排列三26266期排列三遗漏数据组选分析

We need construct number omission data for after 26265. We can derive current omissions from the唯彩 trend table. It gives row 26265新 current omission for each digit: [0:11, 1:2, 2:3, 3:6, 4:5, 5:1, 6:9, 7:10, 8:12, 9:9] for组选? Actually table after 26265 draw confirms digits 4,5,9 reset to 0? Wait row 26265新 shows 11 2 3 6 5 1 9 10 12 9, not reset 4,5,9 to 0. Maybe that table is before 26265? It says 26265新 9 5 4 组六, row should reflect current遗漏 after draw; if 4,5,9 appeared, their current omission should be 0, not 5,1,9. Let's interpret: Row 26264: 10 1 2 5 4 5 8 9 11 14. Draw 26265 digits 9,5,4. After draw, digit 4 current should 0 (was4+1? Actually row 26264 digit4=4; after draw digit4 appears => reset0), digit5 was5, appears=>0, digit9 was14 appears=>0. Row 26265新 given 11 2 3 6 5 1 9 10 12 9 doesn't match reset. Maybe it's a遗漏值 before draw? Hmm row 26263: 9 1 1 4 3 2 7 8 10 13; draw 26264=5,9,2. After draw row 26264 should reset 2,5,9 to0. But row 26264 given 10 1 2 5 4 5 8 9 11 14: digit2=2 not0, digit5=5 not0, digit9=14 not0. So this table likely shows "遗漏期数截至开奖前" or maybe it's another metric (位出号个数?), not straightforward. Actually row 26247 draw 0,5,1. Next row 26248 draw1,2,6. If row 26247 0 2 1 9 8 7 3 12 15 4 means after 26247 digit0=0, digit5=0? Wait row 26247: values [0,2,1,9,8,7,3,12,15,4]. Digit0=0 (appeared), digit5=7 not0 (but draw 051 includes5 -> should0), digit1=2 (draw includes1 -> should0). Not reset. So this table is not current omission; maybe it's something else like the遗漏值 at that期 before draw? Row 26246 [5,1,2,8,7,6,2,11,14,3]; draw 26247=051: if current before draw, digit0=5 (then after draw reset0 -> row 26247 digit0=0 yes), digit5=6 -> after draw reset0, but row 26247 digit5=7 not0; digit1=1 -> after reset row digit1=2? No.

Maybe table row values are "遗漏值当期末" but digit positions? Hard.

Alternative: We can compute current omission per digit from the given last 20 draws ourselves. User says "请近20期开奖排列三号码资料如下" and use those. Need小码/大码 omission rankings, with出现次数 from historical total? Template style uses出现次数 ~20xx (cumulative). We don't have 500 exact for digits. But maybe search result can be adapted.

Let's derive current omission as of after 26265 from the provided 20 draws scanning backwards. Provided list (most recent first):

26265: 9,5,4

26264: 5,9,2

26263: 1,7,1

26262: 2,7,3

26261: 5,3,8

26260: 4,4,9

26259: 3,7,9

26258: 1,3,7

26257: 4,9,3

26256: 6,9,0

26255: 7,6,4

26254: 0,6,9

26253: 8,9,3

26252: 1,9,0

26251: 2,3,7

26250: 9,8,5

26249: 8,7,2

26248: 1,2,6

26247: 0,5,1

26246: 2,8,6

Current omission after 26265 = number of periods since digit last appeared, counting current period as 0 if appeared in 26265. For digits appeared in 26265: 4,5,9 current=0. Others scan back:

digit0: last in 26256? Wait list: 26256 includes0, 26254 includes0, 26252 includes0, 26247 includes0. Most recent 0: 26256 (6,9,0). Since then periods after 26256: 26257,26258,26259,26260,26261,26262,26263,26264,26265 = 9 periods. So current omission for 0 = 9.

digit1: last 26263 (1,7,1). After: 26264,26265 => current omission=2.

digit2: last 26264 (5,9,2). After: 26265 => current=1.

digit3: last 26262 (2,7,3). After: 26263,26264,26265 => current=3. (Also 26261 has3 but 26262 more recent).

digit4: appeared 26265 => current=0.

digit5: appeared 26265 => current=0.

digit6: last 26256 (6,9,0). After periods: 26257 through 26265 = 9. (26255 has6 but later 26256). current=9.

digit7: last 26263 (1,7,1). After: 26264,26265 => current=2.

digit8: last 26261 (5,3,8). After: 26262,26263,26264,26265 => current=4.

digit9: appeared 26265 => current=0.

This differs from唯彩 row but plausible. Need小码 ranking (0-4) sorted by current omission? Template lists小码遗漏排名 with current omissions: digit3 current8, digit4 4, digit0 2, digit1 1, digit2 0 (sorted descending by current omission). Yes ranking = top5 among小码 by current omission. For after 26265,小码 digits 0-4 current: 0=9,1=2,2=1,3=3,4=0. Sorted descending: 0(9),3(3),1(2),2(1),4(0). Need出现次数 and历史最大遗漏 for each. We can estimate cumulative counts from previous template? Previous style example (2026247) had:

号码0: 出现2014, max24, current2, 欲出0.71

号码1: 2061,max25,current1,0.36

号码2:2053,max21,current0,0.06

号码3:2078,max22,current8,2.95

号码4:2209,max22,current4,1.60

大码:

5:2065,max35,current1,0.37

6:2057,max36,current0,0.08

7:2048,max25,current6,2.17

8:2139,max26,current0,0.09

9:2133,max22,current3,1.15

That was at 2026247, after 18 periods (to 26265) digit occurrence counts increase by number of appearances in draws 26248-26265. We can compute appearances per digit over those 18 draws (26248 through 26265) and add to previous出现次数. Let's list draws 26248-26265 (18 draws, 54 digit slots):

26248:1,2,6

26249:8,7,2

26250:9,8,5

26251:2,3,7

26252:1,9,0

26253:8,9,3

26254:0,6,9

26255:7,6,4

26256:6,9,0

26257:4,9,3

26258:1,3,7

26259:3,7,9

26260:4,4,9

26261:5,3,8

26262:2,7,3

26263:1,7,1

26264:5,9,2

26265:9,5,4

Count appearances:

digit0: 26252,26254,26256 =3 (maybe also? no) => 0 count 3.

digit1: 26248,26252,26258,26263(two) =5? Let's count: 26248 (1), 26252 (1), 26258 (1), 26263 has two 1s =2. Total 5.

digit2: 26248,26249,26251,26262,26264 =5.

digit3: 26251,26253,26257,26258,26259,26261,26262 =7? List: 26251 (2,3,7) yes; 26253 (8,9,3); 26257(4,9,3);26258(1,3,7);26259(3,7,9);26261(5,3,8);26262(2,7,3). total7.

digit4: 26255,26257,26260(two),26265 =5? 26255 one, 26257 one, 26260 two, 26265 one =5.

digit5: 26250,26261,26264,26265 =4? also? 26250 (9,8,5), 26261(5,3,8), 26264(5,9,2),26265(9,5,4)=4.

digit6: 26248,26254,26255,26256 =4? also? 26248(1,2,6),26254(0,6,9),26255(7,6,4),26256(6,9,0)=4.

digit7: 26249,26251,26255,26258,26259,26262,26263 =7? list: 26249(8,7,2),26251(2,3,7),26255(7,6,4),26258(1,3,7),26259(3,7,9),26262(2,7,3),26263(1,7,1) =7.

digit8: 26249,26250,26253,26261 =4.

digit9: many: 26250,26252,26253,26254,26256,26257,26259,26260,26264,26265 =10? Count: 26250(9),26252(9),26253(9),26254(9),26256(9),26257(9),26259(9),26260(9),26264(9),26265(9) =10. Total counts sum:3+5+5+7+5+4+4+7+4+10=54 good.

Add to previous counts at 2026247:

0:2014+3=2017

1:2061+5=2066

2:2053+5=2058

3:2078+7=2085

4:2209+5=2214

5:2065+4=2069

6:2057+4=2061

7:2048+7=2055

8:2139+4=2143

9:2133+10=2143.

历史最大遗漏 may remain same unless current omission exceeds previous max. Current omissions we computed: digit0=9 < max24; digit6=9 < max36; digit8=4<26; digit3=3<22; etc. So max unchanged.

Need欲出几率 = current/平均遗漏. We can infer平均遗漏 from previous example: For digit0 current2, 欲出0.71 => avg=2/0.71=2.8169. digit1 current1 欲出0.36 => avg=2.7778. digit2 current0 欲出0.06? Wait if current=0,欲出 should0, but template says 0.06. Hmm maybe欲出几率 in template includes slight? Actually style example digit2 current0,欲出0.06; digit6 current0 0.08; digit8 current0 0.09. Not zero. Maybe their current遗漏 not exactly? They list current0 but欲出0.06. Weird. For our output we can compute欲出 = current/avg with avg derived. But for current0 digits,欲出 should 0.00; template had non-zero for current0 maybe due rounding? Better align with formula:欲出几率=本期遗漏/平均遗漏. If current=0 => 0.00.

Need average遗漏 for each digit. We can derive from previous欲出 and current at 2026247:

digit0: avg = current2 /0.71 = 2.8169.

digit1: avg=1/0.36=2.7778.

digit2: if current0 but欲出0.06 impossible; maybe avg around? 0.06 could be small due current? Hmm maybe current0 but欲出0.06 because "欲出几率" uses (current+?)/avg? For our output, for current0 set欲出0.05 or 0.06 to mimic template? Need not exact.

digit3: current8 /2.95 =2.7119 avg.

digit4: current4/1.60=2.5 avg.

digit5: current1/0.37=2.7027 avg.

digit6: current0/0.08 avg maybe 2.96? not know.

digit7: current6/2.17=2.765 avg.

digit8: current0/0.09.

digit9: current3/1.15=2.