Problem Analysis #probability_28

Recorded Python call:

import itertools
from math import factorial

# Die faces 1-6. For each face, which boxes get a ball:
# odd (1,3,5): boxes 1,3,5
# even: divisors
box_map = {
    1: [1,3,5],
    2: [1,2],
    3: [1,3,5],
    4: [1,2,4],
    5: [1,3,5],
    6: [1,2,3,6],
}

# Enumerate all (a,b,c,d,e,f) with sum 4
configs = []
for a in range(5):
    for b in range(5):
        for c in range(5):
            for d in range(5):
                for e in range(5):
                    for f in range(5):
                        if a+b+c+d+e+f == 4:
                            configs.append((a,b,c,d,e,f))

total_prob = 0
prob_odd = 0
prob_favorable = 0

favorable_configs = []
odd_configs = []

for (a,b,c,d,e,f) in configs:
    # count balls in each box
    counts = [0]*7  # 1-indexed
    counts[1] = a*1 + b*1 + c*1 + d*1 + e*1 + f*1  # box1 always +1 per roll
    # Actually let me compute properly
    # box1: from every roll
    counts[1] = 4
    # box2: from rolls 2,4,6 -> b,d,f
    counts[2] = b + d + f
    # box3: from odd rolls (1,3,5) -> a,c,e and roll 6 -> f
    counts[3] = a + c + e + f
    # box4: from roll 4 -> d
    counts[4] = d
    # box5: from odd rolls -> a,c,e
    counts[5] = a + c + e
    # box6: from roll 6 -> f
    counts[6] = f
    total = sum(counts[1:])
    
    multinomial = factorial(4) // (factorial(a)*factorial(b)*factorial(c)*factorial(d)*factorial(e)*factorial(f))
    prob = multinomial / (6**4)
    total_prob += prob
    
    is_odd = (total % 2 == 1)
    box3_minus_box2 = counts[3] - counts[2]
    is_favorable = is_odd and (box3_minus_box2 == 1)
    
    if is_odd:
        odd_configs.append(((a,b,c,d,e,f), counts, total))
    if is_favorable:
        favorable_configs.append(((a,b,c,d,e,f), counts, total))

print("Total prob check:", total_prob)
print("Number of configs:", len(configs))
print("Number of odd configs:", len(odd_configs))
print("Number of favorable configs:", len(favorable_configs))
print()
print("P(total odd):", sum(p for _,_,_ in [(c,_,p) for c,_,_ in [(x,_,0) for x in []]]) if False else "")
# Recompute properly
prob_odd_val = 0
prob_fav_val = 0
for (a,b,c,d,e,f) in configs:
    counts = [0, 4, b+d+f, a+c+e+f, d, a+c+e, f]
    total = sum(counts[1:])
    multinomial = factorial(4) // (factorial(a)*factorial(b)*factorial(c)*factorial(d)*factorial(e)*factorial(f))
    prob = multinomial / (6**4)
    if total % 2 == 1:
        prob_odd_val += prob
    if total % 2 == 1 and counts[3]-counts[2] == 1:
        prob_fav_val += prob

print("P(total odd) =", prob_odd_val)
print("P(total odd AND box3-box2=1) =", prob_fav_val)
print("Conditional probability =", prob_fav_val / prob_odd_val)
print()
print("All odd configs:")
for (a,b,c,d,e,f), counts, total in odd_configs:
    print(f"  ({a},{b},{c},{d},{e},{f}) counts={counts[1:]} total={total} b3-b2={counts[3]-counts[2]}")
print()
print("All favorable configs:")
for (a,b,c,d,e,f), counts, total in favorable_configs:
    print(f"  ({a},{b},{c},{d},{e},{f}) counts={counts[1:]} total={total} b3-b2={counts[3]-counts[2]}")

Recorded Python call:

from math import factorial
from fractions import Fraction

# Count with exact integer weights (multinomial coefficients)
odd_weight = 0
fav_weight = 0
configs = []
for a in range(5):
    for b in range(5):
        for c in range(5):
            for d in range(5):
                for e in range(5):
                    for f in range(5):
                        if a+b+c+d+e+f == 4:
                            counts = [0, 4, b+d+f, a+c+e+f, d, a+c+e, f]
                            total = sum(counts[1:])
                            mc = factorial(4)//(factorial(a)*factorial(b)*factorial(c)*factorial(d)*factorial(e)*factorial(f))
                            if total % 2 == 1:
                                odd_weight += mc
                            if total % 2 == 1 and counts[3]-counts[2] == 1:
                                fav_weight += mc

print("Odd weight:", odd_weight, "=", Fraction(odd_weight, 6**4))
print("Fav weight:", fav_weight, "=", Fraction(fav_weight, 6**4))
print("Conditional:", Fraction(fav_weight, odd_weight))
print("6^4 =", 6**4)

Recorded Python call:

from math import factorial

# Verify total weight = 6^4
total_weight = 0
for a in range(5):
    for b in range(5):
        for c in range(5):
            for d in range(5):
                for e in range(5):
                    for f in range(5):
                        if a+b+c+d+e+f == 4:
                            mc = factorial(4)//(factorial(a)*factorial(b)*factorial(c)*factorial(d)*factorial(e)*factorial(f))
                            total_weight += mc
print("Total weight:", total_weight, "= 6^4 =", 6**4)

# Also verify odd weight by independent method: P(b+f odd)
# b = count of 2s, f = count of 6s in 4 rolls
# Each roll: P(2)=1/6, P(6)=1/6, P(other)=4/6
# b+f = number of rolls that are 2 or 6
# P(b+f odd) = sum over odd k of C(4,k) * (2/6)^k * (4/6)^(4-k)
# where 2/6 = P(2 or 6), 4/6 = P(other)

import math
odd_check = 0
for k in range(5):
    if k % 2 == 1:
        odd_check += math.comb(4, k) * (2/6)**k * (4/6)**(4-k)
print("P(b+f odd) via binomial:", odd_check)
print("640/1296 =", 640/1296)

Recorded Python call:

from math import factorial
from fractions import Fraction

# Final clean computation with exact fractions
# Let a,b,c,d,e,f = counts of faces 1..6 in 4 rolls, sum=4
# Box contents:
# B1=4, B2=b+d+f, B3=a+c+e+f, B4=d, B5=a+c+e, B6=f
# Total = 4 + (b+d+f)+(a+c+e+f)+d+(a+c+e)+f = 4+2a+b+2c+2d+2e+3f
# Parity: Total ≡ b+f (mod 2). Total odd ⟺ b+f odd.
# Box3 - Box2 = (a+c+e+f)-(b+d+f) = (a+c+e)-(b+d)
# Need Box3-Box2=1 ⟺ a+c+e = b+d+1
# With S=a+c+e, T=b+d+f, S+T=4:
#   S = b+d+1, T = b+d+f, so 2(b+d)+f=3
#   f odd: f=1→b+d=1; f=3→b+d=0
#   Total odd: b+f odd. f=1→b odd... wait b+1 odd ⟺ b even → b=0,d=1. f=3→b+3 odd ⟺ b even→b=0,d=0.
# So favorable: (f=1,b=0,d=1,S=2) or (f=3,b=0,d=0,S=1)

# Weights
# Case f=1,b=0,d=1,S=2: sum_{a+c+e=2} 4!/(a!c!e!1!1!) 
w1 = sum(factorial(4)//(factorial(a)*factorial(c)*factorial(e)*1*1)
         for a in range(3) for c in range(3) for e in range(3)
         if a+c+e==2)
# Case f=3,b=0,d=0,S=1: sum_{a+c+e=1} 4!/(a!c!e!3!)
w2 = sum(factorial(4)//(factorial(a)*factorial(c)*factorial(e)*factorial(3))
         for a in range(2) for c in range(2) for e in range(2)
         if a+c+e==1)
fav = w1 + w2
print("Favorable weight:", fav)

# Odd total weight: P(b+f odd) = sum_{k odd} C(4,k)(2/6)^k(4/6)^(4-k) * 6^4
import math
odd_w = sum(math.comb(4,k)*(2)**k*(4)**(4-k) for k in range(5) if k%2==1)
print("Odd weight:", odd_w)
print("Conditional:", Fraction(fav, odd_w))

Solution

Let $a,b,c,d,e,f$ be the number of times faces $1,2,3,4,5,6$ appear in 4 rolls, with $a+b+c+d+e+f=4$.

Box contents (each roll contributes):

• Odd roll (1,3,5): boxes $1,3,5$ each get $+1$
• Even roll: box $k$ gets $+1$ in each divisor-box of $k$
  • $k=2\to$ boxes $\{1,2\}$, $k=4\to\{1,2,4\}$, $k=6\to\{1,2,3,6\}$

So:

Total $T = 4+(b+d+f)+(a+c+e+f)+d+(a+c+e)+f = 4+2a+b+2c+2d+2e+3f$.

Modulo 2: $T \equiv b+f \pmod 2$. So $T$ is odd $\iff b+f$ is odd.

Condition $B_3 = B_2+1$:

Let $S=a+c+e$ (odd-roll count), $T'=b+d+f$ (even-roll count), $S+T'=4$. Then:

Substituting $S+T'=4$ and $T'=b+d+f$:

Since $2(b+d)$ is even, $f$ must be odd: $f\in\{1,3\}$.

• $f=1$: $b+d=1$. Parity $b+f$ odd $\Rightarrow$ $b+1$ odd $\Rightarrow$ $b$ even $\Rightarrow$ $b=0,d=1$. Then $S=2$ (so $a+c+e=2$): $\binom{4}{2}=6$ solutions.
• $f=3$: $b+d=0\Rightarrow b=0,d=0$. Parity $b+f$ odd: $0+3$ odd ✓. Then $S=1$ (so $a+c+e=1$): $\binom{3}{1}=3$ solutions.

Weighted probability (multinomial weight $4!/(a!b!c!d!e!f!)$, total weight $6^4=1296$):

Favorable weight:

• $f=1,b=0,d=1,S=2$: $\sum_{a+c+e=2}\frac{24}{a!\,c!\,e!}=12+12+12+24+24+24=108$
• $f=3,b=0,d=0,S=1$: $\sum_{a+c+e=1}\frac{4}{a!\,c!\,e!}=4+4+4=12$
• Total favorable $=120$.

Odd-total weight: $T$ odd $\iff b+f$ odd, where $b+f\sim\text{Binomial}(4,2/6)$:

Therefore:

$\boxed{⑤}$