鋤大D有無5隻的機會率有多大??

e個問題極複雜,其實我唔係好睇得明你同花順既答案
我試試淨係計鋤大D抽13隻裡面冇同花順既機會率

Pr(無同花順)=1-Pr(有一個同花順)-Pr(有兩個同花順)

先黎簡單d抽5隻同10隻:
Pr(有一個同花順,抽5隻)=1/(52*51*50*49*48)
Pr(有兩個同花順,抽10隻)=1/(52*51*50*49*48*47*46*45*44*43*42)

鋤大D抽13隻:
Pr(有一個同花順)=13C5* 1/(52*51*50*49*48)
Pr(有兩個同花順)=13C10* 1/(52*51*50*49*48*47*46*45*44*43*42)
點解要乘個nCr係因為13中10有組合
例如:中中中中中中中中中中XXX
中中中中中中中中中XX中X
中中中中中中中中XX中中X....等等總共有13C10個

Pr(無同花順)=1-Pr(有一個同花順)-Pr(有兩個同花順)
=1- 13C5* 1/(52*51*50*49*48) -
13C10* 1/(52*51*50*49*48*47*46*45*44*43*42)

2008-01-10 01:49:10 補充：
全部有*42既都刪除乘多左42

I once saw the answer you want in a web-site... but I lost it ><

anyway, before calculating the probability in夫佬, 花, 金剛, maybe you may want to check if your probability in 同花順 and 蛇 is correct

I think you’ve OVER CALCULATE the probability in both case


For example, in 同花順

You try to count out all possible 同花順 combination, which is 40 x 47C8

40 is different type of 同花順, 47C8 is how to draw other cards, right?


Consider the following cases:

(I) S2 S3 S4 S5 S6 + S7 H8 D8 H9 D9 S9 C9 S10
(II) S3 S4 S5 S6 S7 + S2 H8 D8 H9 D9 S9 C9 S10

Two case are SAME HANDS, but it occupies two spaces in 40 x 47C8

IT should be counted ONCE actually, so it is over-counting

Similar case for 蛇 as well



I may not be correct, so please correct me if I have see it wrong ^^