Independency

If we have:

\[P(A) \cdot P(B) = P(A,B)\]

\[P(A) = P(A|B)\]

We say that \(A\) is independent with \(B\), as \(A \perp B\)

import random

def sample():
    a_done = False
    count = 0
    while not a_done:
        a = random.random() < 0.5
        count += 1
        #print(a, b)
        if count == 1:
            old_a = a
        else:
            if a != old_a: a_done = True
    return count

n = 100000
print(sum(sample() for i in range(n)) / n)

def sample():
    a_done = False
    b_done = False
    count = 0
    while not a_done or not b_done:
        a = random.random() < 0.5
        b = random.random() < 0.5
        count += 2
        #print(a, b)
        if count == 2:
            old_a = a
            old_b = b
        else:
            if a != old_a: a_done = True
            if b != old_b: b_done = True
    return count

n = 1000000
print(sum(sample() for i in range(n)) / n)

def sample():
    a_done = False
    b_done = False
    count = 0
    while not a_done and not b_done:
        a = random.random() < 0.5
        b = random.random() < 0.5
        count += 2
        #print(a, b)
        if count == 2:
            old_a = a
            old_b = b
        else:
            if a != old_a: a_done = True
            if b != old_b: b_done = True
    return count

n = 1000000
print(sum(sample() for i in range(n)) / n)

\(|\Omega| = 52 \times 51 \times 50 \times 49 \times 48\)

\(\(P[X=1] = 0\)\) 123456789abcd 11112 11122 11222 12222

\[P[X=2] = \frac{\binom{52}{1} \binom{51}{1} \binom{50}{1} \binom{49}{1} \binom{48}{1} (2^5 - 2)}{\frac{52!}{(52-5)!}}\]

ABCDE \(\(P[X=5] = \frac{13 * 12 * 11 * 10 * 9 * 4^5}{52 * 51 * 50 * 49 * 48}\)\)

ABCDA ABCDB ABCDC ABCDD

\[P[X=4] = \frac{13*12*11*10*4*4^4}{52 * 51 * 50 * 49 * 48}\]

ABCAA ABCAB ABCAC ABCBA ABCBB ABCBC ABCCA ABCCB ABCCC \(\(P[X=3] = \frac{13*12*11*3^2*4^4}{52 * 51 * 50 * 49 * 48}\)\)