D****6
发帖数: 278 | 1 在别的版看到的。看着简单，那边还没确切答案呢。
We roll a 3 sided dice 100 times. The faces have values 1, 2 and 3. What is
the probability that no 3 consecutive rolls add up to 4? | E*******0
发帖数: 465 | | D****6
发帖数: 278 | 3 连续三个add up to 4的几率是 1/9。怎么考虑这100次呢？ | o******e
发帖数: 1001 | 4 是不是就是等同与问，仍100次，出现连续 1 1 1的概率？
如果这样，另Ｐ(100)为仍１００次出现的个概率，有，
P(100)=2/3*P(99)+1/3(1/3(1/3+2/3*P(97))+2/3*P(98))
=1/27+2/27*P(97)+2/9*P(98)+2/3*P(99)
比如：
P(0)=Ｐ(1)=P(2)=0;
p(3)=1/27;
p(4)=1/27+2/3*1/27=5/81
....
is
【在 D****6 的大作中提到】
: 在别的版看到的。看着简单，那边还没确切答案呢。
: We roll a 3 sided dice 100 times. The faces have values 1, 2 and 3. What is
: the probability that no 3 consecutive rolls add up to 4?
| x*********s
发帖数: 2604 | 5 这个可以类比成某个股票连续100天的交易量，连续三天交易量相加低于某个值的概率
是多少，挺有现实意义的题。。。 | a**j
发帖数: 60 | | a**j
发帖数: 60 | 7 Pr(no 3 consecutive rolls add up to 4)=1-Pr(any 3 consecutive rolls of 100
rolls add up to 4)
(1)For exactly 3 rolls, 3 consecutive rolls add up to 4:
it must be any combination of {1,1,2}s
so Pr (3 consecutive rolls add up to 4,for exactly 3 rolls)=1/9
(2) Let only consider independent cases
Pr (sum of roll 1, 2, 3 =4）=1/9
Pr (sum of roll 4, 5, 6 =4 and sum of roll 1, 2, 3 not equal to 4)=(1/9)*(8/
9)
...
pr (sum of roll 97, 98, 99 =4 and all previous independent 3 rolls not equal
to 4)= (1/9)*(8/9)^32
For all 33 independent consecutive 3 rolls, Pr of any consecutive add up to
4 is:
(1/9)*(1+8/9+(8/9)^2+...+(8/9)^32)=(1/9)*9*(1-(8/9)^33)=0.98
So Pr(any 3 consecutive rolls of 100 rolls add up to 4)>0.98
Pr(no 3 consecutive rolls add up to 4)<0.02 | a**j
发帖数: 60 | 8 result of highly correlated consecutive 3 rolls are hard to calculate
Since for example roll 2,3,4 are dependent on the results of roll 1,2,3 roll
3,4,5 and roll 4,5,6
So it will take a bit of time to compute exact probability.
But in most cases 0 | s*****w
发帖数: 1017 | 9 马农门的数学啊。。。能算出小于2%就算decent的结果了 | a**j
发帖数: 60 | 10 since the 8/9 is the the Probability excluding the prior consecutive 3 rolls
having sum of 4, therefore we can calculate the exact probability of any 3
consecutive rolls of 100 rolls add up to 4
By the previous statement we have:
Pr(any 3 consecutive rolls of 100 rolls add up to 4)
=(1/9)*(1+8/9+(8/9)^2+...+(8/9)^97)=(1/9)*9*(1-(8/9)^98)
=0.9999903
Therefore Pr(no 3 consecutive rolls add up to 4)=1-0.9999903=0.0000097 | a**j
发帖数: 60 | 11 Regular dice should be 6 sided with value of 1, 2,3 4,5,6
in this case rolling 100 times
Pr(any of 3 consecutive rolls add up to 4)
=(1/72)*72*(1-(71/72)^98)=0.746
Pr(no 3 consecutive rolls add up to 4)=0.254
it is geometric series |
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