Tic-Tac-Toyola is a variation of Tic-Tac-Toe with money. Instead of naughts and crosses (0s and Xs for Americans) being placed at the whim of alternating players, players vie for placement control at each turn by bidding some amount in a currency from an account under their control. The high bid decides the placement, with that bid amount being deducted from the corresponding player's account. Bids are in continuous units, so we can neglect the possibility of tie bids as unlikely.

There are two players: C(rosser) and N(aughty). X is placed first, and player C is declared winner if 3 Xs appear in a row before 3 Os appear in a row. If neither player achieves three in a row, N is declared winner. Thus, a game of Tic-Tac-Toyola will always have a winner.

C and N have been playing for some time and have reached the position:

(1) Assuming players are following optimal strategies, show that the outcome is independent of whether bidding is conducted secretly (by sealed bids) or openly (with players offered at each turn the possibility to outbid their opponent).

(2) Show that a player who is following an optimal strategy would not change his bid if he were to learn his opponent's account balance.

(3) Show that N can force a win if and only if he has in excess of twice as much money as C.

Note: Anyone interested in learning more about the Tic-Tac-Toyola family of games is invited to visit http://www.diplom.org/Zine/S2007R/Mayer/tictactoyola.htm

(3) Show that N can force a win if and only if he has in excess of twice as much money as C.

The previous statement is false.

Let us assume that a tie bid is won by Crosser.

To insure Crosser does not win, Naughty bids c+1, winning the bid, and places his naught in the bottom left corner.

To insure Naughty does not win automatically with a "cat" win in the next placement, Crosser must bid [n - (c+1)] with placement of his cross in either remaining corner.

After the two bid-postional controls, Naughty would have a current account of [n - (c+1)] and Crosser would have a current account of [c - [n-(c+1)]] = 2c - n + 1.

If n = (2c - 1) [an amount just less than twice], then Naughty's current account is c - 2; and Crosser's current account then would be 2c - (2c - 1) + 1 = 2. Therefore, if c > 4 then Naughty should win with at least a "cat" win. The "cat" win occuring by Naughty placing his second naught in the other available corner. 

If n = (2c) [an amount equal to twice], then Naughty's current account is c - 1; and Crosser's current account then would be 2c - (2c) + 1 = 1. Therefore, if c > 2 then Naughty should win with at least a "cat" win.

If n = (2c+1) [an amount in excess of twice as much], then Naughty's current account is c; and Crosser's current account then would be 2c - (2c+1) + 1 = 0. Therefore, if c > 0 then Naughty should win with at least a "cat" win.

Thus (3) does not hold true.

Posted by Dej Mar on 2008-04-27 01:18:55