In "minefield mahjong", your goal is to make the highest scoring valid hand out of a group of tiles. I am wondering the possibility of there being no valid hands in the set of tiles.
The makeup of a set:
- 3 suits, each containing 4 copies of the numbers 1-9
- 7 honors, with 4 copies of each
for a total of 136 tiles.
In a minefield, you have 34 tiles. A minefield that cannot create a valid hand includes the following:
- 1 of each honor
- 6 unique tiles from each suit, of which no 3 make a same-suit run (9 valid possibilities per suit*)
- and then either:
- 3 more copies each of 3 different tiles already drawn (making 3 four of a kinds and 22 unique tiles), or,
- 3 copies each of 2 different tiles already drawn and 1 additional copy of 3 different tiles already drawn (making 2 four of a kinds, 3 pairs and 20 unique tiles)
The math I tried never seems to be working out. I know the total number of minefields is $\binom{136}{34}$, but I'm not sure how to calculate the number that fulfill the criteria.
*$[124578, 124589, 124579, 124689, 125689, 134679, 134689, 135689, 235689]$ are the valid combinations in a suit
Best Answer
There are $9$ choices for the non-honor cards in each suit, so we have $9^3=729$ ways to choose which tile values will occur in the hand. In any event there will be $25$ distinct tile vaules. In the first case, there are $\binom{25}{3}$ ways to choose which tiles will be four-of-a-kinds, and $4^{22}$ ways to choose the tiles that occur only once. Of course there's only one way to choose which tiles make up a four-of-a-kind. Therefore, case $1$ gives $$9^3\cdot4^{22}\binom{25}{3}$$ possibilities.
In the second case, we have $\binom{25}{2}$ ways to choose the ranks of the four-of-a-kinds, and then $\binom{23}{3}$ ways to choose the ranks of the pairs. For the pairs, we have $\binom{4}{2}=6$ ways to choose which tiles make up the pair. This gives $$9^3\cdot3^6\cdot4^{20}\binom{25}{2}\binom{23}{3}$$ possibilities for case $2$.
I computed the probability to be $$2.408361017451432\cdot10^{-9}$$