s*********k
发帖数: 1989 | 1 52-card deck. 4 Players
What is the possibility that each has an ace?
What is the P that 3100 (one has three aces, another has one ace and the
left two has none).
What is the P that 2200(one has two aces, another has the other two aces and
the left two has none). |
s*******s
发帖数: 1568 | 2 1)
1*39/51*26/50*13/49
2)
P(4,2) * 13/52*12/51*11/50*13/49
3)
P(4,2) * 13/52*12/51*13/50*12/49
and
【在 s*********k 的大作中提到】
: 52-card deck. 4 Players
: What is the possibility that each has an ace?
: What is the P that 3100 (one has three aces, another has one ace and the
: left two has none).
: What is the P that 2200(one has two aces, another has the other two aces and
: the left two has none).
|
s*********k
发帖数: 1989 | 3 what is 1111 and 2110?
Does your asswer add up P(all)=1?
P(all)=P(4000)+P(3100)+P(2200)+P(2110)+P(1111) |
g**********1
发帖数: 1113 | 4 没有必要这么精确把 13/52， 12/51,...，基本上可以认为是1/4。 |
s***t
发帖数: 49 | 5 P(4000) = 1/64
P(3100) = 12/64
P(2200) = 6/64
P(2110) = 39/64
P(1111) = 6/64
【在 s*********k 的大作中提到】
: what is 1111 and 2110?
: Does your asswer add up P(all)=1?
: P(all)=P(4000)+P(3100)+P(2200)+P(2110)+P(1111)
|
b********u
发帖数: 63 | 6
*4?
*3?
【在 s*******s 的大作中提到】
: 1)
: 1*39/51*26/50*13/49
: 2)
: P(4,2) * 13/52*12/51*11/50*13/49
: 3)
: P(4,2) * 13/52*12/51*13/50*12/49
:
: and
|
b********u
发帖数: 63 | 7 if w/o replacement
P(4000) = 4*13*12*11*10/P(52,4)
P(3100) = 48*13*12*11*13/P(52,4)
P(2200) = 36*13*12*13*12/P(52,4)
P(2110) = 144*13*12*13*13/P(52,4)
P(1111) = 24*13*13*13*13/P(52,4)
sum = 1
【在 s***t 的大作中提到】
: P(4000) = 1/64
: P(3100) = 12/64
: P(2200) = 6/64
: P(2110) = 39/64
: P(1111) = 6/64
|
O***a
发帖数: 121 | 8 还是想讨论一下。
怎么发牌呀？是每人13张，彻底发完了看是否每人有一张A呢？还是每人5张牌啊？还是
每人一张牌，刚好一人拿到一张A呀？ |
b******9
发帖数: 47 | 9
1) the same as swordsman
我做的结果根swordsman 很接近，唯一稍有不同的是
2) P(4,2) * 13/52*12/51*11/50*13/49 *4
3) P(4,2) * 13/52*12/51*13/50*12/49 * 6
P(4,2) is choosing 2 people from 4 for having ace.
while 4 in 2) derived from C(4,3)*C(1,1) because there are combination when
we choose 3 among 4 aces.
6 in 3) is similar from C(4,2)*C(2,2).
What do you think of my result. Thank you.
【在 s*******s 的大作中提到】
: 1)
: 1*39/51*26/50*13/49
: 2)
: P(4,2) * 13/52*12/51*11/50*13/49
: 3)
: P(4,2) * 13/52*12/51*13/50*12/49
:
: and
|
