# Game

wolfgang59
Posers and Puzzles 22 Mar '09 12:38
1. wolfgang59
22 Mar '09 12:38
There are 3 tables, each with a pile of money. Table 1 has \$1,000 on it, table 2 has \$2,000 on it and ... you guessed it, table 3 has \$3,000.

Each player in the game takes turns to sit at a table, once all players are seated the money on the table is divided between those sitting around it.

Assuming all other players play optimally to get the most money for themselves individually.

1. What is your strategy for an unkown number of players?

2. For what numbers of players does it not matter?
2. 22 Mar '09 14:15
Letseee
1 player - 3
2 players - 32
3 players - 233
4 players: Player 1 sits at table 3, player 2 sits at table 3 also, player 3 sits at table 2, now, player 4 could sit at any table and gain \$1000, but he could do so at the expense of other player(s) if he chose to. OP, what choice would the player make in this situation?
3. 22 Mar '09 18:28
nice one I myself got totally confused... but then again i am only 15
4. 23 Mar '09 13:31
For a known number of players, you can track the action of the players from last to first for the optimal way(s) to proceed.

For example, with two players, the second will either take the \$2000 if the first player takes \$3000, or else take the \$3000 if not. So the first play gets \$3000 for selecting the third table, and less at any other table, and would thus pick the third.

For an unknown, you'll have examine patterns within the known player setups in order to try and elicit a pattern and (based on your estimation of how many will be there) pick how to proceed.

May take a look-see into up to 6-8 players later.

There may be a possibility that psychology may be a factor in some circumstances, making into a dilemma whether to take a smaller safe amount, or a larger amount which could become the smaller..
5. PBE6
Bananarama
23 Mar '09 20:32
Pretty interesting game! I've cheated a bit and emailed the Wizard of Odds, I'll let y'all know if he provides any help.
6. uzless
The So Fist
23 Mar '09 20:402 edits
Originally posted by wolfgang59
There are 3 tables, each with a pile of money. Table 1 has \$1,000 on it, table 2 has \$2,000 on it and ... you guessed it, table 3 has \$3,000.

Each player in the game takes turns to sit at a table, once all players are seated the money on the table is divided between those sitting around it.

Assuming all other players play optimally to get the most ...[text shortened]... strategy for an unkown number of players?

2. For what numbers of players does it not matter?
1. I"m assuming the players don't know how much money is on each table. If they do know, then it's just simple math to figure out which table has the most amount of money left over after divided by the number of people already there.

So if you don't know, it makes no difference as long as you sit at a table that has few people already there. It's random and your expected value would be the same each time you sat a different table.

2. It doesn't matter what table you sit at if there are mulitiples of 6 players since 3 would be at the 3000, 2 would be at the 2000 table and 1 left at the 1000 table.

They'd each get 1000.
7. wolfgang59
23 Mar '09 21:33
Originally posted by uzless
1. I"m assuming the players don't know how much money is on each table. If they do know, then it's just simple math to figure out which table has the most amount of money left over after divided by the number of people already there.

So if you don't know, it makes no difference as long as you sit at a table that has few people already there. It's random ...[text shortened]... 3000, 2 would be at the 2000 table and 1 left at the 1000 table.

They'd each get 1000.
2, Correct - for multiples of 6 the problem is trivial.

1. Not sure of your logic. Maybe I should have been clearer:

All players know the number of players involved.
All players will play to optimize their own winnings.
There is no psychology ... just logic!

Do we opt to choose first or last? what is the tactic/strategy?
8. 24 Mar '09 16:40
Originally posted by wolfgang59
2, Correct - for multiples of 6 the problem is trivial.

1. Not sure of your logic. Maybe I should have been clearer:

All players know the number of players involved.
All players will play to optimize their own winnings.
There is no psychology ... just logic!

Do we opt to choose first or last? what is the tactic/strategy?
I took this to be a game with perfect information, which does place it squarely within the realm of games of logic.

That said, that does not mean there won't be psychology involved as well.

Take the four players example. The last player will see 2 players splitting \$3k, 1 player holding \$2k, and an umclaimed \$1k. Whichever he picks, he will have \$1k when he walks away.

Here is where the psychology comes into play. He has to pick one of the tables, and he may use psychology to do that. If he sets as a secondary goal to bring the top player down, he'll pick Table 2. If he decides that it is more important all the money be claimed, he'll pick Table 1.

Now, the first person has this dilemma. Based on where he thinks the last person will go, his best pick could be either Table 2 or Table 3.

If he picks Table 2, and the last person doesn't, then he gets \$2000 instead of \$1000-\$1500. If he is wrong, then he ends up with \$1000 where he could have had \$1500.

If he believes the last person will choose Table 2, then he can pick Table 3 for \$1500 and avoid being left with only \$1000. However, if he is wrong, he has cut himself out of \$500 he could have had, and may even lose another \$500 in addition to this.

On what basis does he choose now? He has no logical reason to believe the last person will or will not choose Table 2, for logic gives it equal status with the other 2 tables.. It must be an assessment of what drives the last player to go, his psychology, unless of course he wishes to give even odds to all 3 possibilities.. That, however, could backfire..
9. wolfgang59
24 Mar '09 16:46
ALL players can see how much is on each table and ALL can see how many currently seated there.

What is your best tactic? Again I stress that psychology is NOT a factor because I have already staed that each player is looking to optimise their own winnings. (No secondary goals!)

What is the best tactic?
10. 24 Mar '09 17:04
It is verily in human nature to have some reason to choose one thing over another.. Left with 2 absolutely equal choices, we would stand in indecision indefinitely. However, even with a tiny difference, we can thus go with one over the other.

A coin toss is good for this..

I am assuming that a player will first and foremost try to maximize what they win, without regard to anything else. However, there is a question of how to proceed when all choices are equal, and how a person proceeds at this point may have implications to those who come before them.

In a 4 person game, the 2nd table is not a guarantee to get the most money. If a person in a moment of indecision uses a fair dice to select which of 3 equal choices to pick, then that's one scenario.

The first person would thus expect \$1666 from Table 2, and \$1333 from Table 3, and would pick Table 2.

However, the indecisive nature of the fourth person's choice combined with natural leanings and complexity of human nature could bring the game into the realm of human psychology, where all 3 choices are not equal, but that the person may rely on other consequences of his/her choice to pick a specific course of action.
11. uzless
The So Fist
24 Mar '09 20:10
Originally posted by wolfgang59
ALL players can see how much is on each table and ALL can see how many currently seated there.

What is your best tactic? Again I stress that psychology is NOT a factor because I have already staed that each player is looking to optimise their own winnings. (No secondary goals!)

What is the best tactic?
Again, mulitiples of 1-6.

If total number of players end back to 1 (1,7,13,19 etc), pick table 1

If total number ends at 2 pick table 2

If total number ends at 3 pick table 1

If total number ends at 4, makes no difference

If total numbers end at 5, it makes no difference.

If total number ends at 6, it makes no difference.
12. forkedknight
Defend the Universe
24 Mar '09 22:23
Originally posted by uzless
Again, mulitiples of 1-6.

If total number of players end back to 1 (1,7,13,19 etc), pick table 1

If total number ends at 2 pick table 2

If total number ends at 3 pick table 1

If total number ends at 4, makes no difference

If total numbers end at 5, it makes no difference.

If total number ends at 6, it makes no difference.
For a number of players n, do the players know what n is? If so, that strategy is not optimal for all players. For instance, if there were 3 players, player 1 would want to choose table 2 instead of 1.
13. uzless
The So Fist
25 Mar '09 17:171 edit
Originally posted by forkedknight
For a number of players n, do the players know what n is? If so, that strategy is not optimal for all players. For instance, if there were 3 players, player 1 would want to choose table 2 instead of 1.
The number of players is known.

If there were 3 players, the first would choose table 1 (3000), the 2nd player would choose table 2 (2000), and the third player would choose table 1 again because he'd get (1500) which would be higher than 1000 at table 2 and 1000 at table 3.

players 4,5,6 can't get anymore than 1000 at any table after that.
14. Palynka
Upward Spiral
25 Mar '09 19:41
Originally posted by uzless
If there were 3 players, the first would choose table 1 (3000), the 2nd player would choose table 2 (2000), and the third player would choose table 1 again because he'd get (1500) which would be higher than 1000 at table 2 and 1000 at table 3.
The first player would choose table 2 and get 2000, because he knows that if he follows your strategy he'll only get 1500. The second player would choose table 1, because even if he splits it with player 3, he still gets more than in the other 2 tables. Player 3 has a choice of 1500, 1000, 1000 and therefore chooses Table 1 and 1500.
15. uzless
The So Fist
25 Mar '09 20:52
Originally posted by Palynka
The first player would choose table 2 and get 2000, because he knows that if he follows your strategy he'll only get 1500. The second player would choose table 1, because even if he splits it with player 3, he still gets more than in the other 2 tables. Player 3 has a choice of 1500, 1000, 1000 and therefore chooses Table 1 and 1500.
I stand corrected