Tarock in numbers
Tarock probabilities explained: trumps in your hand and the talon, trula, the called king, announcements and player statistics.
Quick answers
- How many points is the Mond worth?
- The Mond (XXI) has a basic value of 5 points, as do Pagat (I) and Škis. Count cards in groups of three and subtract 2 per complete group; the 54-card deck totals 70 points.
- How likely is a trump in the talon?
- Before seeing your hand, a random six-card talon contains at least one trump with probability 96.49%. This assumes 22 trumps in a uniformly shuffled 54-card deck. Your known hand changes this probability.
- How likely is the complete trula in the starting hand?
- Before using the talon, the probability of holding Pagat, Mond and Škis together is 0.89% in a 12-card hand with four players, or 2.26% in a 16-card hand with three players. Holding them does not guarantee capturing them.
- What is the probability that the called king is in the talon?
- With four players, if the king is absent from your 12 cards and no further information is known, its probability of being in the six-card talon is 6 / 42 = 14.29%. Under the portal rules, the declarer then plays alone.
- Does a 60% win rate after ten games prove playing strength?
- No. Six wins in ten games is a small sample. In a simplified model of comparable independent games, the approximate 95% Wilson interval is 31.3–83.2%. Consider sample size, opponents, partners and game variant.
01 What do the calculations assume?
The deck has 22 trumps and 32 suit cards: eight in each of four suits. We assume a fair shuffle and random dealing without replacement. The tables describe the initial deal before selecting from the talon and discarding.
| Variant | Cards in hand | Talon | Total |
|---|---|---|---|
| 4 players | 12 | 6 | 54 |
| 3 players | 16 | 6 | 54 |
02 Card values and counting
Distinguish card points, game points and player rating. Count cards in groups of three: add their basic values and subtract two. A king, queen and ordinary card yield 5 + 4 + 1 − 2 = 8 points. The complete deck contains 70 points.
| Card | Basic value |
|---|---|
| Škis — Sküs | 5 |
| Mond — XXI | 5 |
| Pagat — I | 5 |
| King | 5 |
| Queen | 4 |
| Cavalier | 3 |
| Jack | 2 |
| Other trumps and ordinary suit cards | 1 |
03 How many trumps are usual in a hand?
Eight trumps are rare with four players but much more common with three, because each player receives more cards. The average describes many deals, not a promise about your next hand.
| Measure | 4 players · 12 cards | 3 players · 16 cards |
|---|---|---|
| Average number of trumps | 4.89 | 6.52 |
| Probability of at least 8 trumps | 4.17 % | 27.45 % |
| Probability of no trumps | 0.0658 % | 0.00285 % |
| No trumps approximately once per this many deals | 1,519 | 35,095 |
04 How many trumps are in the talon?
Before you see your hand, a random six-card talon has the following distribution. Two trumps are the most common outcome, with three almost as common.
| Trumps in the talon | Probability |
|---|---|
| 0 | 3.51 % |
| 1 | 17.15 % |
| 2 | 32.16 % |
| 3 | 29.58 % |
| 4 | 14.05 % |
| 5 | 3.26 % |
| 6 | 0.29 % |
05 Your hand changes the talon probability
With four players, 42 cards remain unknown after you receive 12. The more trumps you hold, the fewer remain outside your hand. This table uses only your cards, with no additional information from play.
| Trumps in your hand | At least one trump in talon | Expected trumps in talon |
|---|---|---|
| 2 | 98.58 % | 2.86 |
| 4 | 97.43 % | 2.57 |
| 6 | 95.61 % | 2.29 |
| 8 | 92.82 % | 2.00 |
| 10 | 88.68 % | 1.71 |
06 How rare is the trula?
The trula consists of Pagat, Mond and Škis. The table counts these cards in your initial hand before using the talon. All three arrive about once every 113 deals with four players, or every 44 with three.
| Trula cards in hand | 4 players | 3 players |
|---|---|---|
| None | 46.28 % | 34.01 % |
| One | 41.65 % | 45.35 % |
| Two | 11.18 % | 18.38 % |
| All three | 0.89 % | 2.26 % |
07 Where is the called king?
In a four-player game, the king you call is absent from your 12 cards. Without further information it is one of 42 unknown cards, so use 6 / 42 rather than 6 / 54.
| Location of the called king | Probability |
|---|---|
| With the first other player | 28.57 % |
| With the second other player | 28.57 % |
| With the third other player | 28.57 % |
| In the talon | 14.29 % |
08 How often is a suit missing?
Each suit has eight cards. Distinguish having no cards of a predetermined suit, such as hearts, from being void in any suit. These possibilities overlap: you cannot simply multiply the first probability by four.
| Event in the initial hand | 4 players | 3 players |
|---|---|---|
| Void in a specific suit | 11.34 % | 4.70 % |
| Void in at least one of four suits | 40.74 % | 18.17 % |
09 When is an announcement worthwhile?
A simplified example: an unannounced success scores +10 and failure 0; an announced success scores +20 and failure −20. Assume the same success probability either way, without counters or other multipliers. The table shows expected points, not a single game's result.
| Success probability | Without announcement | With announcement |
|---|---|---|
| 50 % | +5 | 0 |
| 60 % | +6 | +4 |
| 70 % | +7 | +8 |
| 80 % | +8 | +12 |
10 Why are ten games not enough?
All three examples show 60% wins, but different uncertainty. The ranges are approximate 95% Wilson intervals for comparable independent games. These are teaching examples, not portal statistics or guarantees of future results.
| Wins / games | Observed win rate | 95% interval |
|---|---|---|
| 6 / 10 | 60 % | 31.3–83.2 % |
| 60 / 100 | 60 % | 50.2–69.1 % |
| 600 / 1 000 | 60 % | 56.9–63.0 % |
How are the probabilities calculated?
We use the hypergeometric distribution. N is the unknown card pool, K the desired cards, n the sample size and k the desired cards drawn. C(a, b) counts combinations. For an unseen talon, N = 54, K = 22, n = 6; after seeing your hand, adjust N and K. E[X] is the expected count. The mathematical tables are our calculations; the sources below explain rules, which can differ between variants.
P(X = k) = C(K, k) × C(N − K, n − k) / C(N, n)
E[X] = n × K / N