Key takeaways

  • Build the union of effective tiles and remove every visible copy before obtaining live count K.
  • The next-draw probability is K/N, where N is the unknown draw pool defined by the model.
  • Multiple draws are sampling without replacement; use one minus the probability of missing all draws.
  • Multiplying one-draw probability by the number of draws is only a rough approximation and can overstate the result.
  • If the hand, discard, or wait changes after a draw, fixed-K formulas are insufficient; expand the states.

First define denominator N

“Eight effective tiles” is not a probability until the draw pool is stated.

Two questions are often confused:

  1. From your information set: tiles not visible to you may be in the wall or opponents’ hands.
  2. A simplified live-wall model: assume a known draw pool of N tiles contains K effective tiles.

An article or tool must state which model it uses. Treating every unknown tile as physically in the wall ignores opponents’ concealed tiles. It can still serve as a conditional estimate when clearly labeled, not as known fact.

Then build effective set K

The block 45m theoretically accepts four 3m and four 6m. Suppose one 3m is in your hand, two 6m are discarded, and an open group contains another 3m. Four effective copies are visible, leaving four unknown.

If another block also accepts 6m, those same copies cannot be added again. Build the union of effective tile types for the whole hand, then subtract visible copies by type.

One-draw probability

Under a simplified draw-pool model:

next-draw hit probability = K / N

For N = 60 and K = 8:

8 / 60 = 0.1333…

The next draw is about 13.3%. This does not mean the tile is guaranteed within an average of eight turns, and it does not include an opponent discarding a tile that can be claimed.

At least one hit in two draws

The first miss probability is (N-K)/N. After a miss is removed, the conditional second miss probability is (N-K-1)/(N-1).

at least one hit in two draws
= 1 - [(N-K)/N × (N-K-1)/(N-1)]

For N = 60 and K = 8:

1 - (52/60 × 51/59)
≈ 25.1%

Simply doubling 13.3% produces 26.7%, a slight overestimate because the samples are not independent.

At least one hit in n draws

If the effective set remains fixed:

at least one hit
= 1 - C(N-K, n) / C(N, n)

C(a,b) is the number of combinations of b objects selected from a. The ratio is the probability that all n draws come from the non-effective tiles.

The same calculation can be written as a product of miss probabilities:

1 - [(N-K)/N
     × (N-K-1)/(N-1)
     × …
     × (N-K-n+1)/(N-n+1)]

NIST describes the hypergeometric distribution as sampling from a finite population without replacement, which is the mathematical basis for this fixed-set calculation.

Why not multiply by the number of turns?

n × K/N ignores:

  1. Removed draws changing the denominator.
  2. A path that hits on the first draw and should not be counted again as a first hit.
  3. A successful Mahjong draw changing the hand and therefore changing K.

It can be a mental approximation when K/N and n are small. Published calculations should use the without-replacement formula or explicit state expansion.

Two draws to tenpai is not two draws to win

A one-shanten hand usually reaches tenpai after the first successful improvement. The second draw must then hit a new completion set. Keeping the original one-shanten K fixed for two draws does not calculate two-draw win probability.

A correct tree:

  1. Lists every shanten-reducing first draw.
  2. Selects the new discard after each draw.
  3. Recalculates the resulting wait and live copies.
  4. Weights each second-draw route by its first-draw probability.

This is why finite-horizon self-draw probability is more involved than multiplying ukeire.

Self-draw, discard win, and calls differ

Self-draw counts your own wall draw. A discard win depends on opponents’ concealed tiles, behavior, furiten, and rules. Chow and pung further depend on seat, priority, and strategy.

Therefore:

  • The next self-draw can use a draw model.
  • “Someone discards it this turn” needs a behavior model.
  • Final hand win rate also needs competition, draws, defense, and rules.

One K/N cannot answer all three.

Which visible tiles adjust the model?

Count:

  • Your own hand
  • Every discard row
  • Chows, pungs, and exposed kongs
  • Publicly revealed concealed kongs
  • Dora indicators in Riichi
  • Any other identifiable public tile under the rules

Unknown tiles can be in the wall or opponents’ hands. Visible counting removes impossibilities but does not locate the unknown copies.

Common errors

  • Using tile-type count instead of remaining copies.
  • Counting overlapping acceptance twice.
  • Claiming every unknown tile is in the wall.
  • Multiplying a one-draw rate by turns.
  • Failing to recalculate waits after leaving one-shanten.
  • Treating self-draw probability as total win probability including calls.

Scope and limitations

The fixed-K hypergeometric model answers a conditional sampling question. Discards, opponents’ draws, calls, kongs, and route changes alter the real state. Every percentage should be presented with N, K, draws, rules, and assumptions rather than as a context-free fact.

Understanding finite horizons in Tingpai Lab

Tingpai Lab reevaluates the new hand after different draws instead of extending one fixed acceptance count. Practice by calculating the next draw’s K/N, then inspect how new waits, overlap, and discard choices change a two- or three-draw result.

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