From Matt Yeager comes a game that is immensely popular with his fourth-grade class:
Three of Matt’s students — Players A, B and C — are engaged in a game of veinte. In each round, players take turns saying numbers in order (Player A, then B, then C, then A again, etc.). The first player to go says the number “1.” Each number must be either one, two, three or four more than the number said by the previous player. When someone says “20,” the round is over and the next person is eliminated, with the following person beginning the subsequent round. For example, if Player A says “20,” then Player B is eliminated, while Player C begins the next round by saying “1.” At no point can anyone say a number greater than 20.
All three players want to be the winner (i.e., the only player remaining) after the two rounds. But if they realize they can’t win, then they will prioritize making it to the second round.
Player A starts things off by saying “1.” Which player will win?
Extra credit: Instead of three players, now suppose there are four — Players A, B, C and D — all of whom want to make it through as many rounds of the game as possible. Again, Player A starts things off by saying “1.” Which player will win?
The second player in a two-player game of veinte is guaranteed to win. This occurs by simply subtracting the first players' number choice from 5. The sum of the number pairs will thus be 5, 10, 15, and 20, ensuring the second player wins.
Similarly, for a three-player game, two players can team up such that the sum of all numbers chosen for a round is 9. We will call these two players Winner and Supporter. Loser is the player elimated after the Winner says the number to achieve a sum of 20. The Loser will play any number from 1 to 4, and the Supporter and Winner will add on to that such that the sum is 9.
Working backwards yields the following table. Entries represent total sum:
| Loser | Winner | Supporter |
|---|---|---|
| -- | 20 | 16 to 19 |
| 2 to 15 | 11 | 7 to 10 |
| 3 to 6 | 2 | 1 |
Reversing yields:
| A | B | C |
|---|---|---|
| 1 | 2 | 3 to 6 |
| 7 to 10 | 11 | 12 to 15 |
| 16 to 19 | 20 | -- |
Note that the current setup has A as Supporter, B as Winner, and C as Loser. If A deviates, B will be Supporter, C will be Winner, and A will become the Loser.
Two additional solutions exist with the same end result.
One is if the sum is 6:
| A | B | C |
|---|---|---|
| 1 | 2 | 3 to 6 |
| 4 to 7 | 8 | 9 to 12 |
| 10 to 13 | 14 | 15 - 18 |
| 16 to 19 | 20 | -- |
The other is if the sum is 8:
| A | B | C |
|---|---|---|
| 1 | 4 | 5 to 8 |
| 8 to 11 | 12 | 13 to 16 |
| 16 to 19 | 20 | -- |
Again, if A deviates, they will become the Loser. Since C is the Loser in all cases, A starts the second round, and B wins.