## Euphrat & Tigris 1.0 Solution

Last month's puzzle was a bit troubling for some, not the least of which was because the best move was not all that certain of victory.

The correct solution is:

- First action: Red leader A6 to B12 (This causes an internal conflict in red which you win if your opponent has less than 2 red tiles in hand.)
- Second action: Place a blue tile at C10 and execute the external conflicts in this order: blue, black then green. You intentionally lose the blue battle giving your opponent 4 blue VPs, but in the process you sever columns 3, 4, 5 and 6 from his green and black leaders. You will win the black conflict if your opponent has less than 4 black tiles in hand and you'll also win the green conflict if he has less than 4 green tiles in hand.

Overall, this plan succeeds if the opponent's hand contains less than 2 red tiles (probability about 50%), less than 4 black tiles (very high probability), and less than 4 green tiles (very high probability). Your final score would be 11/11/12/12 to his 11/11/11/14.

Here is the detailed argument of why this plan is best:

- With the score 10/11/11/11 to 10/10/10/10, it is clear that if you gain only 2 VPs (of the same or different colors), you will still lose. Therefore, you cannot have both actions be disaster placement and/or simple tile placement and/or internal conflict, since each such action only earn 0 or 1 VP. In other words, you need to start an external conflict using one or both actions, since only an external conflict is the only action that can earn more than 1 VP per action.
- There are several ways to attempt to gain a bunch of VPs in one color, e.g., to gain green VPs, you may attempt: Disaster to B3 (eliminating one of opponent's green supporter tiles), followed by black tile to B11 and then resolving the conflict in green. On your side you have 4 supporters + 1 tile in hand (denoted 4+1), and on your opponent's side he has 4 supporters, so you will win if (and only if) your opponent does not have any green tiles in hand. However, even if you win those 5 green VPs, your score will only be 10/10/11/14, which still loses to opponent's 10/11/11/11. The conclusion is that you need to gain VPs in 2 colors, in order to win the game.
- A slight variation of the above example is to do Disaster to B3, black tile toB11 as before, but first play the red conflict (your 3+2 versus his 4). You will win if and only if the opponent has no red tile in hand. After winning, you can play the green conflict (4+1 versus 4), and again you will win if and only if the opponent has no green tile in hand. If you do win both, then your score of 10/11/11/14 will win the game. However, for this to succeed you need the opponent to have no red nor green tiles in hand.
- Yet another variation is to play Disaster to B3, red tile to A11 and then play green conflict (4+1 versus 4). If you win that, it will isolate the black tile at A4, and you can start the black conflict (4+2 versus 5). To win the game, you need to win both, and that only happens if the opponent has neither green nor black tiles in his hand.
- In either solution 3 or 4 (or other similar solutions), success requires that the opponent's hand contains no tiles of two specific colors (red and green in solution 3, green and black in solution 4). The probability of this event is rather low: (1/2)^6 = 1/64 = 1.6%. The proposed "correct" solution has a much higher probability than this.
- The key insight is that if you gain enough in two colors, then
it doesn't matter if you sacrifice a third color to the opponent!
Specifically, the correct move attempts to sever his green and
black leaders from the huge support base in columns 3-6. The way
to achieve that is to intentionally
**lose**the blue external conflict (which would empty the squares C7 and B7) and also somehow sever the remaining A7 link. - One approach (which almost works but not quite) is to play Disaster to A7, black tile to C10, then play blue conflict first (3+0 versus 4). You will lose the blue conflict, raising the opponent's score to 11/11/11/14, but then columns 8-10 are severed from columns 3-6, and it is much easier to win first the black conflict (4+2 versus 2) and then the green conflict (4+1 versus 1). However, at the end, your VPs will be: 9 red, 9 blue, 11 green, 12 black and 4 treasures, for a score will 11/11/11/12, which still loses to the opponent's 11/11/11/14.
- The correct solution is a minor variation on solution 7. You need to somehow gain 1 more VP, instead of a non-VP-generating action of Disaster to A7. The solution is: red leader to B12 as first action! This is an internal conflict (1+2 versus 1) which you can win if the opponent has only 0 or 1 red tile. On average, your opponent's six tiles contain 1.5 red tiles, so having less than two red tiles is a rather likely event. (An exact calculation gives the probability as 2187/4096 = 53.4%). Assuming you win, the second action is blue tile to C10, followed by external conflicts in blue (3+0 versus 4, intentional loss), green (4+1 versus 1, very high probability win) and black (4+2 versus 2, very high probability win). If you win the red internal conflict, and both green and black external conflicts, your final VPs will be: 9 blue, 10 red, 11 green, 12 black, 4 treasures for 11/11/12/12, which just beats your opponent's 11/11/11/14. For this plan to succeed, your opponent's hand must contain less than 2 red tiles, less than 4 green tiles and less than 4 black tiles. An exact calculation shows that this plan will succeed with probability 1909/4096 = 46.6%.

Congratulation to Dave Arnott whose name was randomly drawn from all correct entries!

- Anthony Kam

GGA - Please note that I inadvertently created an alternate winning move when I created the puzzle graphics. Instead of moving your red leader from A6 to B12 (causing a red internal conflict) you could move your red leader from A6 to D9 and conduct a red battle as the first part of the external conflict. Both options give an equal likelihood of success. I apologize to Anthony for introducing this added possibility.