1組6來吃,看能有多少跳技期望值
1組5:669.15072
1組6:779.2512
1組7:810.708768
1組8:801.271296
1組9:778.5171968
1組10:781.4532096
所以如果模型是對的,效益由大到小應是
7 > 8 > 10 > 6 > 9 > 5