One of the three players will be Dizhu (meaning landlord) and will utimately have 20 cards at the end of the bidding phase. The rest two are called farmers who will have 17 cards each. A game ends if a player had played all of its cards, given several categories of playable hands.

I found the most complex and confusing category of playable hands to be

*Suits do not matter*in this game. So from now on I will treat all cards of the same rank equivalently. This entails that the numbers of cards for each rank become the only relevant information.- The "Rocket" category (two Jokers) and the "Bomb" category (four-of-a-rank) predominate over all other categories. (Rocket domaintes all bombs, and within bombs the card ranks decide the order).
- During a trick a player (except the first one) can only play the hands of higher rank within the same category of the previous plays.
- For some categories, there will be a
*primal*and a*secondary*components of the hand. To compare two hands of the same category, it will be based only on the order of primals. - The order for solo cards are ColoredJoker>BlackWhiteJoker>2>A>K>Q>J>10>9...>4>3. So there are 15 different solo ranks possible. However for categories that involve
*consecutive*components, 2-s and Jokers are forbidden. For example, 3-4-5-6-7 is playable, while J-Q-K-A-2 is not.

For example, it says the hand 333-444 + 5-6 is the lowerest rank of the shortest chain length (2 consecutive trios). While 101010-JJJ-QQQ-KKK-AAA + 7-8-9-2-colored joker is the highest rank of the longest chain possible (since the maximum number of hands a player can have is 20). Here the consecutive parts are the primals while the kickers are the secondaries.

Also it worth pointing out that the order among the solo cards does not matter: 333-444 + 5-6 and 333-444 + 6-5 are the same, so it really didn't specify which trio "carries" which solo card.

However, Wikipedia also suggests that pair cannot be kickers. For example, 333-444 + 5-5 is not allowed under this rule. This contradicts to the local playing custom I am familaring with. And also to make my calculation more fun, I will use the following rule for computing the answer:

- Pair can be kickers.

The first question is: can \(X\) or \(Y\) be one of B or C? If so then it will make it a

Therefore, given a consecutive trios of length 2, there are \(\binom{13}{2}-1\) possible \(\{X, Y\}\). This is because we should subtract the 2 different ranks within the trios out of the 15. And we should also subtract the case where \(\{X, Y\}=\{ColoredJoker, BlackWhiteJoker\}\).

For pattern \(\{X, X\}\), there are 11 possibilities, since 2 of the 13 ranks (excluding Jokers) had already been taken by the trios.

Therefore, the number of possible combinations for consecutive trios of length 2 is \[ 11\times\left(\binom{13}{2}-1+11\right)=968\]

For pattern \(\{X, Y, Z\}\), there are \(\binom{12}{3}-10\) possibilities, where in \(-10\) we eliminate the cases of \(\{X, ColoredJoker, BlackWhiteJoker\}\).

For pattern \(\{X, X, Y\}\), there are \(10\times 11\), where 10 corresponds to if we first enumerate \(X\) (exclude Jokers) and 11 corresponds to then we enumerate \(Y\) (including Jokers).

Now here comes the tricky part: do we need to consider pattern \(\{X, X, X\}\)? For example if we play 666-777-888-A-A-A, that probably should be acceptable. Although playing A-A-A as the carried cards looks like a waste, if A-A-A are the only cards remaining, why not playing it?

However, how about 666-777-888-9-9-9? Now here is a confusion: we can actually also interpret it as 777-888-999-6-6-6. Furthermore, it can also be interpreted as the TrioChain category: 666-777-888-999.

Now, in real human plays, people can definitely resolve the confusion by explicitly saying which ones they want to play. However if we were to write a computer program, some pre-determined tie-breaking rules should be established. In the few computer Dou Dizhu programs I examined on (with graphical interfaces), they tends to interpret it as 666-777-888-999 if it was first played in a trick. But if it is played during the middle of a trick and the previous plays were airplane + solo, it is unclear which one it was interpreted to.

However, if you think about it, 666-777-888-9-9-9 and 777-888-999-6-6-6 are strategically equivalent during the middle of a trick: the hands that

So, are we going to view 666-777-888-9-9-9 and 777-888-999-6-6-6 as the same primitive action? Well, I would argue that depends on the endgoal of you application. If you are going to build an effective AI bot: sure, incorporating such inductive biases typically will speed up the learning process. In fact, such technique is called

However, if you are going to implement a game engine: probably not. You should still treat them as two different primitive actions. Although they are

For my computation, I decide to treat 666-777-888-9-9-9 and 777-888-999-6-6-6 as

Then, for pattern \(\{X, X, X\}\), there should be 10 possibilities, after excluding the three ranks in the airplane and the two Jokers. That makes the answer for this part \[ 10\times \left(\binom{12}{3}-10+10\times 11+10\right)=3300\]

My answer for this part is \[ 9\times \left(\binom{11}{4}-\binom{9}{2}+\binom{9}{1}\cdot\left(\binom{10}{2}-1\right)+\binom{9}{2}+9\times10\right)=7344\] Where each terms within the most outer bracket corresponds to each patterns.

- Rocket/Bombs patterns cannot appear in other categories.
- 333-444-555-6-6-6, 444-555-666-3-3-3, 333-444-555-666 are all different actions.

- Pairs and trios can be kickers, as long as they do not violate other rules.