BMAD 64 Posted June 1, 2013 Report Share Posted June 1, 2013 (edited) Fearless Frank decided to play a fair coin-flip game with probability 1/2 of winning each bet, and risked 1/m of his fortune (originally A dollars, m>1) at every flip. After 2n games, Frank has won n games and lost n. Choose and explain the correct answer from this list: a) Frank has broken even; he still has his A dollars b) Frank is predictably ahead by a certain amount c) Frank is predictably behind by a certain amount (in case b or c give the exact formula in terms of m and n) d) Frank is now ahead, behind, or even, depending on the order in which the wins and losses occurred e) He is ahead, behind, or even, depending on m and n Edited June 1, 2013 by BMAD Quote Link to post Share on other sites

0 Solution phil1882 13 Posted June 1, 2013 Solution Report Share Posted June 1, 2013 let's say m is 3. A = 243. let's say n is 4. let's futher take the most extreame example, winning all four then losing all four, 243+81 = 324 +108 = 432 +144 = 576 +192 = 768 768-256= 512 -170.66 = 341.34 -113.78 =227.56 -75.85 =151.71 so it looks to me like the answer is c. let's try the oppisite way, losing 4 then winning 4. 243 -81 = 162 -54 = 108 -36 = 72 -24 = 48 48 +16 = 64 +21.33 = 85.33 +28.44 =113.77 +37.92 =151.69. so now for the formula.... A*(m-1)/m)^n*(1+1/m)^n A*((m-1)/m*(m+1)/m)^n A*((m^2 -1)/m^2)^n A*(1 -1/m^2)^n Quote Link to post Share on other sites

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