A bag contain 5 dices and each have six faces with probability $p_1$=$p_2$=$p_3$=$2p_4$=$2p_5$=$3*p_6$. What is the probability of selecting two dice with face 4, and three dice with face 1?
Someone have try the codes for this problem in r shown in picture but I not understand how the probability is obtained. Kindly explain me the answer for this problem.
First, find p_is values based on the given equalities and the following assumption (replace all p_is based on their relations with p_6 to find it, then find the value of each p_i):
p_1 + p_2 + p_3 + p_4 + p_5 + p_6 = 1
To find the probability, we need to select 2 out of 5 of dices with face 4 which its probability is p_4^2 and for other dices, they should have face 1 that it's probability is p_1^3.
Now, to understand the last step, you should read the explanation of the dmultinom(x, size, prob) function (from this post):
Generate multinomially distributed random number vectors and compute multinomial probabilities. If x is a K-component vector, dmultinom(x, prob) is the probability
P(X[1]=x[1], … , X[K]=x[k]) = C * prod(j=1 , …, K) p[j]^x[j]. where C is the ‘multinomial coefficient’ C = N! / (x[1]! * … * x[K]!) and N = sum(j=1, …, K) x[j].
By definition, each component X[j] is binomially distributed as Bin(size, prob[j]) for j = 1, …, K.
Therefore, dmultinom(j, n, p) means C(5,2) * p_1^3 * p_4^2, as j = (3, 0, 0, 2, 0), n=5, and p = (p_1, p_2, p_3, p_4, p_5, p_6).
Related
I have probabilities of 110 independent events.
I want to calculate for every number of events, that exactly that number of distinct events will appear.
Example, if we have only three events
A = 0.45
B = 0.65
C = 0.73
# Probability of none event
P[0] = (1-A)*(1-B)*(1-C)
# Probability of exaclty one event
P[1] = A*(1-B)*(1-C)+(1-A)*B*(1-C)+(1-A)*(1-B)*C
# Probability of exactly two events
P[2] = A*B*(1-C)+A*(1-B)*C+(1-A)*B*C
# Probability of exactly three events
P[3] = A*B*C
Is realistic to compute it for 110 events under 1 hour?
If yes, how to do it in any programming language?
Let the probabilities be p_1, p_2, ..., p_n. You're in essence trying to expand the polynomial
(1 - p_1 + p_1 x) (1 - p_2 + p_2 x) ... (1 - p_n + p_n x),
where the probability of getting m events is the coefficient of x^m. Rather than computing all 2^n monomials and summing them, you can simplify after each multiplication. In Python:
def f(ps):
coefs = [1]
for p in ps:
coefs.append(0)
for i in range(len(coefs) - 1, 0, -1):
coefs[i] = coefs[i] * (1 - p) + coefs[i - 1] * p
coefs[0] *= 1 - p
return coefs
Sample execution (note the floating point error).
>>> f([0.45, 0.65, 0.73])
[0.05197500000000001, 0.279575, 0.454925, 0.21352500000000002]
I am solving this problem where we need to reach from X=0 to X=N.We can only take a step of 2 or 3 at a time.
For each step of 2 we have a probability of 0.2 and for each step of 3 we have a probability of 0.8.How can we find the total probability to reach N.
e.g. for reaching 5,
2+3 with probability =0.2 * 0.8=0.16
3+2 with probability =0.8 * 0.2=0.16 total = 0.32.
My initial thoughts:
Number of ways can be found out by simple Fibonacci sequence.
f(n)=f(n-3)+f(n-2);
But how do we remember the numbers so that we can multiply them to find the probability?
This can be solved using Dynamic programming.
Lets call the function F(N) = probability to reach 0 using only 2 and 3 when the starting number is N
F(N) = 0.2*F(N-2) + 0.3*F(N-3)
Base case:
F(0) = 1 and F(k)= 0 where k< 0
So the DP code would be somthing like that:
F[0] = 1;
for(int i = 1;i<=N;i++){
if(i>=3)
F[i] = 0.2*F[i-2] + 0.8*F[i-3];
else if(i>=2)
F[i] = 0.2*F[i-2];
else
F[i] = 0;
}
return F[N];
This algorithm would run in O(N)
Some clarifications about this solution: I assume the only allowed operation for generating the number from 2s and 3s is addition (your definition would allow substraction aswell) and the input-numbers are always valid (2 <= input). Definition: a unique row of numbers means: no other row with the same number of 3s and 2s in another order is in scope.
We can reduce the problem into multiple smaller problems:
Problem A: finding all sequences of numbers that can sum up to the given number. (Unique rows of numbers only)
Start by finding the minimum-number of 3s required to build the given number, which is simply input % 2. The maximum-number of 3s that can be used to build the input can be calculated this way:
int max_3 = (int) (input / 3);
if(input - max_3 == 1)
--max_3;
Now all sequences of numbers that sum up to input must hold between input % 2 and max_3 3s. The 2s can be easily calculated from a given number of 3s.
Problem B: calculating the probability for a given list and it's permutations to be the result
For each unique row of numbers, we can easily derive all permutations. Since these consist of the same number, they have the same likeliness to appear and produce the same sum. The likeliness can be calculated easily from the row: 0.8 ^ number_of_3s * 0.2 ^ number_of_2s. Next step would be to calculate the number of different permuatations. Calculating all distinct sets with a specific number of 2s and 3s can be done this way: Calculate all possible distributions of 2s in the set: (number_of_2s + number_of_3s)! / (number_of_3s! * numer_of_2s!). Basically just the number of possible distinct permutations.
Now from theory to praxis
Since the math is given, the rest is pretty straight forward:
define prob:
input: int num
output: double
double result = 0.0
int min_3s = (num % 2)
int max_3s = (int) (num / 3)
if(num - max_3 == 1)
--max_3
for int c3s in [min_3s , max_3s]
int c2s = (num - (c3s * 3)) / 2
double p = 0.8 ^ c3s * 0.2 * c2s
p *= (c3s + c2s)! / (c3s! * c2s!)
result += p
return result
Instead of jumping into the programming, you can use math.
Let p(n) be the probability that you reach the location that is n steps away.
Base cases:
p(0)=1
p(1)=0
p(2)=0.2
Linear recurrence relation
p(n+3)=0.2 p(n+1) + 0.8 p(n)
You can solve this in closed form by finding the exponential solutions to the linear recurrent relation.
c^3 = 0.2 c + 0.8
c = 1, (-5 +- sqrt(55)i)/10
Although this was cubic, c=1 will always be a solution in this type of problem since there is a constant nonzero solution.
Because the roots are distinct, all solutions are of the form a1(1)^n + a2((-5+sqrt(55)i) / 10)^n + a3((-5-sqrt(55)i)/10)^n. You can solve for a1, a2, and a3 using the initial conditions:
a1=5/14
a2=(99-sqrt(55)i)/308
a3=(99+sqrt(55)i)/308
This gives you a nonrecursive formula for p(n):
p(n)=5/14+(99-sqrt(55)i)/308((-5+sqrt(55)i)/10)^n+(99+sqrt(55)i)/308((-5-sqrt(55)i)/10)^n
One nice property of the non-recursive formula is that you can read off the asymptotic value of 5/14, but that's also clear because the average value of a jump is 2(1/5)+ 3(4/5) = 14/5, and you almost surely hit a set with density 1/(14/5) of the integers. You can use the magnitudes of the other roots, 2/sqrt(5)~0.894, to see how rapidly the probabilities approach the asymptotics.
5/14 - (|a2|+|a3|) 0.894^n < p(n) < 5/14 + (|a2|+|a3|) 0.894^n
|5/14 - p(n)| < (|a2|+|a3|) 0.894^n
f(n, p) = f(n-3, p*.8) + f(n -2, p*.2)
Start p at 1.
If n=0 return p, if n <0 return 0.
Instead of using the (terribly inefficient) recursive algorithm, start from the start and calculate in how many ways you can reach subsequent steps, i.e. using 'dynamic programming'. This way, you can easily calculate the probabilities and also have a complexity of only O(n) to calculate everything up to step n.
For each step, memorize the possible ways of reaching that step, if any (no matter how), and the probability of reaching that step. For the zeroth step (the start) this is (1, 1.0).
steps = [(1, 1.0)]
Now, for each consecutive step n, get the previously computed possible ways poss and probability prob to reach steps n-2 and n-3 (or (0, 0.0) in case of n < 2 or n < 3 respectively), add those to the combined possibilities and probability to reach that new step, and add them to the list.
for n in range(1, 10):
poss2, prob2 = steps[n-2] if n >= 2 else (0, 0.0)
poss3, prob3 = steps[n-3] if n >= 3 else (0, 0.0)
steps.append( (poss2 + poss3, prob2 * 0.2 + prob3 * 0.8) )
Now you can just get the numbers from that list:
>>> for n, (poss, prob) in enumerate(steps):
... print "%s\t%s\t%s" % (n, poss, prob)
0 1 1.0
1 0 0.0
2 1 0.2
3 1 0.8
4 1 0.04
5 2 0.32 <-- 2 ways to get to 5 with combined prob. of 0.32
6 2 0.648
7 3 0.096
8 4 0.3856
9 5 0.5376
(Code is in Python)
Note that this will get you both the number of possible ways of reaching a certain step (e.g. "first 2, then 3" or "first 3, then 2" for 5), and the probability to reach that step in one go. Of course, if you need only the probability, you can just use single numbers instead of tuples.
I have a data set of the form:
[9.1 5.6 7.4] => 8.5, [4.1 4.4 5.2] => 4.9, ... , x => y(x)
So x is a real vector of three elements and y is a scalar function.
I'm assuming a weighted average model of this data:
y(x) = (a * x[0] + b * x[1] + c * x[2]) / (a+b+c) + E(x)
where E is an unknown random error term.
I need an algorithm to find a,b,c, that minimizes total sum square error:
error = sum over all x of { E(x)^2 }
for a given data set.
Assume that the weights are normalized to sum to 1 (which happily is without loss of generality), then we can re-cast the problem with c = 1 - a - b, so we are actually solving for a and b.
With this we can write
error(a,b) = sum over all x { a x[0] + b x[1] + (1 - a - b) x[2] - y(x) }^2
Now it's just a question of taking the partial derivatives d_error/da and d_error/db and setting them to zero to find the minimum.
With some fiddling, you get a system of two equations in a and b.
C(X[0],X[0],X[2]) a + C(X[0],X[1],X[2]) b = C(X[0],Y,X[2])
C(X[1],X[0],X[2]) a + C(X[1],X[1],X[2]) b = C(X[1],Y,X[2])
The meaning of X[i] is the vector of all i'th components from the dataset x values.
The meaning of Y is the vector of all y(x) values.
The coefficient function C has the following meaning:
C(p, q, r) = sum over i { p[i] ( q[i] - r[i] ) }
I'll omit how to solve the 2x2 system unless this is a problem.
If we plug in the two-element data set you gave, we should get precise coefficients because you can always approximate two points perfectly with a line. So for example the first equation coefficients are:
C(X[0],X[0],X[2]) = 9.1(9.1 - 7.4) + 4.1(4.1 - 5.2) = 10.96
C(X[0],X[1],X[2]) = -19.66
C(X[0],Y,X[2]) = 8.78
Similarly for the second equation: 4.68 -13.6 4.84
Solving the 2x2 system produces: a = 0.42515, b = -0.20958. Therefore c = 0.78443.
Note that in this problem a negative coefficient results. There is nothing to guarantee they'll be positive, though "real" data sets may produce this result.
Indeed if you compute weighted averages with these coefficients, they are 8.5 and 4.9.
For fun I also tried this data set:
X[0] X[1] X[2] Y
0.018056028 9.70442075 9.368093544 6.360312244
8.138752835 5.181373099 3.824747424 5.423581239
6.296398214 4.74405298 9.837741509 7.714662742
5.177385358 1.241610571 5.028388255 4.491743107
4.251033792 8.261317658 7.415111851 6.430957844
4.720645386 1.0721718 2.187147908 2.815078796
1.941872069 1.108191586 6.24591771 3.994268819
4.220448549 9.931055481 4.435085917 5.233711923
9.398867623 2.799376317 7.982096264 7.612485261
4.971020963 1.578519218 0.462459906 2.248086465
I generated the Y values with 1/3 x[0] + 1/6 x[1] + 1/2 x[2] + E where E is a random number in [-0.1..+0.1]. If the algorithm is working correctly we'd expect to get roughly a = 1/3 and b = 1/6 from this result. Indeed we get a = .3472 and b = .1845.
OP has now said that his actual data are larger than 3-vectors. This method generalizes without much trouble. If the vectors are of length n, then you get an n-1 x n-1 system to solve.
I have number A (build from digits 0,1,2,3). I want to find the smallest x and y, that if I do x^y i got the biggest number smaller than A
x^y <= A x^y is maximal
Plus x and y must not be decimal numbers, only "integers"
For example:
A = 7 => x^y = 2^2
A = 28 => x^y = 3^3
A = 33 => x^y = 2^5
etc
Edit:
As izomorphius suggested in comment, it will have always solution for x = A and y = 1. But that is not desirable result. I want x and y to be as much close numbers, as it can be.
A naive solution could be:
The "closest yet not higher" number to A by doing a^y for some constant a is:
afloor(log_a(A)) [where log_a(A) is the logarithm with base a of A, which can be calculated as log(A)/log(a) in most programming languages]
By iterating all as in range [2,A) you can find this number.
This solution is O(A * f(A)) where f(A) is your pow/log complexity
P.S. If you want your exponent (y) be larger then 1, you can simply iterate in range [2,sqrt(A)] - it will reduce the time complexity to O(sqrt(A) * f(A)) - and will get you only numbers with an exponent larger then 1.
It is not clear what you are asking, but I will try to guess.
We first solve the equation z^z = a for a real number z. Let u and v be z rounded down and up, respectively. Among the three candidates (u,u), (v,u), (u,v) we choose the largest one that does not exceed a.
Example: Consder the case a = 2000. We solve z^z = 2000 by numerical methods (see below) to get an approximate solution z = 4.8278228255818725. We round down an up to obtain u = 4 and v = 5. We now have three candidates, 4^4 = 256, 4^5 = 1023 and 5^4 = 625. They are all smaller than 2000, so we take the one that gives the largest answer, which is x = 4, y = 5.
Here is Python code. The function solve_approx does what you want. It works well for a >= 3. I am sure you can cope with the cases a = 1 and a = 2 by yourself.
import math
def solve(a):
""""Solve the equation x^x = a using Newton's method"""
x = math.log(a) / math.log(math.log(a)) # Initial estimate
while abs (x ** x - a) > 0.1:
x = x - (x ** x - a) / (x ** x * (1 + math.log(x)))
return x
def solve_approx(a):
""""Find two integer numbers x and y such that x^y is smaller than
a but as close to it as possible, and try to make x and y as equal
as possible."""
# First we solve exactly to find z such that z^z = a
z = solve(a)
# We round z up and down
u = math.floor(z)
v = math.ceil(z)
# We now have three possible candidates to choose from:
# u ** zdwon, v ** u, u ** v
candidates = [(u, u), (v, u), (u, v)]
# We filter out those that are too big:
candidates = [(x,y) for (x,y) in candidates if x ** y <= a]
# And we select the one that gives the largest result
candidates.sort(key=(lambda key: key[0] ** key[1]))
return candidates[-1]
Here is a little demo:
>>> solve_approx(5)
solve_approx(5)
(2, 2)
>>> solve_approx(100)
solve_approx(100)
(3, 4)
>>> solve_approx(200)
solve_approx(200)
(3, 4)
>>> solve_approx(1000)
solve_approx(1000)
(5, 4)
>>> solve_approx(1000000)
solve_approx(1000000)
(7, 7)
This problem is taken from interviewstreet.com
Given array of integers Y=y1,...,yn, we have n line segments such that
endpoints of segment i are (i, 0) and (i, yi). Imagine that from the
top of each segment a horizontal ray is shot to the left, and this ray
stops when it touches another segment or it hits the y-axis. We
construct an array of n integers, v1, ..., vn, where vi is equal to
length of ray shot from the top of segment i. We define V(y1, ..., yn)
= v1 + ... + vn.
For example, if we have Y=[3,2,5,3,3,4,1,2], then v1, ..., v8 =
[1,1,3,1,1,3,1,2], as shown in the picture below:
For each permutation p of [1,...,n], we can calculate V(yp1, ...,
ypn). If we choose a uniformly random permutation p of [1,...,n], what
is the expected value of V(yp1, ..., ypn)?
Input Format
First line of input contains a single integer T (1 <= T <= 100). T
test cases follow.
First line of each test-case is a single integer N (1 <= N <= 50).
Next line contains positive integer numbers y1, ..., yN separated by a
single space (0 < yi <= 1000).
Output Format
For each test-case output expected value of V(yp1, ..., ypn), rounded
to two digits after the decimal point.
Sample Input
6
3
1 2 3
3
3 3 3
3
2 2 3
4
10 2 4 4
5
10 10 10 5 10
6
1 2 3 4 5 6
Sample Output
4.33
3.00
4.00
6.00
5.80
11.15
Explanation
Case 1: We have V(1,2,3) = 1+2+3 = 6, V(1,3,2) = 1+2+1 = 4, V(2,1,3) =
1+1+3 = 5, V(2,3,1) = 1+2+1 = 4, V(3,1,2) = 1+1+2 = 4, V(3,2,1) =
1+1+1 = 3. Average of these values is 4.33.
Case 2: No matter what the permutation is, V(yp1, yp2, yp3) = 1+1+1 =
3, so the answer is 3.00.
Case 3: V(y1 ,y2 ,y3)=V(y2 ,y1 ,y3) = 5, V(y1, y3, y2)=V(y2, y3, y1) =
4, V(y3, y1, y2)=V(y3, y2, y1) = 3, and average of these values is
4.00.
A naive solution to the problem will run forever for N=50. I believe that the problem can be solved by independently calculating a value for each stick. I still need to know if there is any other efficient approach for this problem. On what basis do we have to independently calculate value for each stick?
We can solve this problem, by figure out:
if the k th stick is put in i th position, what is the expected ray-length of this stick.
then the problem can be solve by adding up all the expected length for all sticks in all positions.
Let expected[k][i] be the expected ray-length of k th stick put in i th position, let num[k][i][length] be the number of permutations that k th stick put in i th position with ray-length equals to length, then
expected[k][i] = sum( num[k][i][length] * length ) / N!
How to compute num[k][i][length]? For example, for length=3, consider the following graph:
...GxxxI...
Where I is the position, 3 'x' means we need 3 sticks that are strictly lower then I, and G means we need a stick that are at least as high as I.
Let s_i be the number of sticks that are smaller then the k th the stick, and g_i be the number of sticks that are greater or equal to the k th stick, then we can choose any one of g_i to put in G position, we can choose any length of s_i to fill the x position, so we have:
num[k][i][length] = P(s_i, length) * g_i * P(n-length-1-1)
In case that all the positions before I are all smaller then I, we don't need a greater stick in G, i.e. xxxI...., we have:
num[k][i][length] = P(s_i, length) * P(n-length-1)
And here's a piece of Python code that can solve this problem:
def solve(n, ys):
ret = 0
for y_i in ys:
s_i = len(filter(lambda x: x < y_i, ys))
g_i = len(filter(lambda x: x >= y_i, ys)) - 1
for i in range(n):
for length in range(1, i+1):
if length == i:
t_ret = combination[s_i][length] * factorial[length] * factorial[ n - length - 1 ]
else:
t_ret = combination[s_i][length] * factorial[length] * g_i * factorial[ n - length - 1 - 1 ]
ret += t_ret * length
return ret * 1.0 / factorial[n] + n
This is the same question as https://cs.stackexchange.com/questions/1076/how-to-approach-vertical-sticks-challenge and my answer there (which is a little simpler than those given earlier here) was:
Imagine a different problem: if you had to place k sticks of equal heights in n slots then the expected distance between sticks (and the expected distance between the first stick and a notional slot 0, and the expected distance between the last stick and a notional slot n+1) is (n+1)/(k+1) since there are k+1 gaps to fit in a length n+1.
Returning to this problem, a particular stick is interested in how many sticks (including itself) as as high or higher. If this is k, then the expected gap before it is also (n+1)/(k+1).
So the algorithm is simply to find this value for each stick and add up the expectation. For example, starting with heights of 3,2,5,3,3,4,1,2, the number of sticks with a greater or equal height is 5,7,1,5,5,2,8,7 so the expectation is 9/6+9/8+9/2+9/6+9/6+9/3+9/9+9/8 = 15.25.
This is easy to program: for example a single line in R
V <- function(Y){(length(Y) + 1) * sum(1 / (rowSums(outer(Y, Y, "<=")) + 1) )}
gives the values in the sample output in the original problem
> V(c(1,2,3))
[1] 4.333333
> V(c(3,3,3))
[1] 3
> V(c(2,2,3))
[1] 4
> V(c(10,2,4,4))
[1] 6
> V(c(10,10,10,5,10))
[1] 5.8
> V(c(1,2,3,4,5,6))
[1] 11.15
As you correctly, noted we can solve problem independently for each stick.
Let F(i, len) is number of permutations, that ray from stick i is exactly len.
Then answer is
(Sum(by i, len) F(i,len)*len)/(n!)
All is left is to count F(i, len). Let a(i) be number of sticks j, that y_j<=y_i. b(i) - number of sticks, that b_j>b_i.
In order to get ray of length len, we need to have situation like this.
B, l...l, O
len-1 times
Where O - is stick #i. B - is stick with bigger length, or beginning. l - is stick with heigth, lesser then ith.
This gives us 2 cases:
1) B is the beginning, this can be achieved in P(a(i), len-1) * (b(i)+a(i)-(len-1))! ways.
2) B is bigger stick, this can be achieved in P(a(i), len-1)*b(i)*(b(i)+a(i)-len)!*(n-len) ways.
edit: corrected b(i) as 2nd term in (mul)in place of a(i) in case 2.