- A Betting Game Involves 3 Players That Start The Game With Amounts Of Money X Y Z All 0 Respectively At Each 1 (161.42 KiB) Viewed 95 times
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A betting game involves 3 players, that start the game with amounts of money $x, $y, $z (all > 0) respectively. At each round n € N of the game, one player (the giver) is chosen uniformly at random to give some money to one of the other players (the receiver) chosen uniformly at random (independent of previous rounds). If these two chosen players had $V and $W at the beginning of the round, then the giver must give the receiver min{$V, $W}, and the round ends. (For those of you who may be familiar with e.g. no limit poker, you can think of this as having two players doing an “all in” bet in each round). The first player to reach $0 in this game is called the loser. After a loser has been determined the remaining two players continue until one of those two players has all the money. The player with all of the money at the end is called the winner. Let the amounts of money at time n (i.e. after n rounds) of the 3 players be Xn, Yn, and Zn respectively (so X 1, Y0 = y, Z = z). Let T1 = inf{n> 1 : min{Xn, Yn, Zn} = 0} and T2 = infín 1: max{Xn, Yn, Zn} = x+y+z}. (a) Explain in words what these times T1 and T2 represent in this game. (b) Show that E[T1] < 2 and E[T2] < 4. (c) Using Martingale theory (see graduate course in probability) one can prove that in this game E[XT2] = E[Xo] = x (XT, represents the amount of money of player 1 at the random time T2). Use this fact to find the probability that Player 1 is the winner of the game. (d) Find the probability that Player 1 is the loser of the game if (x, y, z) = (1,2,3). (e) Find the probability that Player 1 is the loser if (2, 4, z) = (12, 24, 36).