[bonanova]

There's an old con game played with the 24 cards of rank A-6 where opponents choose cards in turn from 6 face-up piles.

The object is to make the running total exactly 31 with your card, or force opponent to exceed 31 with his.

There is a simple formula to calculate a winning position, but care must be taken at the start.

The "con" is to let opponent see the formula, then make him run out of needed cards at the end.

When played with a single die, however, there is no limit to the number of times a particular value 1-6 can appear.

Here's how it works.

Player 1 throws a fair die, exposing a number from 1-6. Say it's a 2.

Next player can then play any number that shows on the four exposed vertical sides.

That is, he may play anything but 2 [the top face] or 5 [bottom face], by rotating the die 90^o, thus putting one of side faces on top.

Players alternate turning the die in this manner until the running total becomes 31 [player wins] or exceeds 31 [player loses].

If you're rolling the die, what number or numbers would be beneficial to roll?

[Guest]

you want to roll 1 or 6. The fastest way to make 31 is 5 x 6 + 1. Since 1 and 6 are on opposite sides of the die, you can always move to 6 or to 1 after the opponent makes his move.

[plainglazed]

4?

[bonanova]

Yes. That's it. Care to elaborate?

[plainglazed]

I worked backwards to find die positions and tallies that always win. Starting with you turning the die to 1 or 6 and tally = 30. If, after your turn, the tally is 29, you loose to an optimal strategy. The next player can either turn the die to 1 with a tally total = 30 or turn the die to 2 totalling 31. If you turn the die to 3 or 4 and the tally totals 28 or 27, you cannot be beat. If you turn the die to 2 or 5 and the tally is 26, you cannot be beat. If, after your turn, the tally is 25 or 24, you loose to an optimal strategy:

so here are the automatic winners of upturned die position and tally:

1 - 30

6 - 30

3 - 28

4 - 28

3 - 27

4 - 27

2 - 26

5 - 26

now all other die positions and totals that can always be adjusted to one of the above are also automatic winners and all subsequent automatic winners are added to the list:

3 - 23

4 - 23

1 - 22

2 - 22

3 - 22

4 - 22

5 - 22

6 - 22

3 - 19

4 - 19

3 - 18

4 - 18

2 - 17

5 - 17

3 - 14

4 - 14

1 - 13

2 - 13

3 - 13

4 - 13

5 - 13

6 - 13

3 - 10

4 - 10

3 - 9

4 - 9

2 - 8

5 - 8

3 - 5

4 - 5

1 - 4

2 - 4

3 - 4

4 - 4

5 - 4

6 - 4

If you start with a 1,2 or 3; your opponent can total 4 and be in command. If you start with a 5 or 6, your opponent can turn up a 4 or 3 totalling 9 and be in command. You can see a pattern of winning positions after a bit (groups of 9 when ordered and listed as above) and there may be an easy way to calculate your next move but havent studied that aspect.

A most enjoyable puzzle, bonanova!

Edited March 13, 2011 by plainglazed

[plainglazed]

I had made some wrong assumptions on my first go 'round. Hoping the following is better.

Working backwards to find die positions and tallies that always win. Starting with turning the die to 1 or 6 and tally = 30. If, after your turn, the tally is 29, you loose to an optimal strategy. The next player can either turn the die to 1 with a tally total = 30 or turn the die to 2 totalling 31. If you turn the die to 3 or 4 and the tally totals 28 or 27, you cannot be beat. If you turn the die to 2 or 5 and the tally is 26, you cannot be beat. If, after your turn, the tally is 25 or 24, you loose to an optimal strategy:

so here are the automatic winners of upturned die position and tally:

1 - 30

6 - 30

3 - 28

4 - 28

3 - 27

4 - 27

2 - 26

5 - 26

if you can achieve any of the above positions of "die value" - "tally", you win.

3 - 23

4 - 23

1 - 22

2 - 22

3 - 22

4 - 22

5 - 22

6 - 22

and whoever gets to 22 wins - the opponent's options are 1 - 23 followed by 4 - 27; 2 - 24 followed by 3 - 27; and 3 - 25, 4 - 26, 5 - 27, 6 - 28 allowing 31 to be reached directly.

or wohoever gets to 23 with a 3 or 4 wins - the opponent's options are 1 - 24 followed by 3 - 27 or 6 - 30; and 2 - 25, 5 - 28, 6 - 29 allowing 31 to be reached directly.

3 - 18

4 - 18

3 - 14

4 - 14

1 - 13

2 - 13

3 - 13

4 - 13

5 - 13

6 - 13

also, whoever gets to 13 or 14 with a 3 or 4 can always win -

from 13 the opponent's options are 3 - 16, 4 - 17, 5 - 18, 6 - 19 which can all be followed with a move of the die tallying 22; or 1 - 14 and 2 - 15 followed by 3 - 18 or 4 - 18.

from there their options are 1 - 19, 2 - 20 which can be followed by 4 - 23 or 3 - 23; or 5 - 23, 6 - 24 which can be followed by 4 - 27 or 3 - 27; all already established as winning positions.

from 14 with a 3 or 4 the opponent's options are 1 - 15 which can be followed by 3 - 18 and then as above; or 2 - 16, 5 - 19, 6 - 20 which can all be followed with a move of the die tallying 22.

3 - 9

4 - 9

3 - 5

4 - 5

1 - 4

2 - 4

3 - 4

4 - 4

5 - 4

6 - 4

similarly, whoever gets to 4 or 5 with a 3 or 4 can always win.

so if you start with a 1 or 3; your opponent can total 4 and be in command.

if you start with a 2, your opponent can turn up a 3 tallying 5 and be in command.

if you start with a 5 or 6, your opponent can turn up a 4 or 3 totalling 9 and be in command. your options are then 1 - 10, 2 - 11 which allows them to reach 4 - 14 or 3 - 14; or 5 - 14, 6 - 15 which allows them to reach 4 - 18 or 3 - 18.

overall strategy more succinctly put: starting with a 4 always make the tally a multiple of 9 or a {(multiple of 9) - 4} by turning the die to a 3 or 4; otherwise make the tally either 13 or 22.

[Guest]

Can anyone help me with my english skills..

I read this puzzle for at least 10 times and i still dont have a clue what it says...

I think I understand all the words, but i dont get the idea and the form of the gameplay.

can anyone post a simpler version of how exactly the game goes?