- 22 Mar '09 12:38There 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? - 22 Mar '09 14:15Letseee

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? - 23 Mar '09 13:31For 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.. - 23 Mar '09 20:40 / 2 edits

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.*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?

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. - 23 Mar '09 21:33

2, Correct - for multiples of 6 the problem is trivial.*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.

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? - 24 Mar '09 16:40

I took this to be a game with perfect information, which does place it squarely within the realm of games of logic.*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?

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.. - 24 Mar '09 16:46ALL 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? - 24 Mar '09 17:04It 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. - 24 Mar '09 20:10

Again, mulitiples of 1-6.*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?

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. - 24 Mar '09 22:23

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.*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. - 25 Mar '09 17:17 / 1 edit

The number of players is known.*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.**

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. - 25 Mar '09 19:41

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.*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.** - 25 Mar '09 20:52

I stand corrected*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.**