[Guest]

Tom was playing 4 square and made an observation that if you are in square 3, to go to square 4 there is only one possible person who can lose. If you are in square 2 to go to square 3 any of 2 people can lose. and if you are in square 1, you advance as long as you dont lose. It gets harder to advance with each higher square thought Tom and he began to wonder which square a player spends the most time on average.

[Guest]

Tom was playing 4 square and made an observation that if you are in square 3, to go to square 4 there is only one possible person who can lose. If you are in square 2 to go to square 3 any of 2 people can lose. and if you are in square 1, you advance as long as you dont lose. It gets harder to advance with each higher square thought Tom and he began to wonder which square a player spends the most time on average.

Could you clarify please? Do I assume that the four players are equally competent? Do I assume that receiving from a server is just as difficult as receiving from a receiver? Lastly, is serving difficult, or can one assume that a server is always considered to succeed?

I have never played or even seen this fascinating game.

Edited September 1, 2009 by jerbil

[Guest]

you can assume all players are equally competent or not. it doesn't affect the answer it's really more of a logic problem but i framed it like a probability question just to throw you off.

[Guest]

The players spend the most time in the big square that encases the four other squares...there are really 5 squares.

Edited September 1, 2009 by futballa16

[Guest]

The players spend the most time in the big square that encases the four other squares...there are really 5 squares.

lol i guess that's technically right but that's not the answer i was looking for. It's not that type of trick question. Good thinking though.

[Guest]

i would think square three as even if you lose you stay in that square.

but when i ran a cpu simulation i found they were about equal.

here's a matrix of my results.

[[24969, 25556, 24747, 24729], [24967, 24841, 24915, 25278], [25034, 24719, 25198, 25050], [25031, 24885, 25141, 24944]]

player 0 was in square 0 (the starting square) 24969 times out of 100001, in square 1 25556 times, etc.

surprisingly, player 3 who started in the last square got to square 0 more often than the starting player.

Edited September 1, 2009 by phillip1882

[Guest]

If there are 4 players, then I believe the average time spent in each square has to be the same.

[bonanova]

Suppose there are 4 players.

At all times each of the squares is occupied by one of the four players.

The lowest to highest player leaves his square with respective probability of .75 .75 .50 and .25.

No square being permanently occupied, symmetry demands each square be occupied with equal likelihood by all players; namely 1/4.

Suppose there are N>4 players.

The equal result still obtains; just that now only 4/N of the players occupy a square.

That reduces occupancy time by each player by that factor from 1/4 to 1/N of the time for each square.

The nice red herring that the occupant of the highest square seems more permanent

is balanced by a lower initial probability of entering the square.

[Guest]

bonanova and ogden_tbsa got it!