A Question About Cheat (game) From the game Cheat : https://en.wikipedia.org/wiki/Cheat_(game)
Assume that we have 13 cards from 1 to 13 and that we play with 4 players. Each player has at least 1 card and knows where are each of the 13 cards at every moment of the game (complete information).
The first play of the game must call 1; subsequent calls must be exactly one rank higher, with 13 being followed by 1 and continuing again. The player can't use a strategy they can only play their card or pick the cards that are on the table if they can't play.
The objective of the game is to be the first player to get rid of all their cards.
What is the longest possible game you can have (in term of turns) ?
 A: I don't see how to solve this without a computer. Since there are only
$$\sum_{k=0}^4(-1)^k\binom4k(4-k)^{13}=60780720$$
different initial configurations, they're easily checked by computer. Here's code that does so. The result is that the game can take up to $102$ turns, and there are $1537$ initial configurations that lead to this game length, so the probability for a uniformly randomly drawn configuration to lead to this game length is about $25$ in a million. These maximal initial configurations vary greatly, but there are some patterns. The $13$ is always on the last player's hand, and the positions of the higher cards are less variable than those of the lower cards. Only six different distributions of the top six cards occur:
9
12
8 10 11
13

----
9 12
8 10 11
13

----
9 10 12
8
11 13

9
12
8 10
11 13

----
9 12
8 10
11 13

9
10
8 11
12 13

Here's a random sample of maximal initial configurations:
1 3 9
2 4 5 12
7 8 10 11
6 13

3 9
1 12
5 7 8 10 11
2 4 6 13

7
1 9 12
2 5 6 8 10 11
3 4 13

1 4 5 7
6 9 10 12
2 8
3 11 13

1 3
4 9 10 12
5 8
2 6 7 11 13

