A gambler repeatedly plays a game where in each round he win

A gambler repeatedly plays a game where in each round, he wins a dollar with probability 3/11 and loses a dollar with probability 8/11. His strategy is \"quit when he is ahead S3\" though some suspect he is a gambling addict anyway. Suppose that he starts with a million dollars. What is maximal probability that he can achieve his strategy goal if he try his best?

Solution

expected gain = ( 1* 3 /11) + ( -1 * 8 /11) = -5 /11.......
variance of gain = [ 1*( 3/11) + ( 1 * 8 /11) ] - ( - 5/11) ^2 = 96 / 121......

P [ gain >= 3 ] = P [ Z >= ( 3 - (-5/11) ) / sqrt ( 96 /121) ] = P [ Z >= 3.878359 ] = 1 - P [ Z < 3.878359 ]
= 0.0000526......
maximum probability = 0.0000525.....