问题
I have a probability problem, which I need to simulate in a reasonable amount of time. In simplified form, I have 30 unfair coins each with a different known probability. I then want to ask things like "what is the probability that exactly 12 will be heads?", or "what is the probability that AT LEAST 5 will be tails?".
I know basic probability theory, so I know I can enumerate all (30 choose x) possibilities, but that's not particularly scalable. The worst case (30 choose 15) has over 150 million combinations. Is there a better way to approach this problem from a computational standpoint?
Any help is greatly appreciated, thanks! :-)
回答1:
You can use a dynamic programming approach.
For example, to calculate the probability of 12 heads out of 30 coins, let P(n, k) be the probability that there's k heads from the first n coins.
Then P(n, k) = p_n * P(n - 1, k - 1) + (1 - p_n) * P(n - 1, k)
(here p_i is the probability the i'th coin is heads).
You can now use this relation in a dynamic programming algorithm. Have a vector of 13 probabilities (that represent P(n - 1, i) for i in 0..12). Build a new vector of 13 for P(n, i) using the above recurrence relation. Repeat until n = 30. Of course, you start with the vector (1, 0, 0, 0, ...) for n=0 (since with no coins, you're sure to get no heads).
The worst case using this algorithm is O(n^2) rather than exponential.
回答2:
This is actually an interesting problem. I was inspired to write a blog post about it covering in detail fair vs unfair coin tosses all the way to the OP's situation of having a different probability for each coin. You need a technique called dynamic programming to solve this problem in polynomial time.
General Problem: Given C, a series of n coins p1 to pn where pi represents the probability of the i-th coin coming up heads, what is the probability of k heads coming up from tossing all the coins?
This means solving the following recurrence relation:
P(n,k,C,i) = pi x P(n-1,k-1,C,i+1) + (1-pi) x P(n,k,C,i+1)
A Java code snippet that does this is:
private static void runDynamic() {
long start = System.nanoTime();
double[] probs = dynamic(0.2, 0.3, 0.4);
long end = System.nanoTime();
int total = 0;
for (int i = 0; i < probs.length; i++) {
System.out.printf("%d : %,.4f%n", i, probs[i]);
}
System.out.printf("%nDynamic ran for %d coinsin %,.3f ms%n%n",
coins.length, (end - start) / 1000000d);
}
private static double[] dynamic(double... coins) {
double[][] table = new double[coins.length + 2][];
for (int i = 0; i < table.length; i++) {
table[i] = new double[coins.length + 1];
}
table[1][coins.length] = 1.0d; // everything else is 0.0
for (int i = 0; i <= coins.length; i++) {
for (int j = coins.length - 1; j >= 0; j--) {
table[i + 1][j] = coins[j] * table[i][j + 1] +
(1 - coins[j]) * table[i + 1][j + 1];
}
}
double[] ret = new double[coins.length + 1];
for (int i = 0; i < ret.length; i++) {
ret[i] = table[i + 1][0];
}
return ret;
}
What this is doing is constructing a table that shows the probability that a sequence of coins from pi to pn contain k heads.
For a deeper introduction to binomial probability and a discussion on how to apply dynamic programming take a look at Coin Tosses, Binomials and Dynamic Programming.
回答3:
Pseudocode:
procedure PROB(n,k,p)
/*
input: n - number of coins flipped
k - number of heads
p - list of probabilities for n-coins where p[i] is probability coin i will be heads
output: probability k-heads in n-flips
assumptions: 1 <= i <= n, i in [0,1], 0 <= k <= n, additions and multiplications of [0,1] numbers O(1)
*/
A = ()() //matrix
A[0][0] = 1 // probability no heads given no coins flipped = 100%
for i = 0 to k //O(k)
if i != 0 then A[i][i] = A[i-1][i-1] * p[i]
for j = i + 1 to n - k + i //O( n - k + 1 - (i + 1)) = O(n - k) = O(n)
if i != 0 then A[i][j] = p[j] * A[i-1][j-1] + (1-p[j]) * A[i][j-1]
otherwise A[i][j] = (1 - p[j]) * A[i][j-1]
return A[k][n] //probability k-heads given n-flips
Worst case = O(kn)
来源：https://stackoverflow.com/questions/3519395/probability-of-outcomes-algorithm