Chromatic Orb Probability Formula in Path of Exile

Per-Socket Color Probability

Each socket rolls its color independently based on the item’s attribute requirements:

P(Red)   = (Strength + 10)     / (STR + DEX + INT + 30)
P(Green) = (Dexterity + 10)    / (STR + DEX + INT + 30)
P(Blue)  = (Intelligence + 10) / (STR + DEX + INT + 30)

The +10 floor ensures no color has zero probability even on items with zero requirements in that attribute. The three probabilities always sum to 1.

Example: 100 STR, 0 DEX, 0 INT

Total = 100 + 0 + 0 + 30 = 130
P(Red)   = 110 / 130 = 84.6%  ← wrong, let me recalc:

Total = 130
P(Red)   = (100+10)/130 = 110/130 = 84.6%

Wait — example in main MDX uses STR=100, DEX=0, INT=0: P(Red) = 110/130 = 84.6%, P(Green) = 10/130 = 7.7%, P(Blue) = 10/130 = 7.7%

Note: items typically have requirements in multiple attributes. A body armour with 100 STR, 50 DEX: P(Red) = 110/190 = 57.9%, P(Green) = 60/190 = 31.6%, P(Blue) = 20/190 = 10.5%

Multinomial Probability for Full Configuration

A chromatic orb recolors all sockets simultaneously. The probability of getting exactly r Red, g Green, and b Blue sockets in an N-socket item:

P(r,g,b) = [N! / (r! × g! × b!)] × P(Red)^r × P(Green)^g × P(Blue)^b

This is the multinomial probability. The coefficient N!/(r!×g!×b!) counts the number of ways to arrange the colors across positions.

Example: 4-socket item, 100 STR, want 1R 1G 2B

pR = 110/130 = 0.846, pG = 10/130 = 0.077, pB = 10/130 = 0.077

P = [4!/(1!×1!×2!)] × 0.846^1 × 0.077^1 × 0.077^2
  = 12 × 0.846 × 0.077 × 0.00593
  = 12 × 0.000387
  = 0.00464  →  0.464%

Expected orbs = 1 / 0.00464 ≈ 216

Geometric Distribution and Percentiles

Each chromatic roll is an independent Bernoulli trial with success probability p. The number of rolls until first success follows a geometric distribution.

Expected rolls: 1/p

Percentile k (probability of succeeding within k rolls = q):

k = ceil(log(1 - q) / log(1 - p))

For q = 0.90 (90th percentile), q = 0.99 (99th percentile).

For p = 0.00464 (example above):

• Expected: 216 orbs
• 90th percentile: ceil(log(0.10)/log(0.99536)) = ceil(2.303/0.00465) = 496 orbs
• 99th percentile: ceil(log(0.01)/log(0.99536)) = ceil(4.605/0.00465) = 991 orbs

The ratio of 99th to expected is roughly 4.6x for geometric distributions — off-color crafts are highly variance-prone.

Vorici Bench Hybrid Method

When using the bench to guarantee k sockets of color X, then rolling remaining sockets freely:

Total expected = bench_cost[k] + (1 / P_remaining)

P_remaining = multinomial probability for remaining (N-k) sockets
              with target = (r - k_R, g - k_G, b - k_B)
              using the same per-socket probabilities

If the remaining socket count is 0 (bench covers all desired sockets): total = bench_cost only.

The bench is efficient when P_remaining is significantly higher than P_full, meaning the guaranteed sockets remove the hardest-to-hit constraint.